Mass Moment Of Inertia - NetBadi.com
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MASS MOMENT OF INERTIA
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ROTATIONAL EFFECT OF A FORCE: Before introducing any new physical quantity, let us analyze the following case:
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An uniform rod of mass m and length l is placed on a smooth horizontal surface with one of its ends pivoted at a point 0 about which the rod can rotate freely. Now, if we apply a force F at the other end of the rod then what would happen? In our daily life we come across various similar cases. We know that the rod would not accelerate translationally along F because it is pivoted at O but the rod may rotate. Hence, along with producing a translational acceleration a force can produce rotational (or angular) acceleration as well. The effect of force causing angular acceleration in a body is defined as torque of that force. In figure 7.10 (a), F is applied along the length of the rod. In this case F does not produce rotation hence torque of F is zero.
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In figure 7.10 (b), F is applied perpendicular to the length of the rod. In this case F produces rotation in the rod, hence, torque of F is nonzero.
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In Figure 7.10 (c), F is applied perpendicular to the length of the rod but it is applied at the mid-point of the rod not at the free end. F produces angular acceleration in this case too but angular acceleration produced in this case is exactly half of that produced in the case (b). Hence, torque of F in this case is half of that in case (b).
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In figure 7.10(d), F is applied at the pivoted end of the rod. In this case no rotation is produced, hence, torque of F is zero.
Summary of all four is given in figure 7.10 (e).F is applied at a distance r from the pivot and at an angle T with the length of the rod. It is obvious that only perpendicular component of the force, FA (= F sin T ), causes rotation and rotation caused (i.e., angular acceleration D ) is directly proportional to the magnitude of FA .Again, it is also learnt that D is also directly proportional to r. Hence, torque of F is directly proportional to r and FA .Therefore, torque of F, denoted by T, can be defined as 1
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* = r.. FA
………………. (7.16 a)
………………. (7.16 b) = r. F sin T Equation 7.16 (b) can be rearranged to obtain, * = (r sin T ). F = rA . F
………………. (7.16 c)
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Hence, torque of a force can be calculated using either of two ways: Multiply perpendicular component of the force with the distance of point of application of force from 0 or multiply force with the perpendicular distance of 0 from the line of action of force.
Hence, a force can not produce a torque about a point its line of action passes through that point (i.e., rA = 0).
& )& ruF
………… (7.17)
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)& *
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Torque is a rotational analogue of force. If you apply a net force on a body, body gets an acceleration along the direction of the net force and the magnitude of acceleration is directly proportional to the magnitude of the net force. Similarly, when we apply a net torque on a body, body gets an angular acceleration along the direction of the net torque and the magnitude of the angular acceleration is directly proportional to the magnitude of the net torque. Figure7.lO (e) is redrawn in figure 7.12. When F is applied, as shown in figure, the rod rotates in anti clockwise direction in the given plane. As the torque and angular acceleration should have same directions, the torque is also in the anticlockwise direction. Using right hand rule it can be more appropriately said that torque is on the axis of rotation and is pointing out of the plane of rotation, denoted by symbol (A direction perpendicular to the plane of paper and pointing inwards is denoted by
symbol). Here direction of torque with the direction & )& of r u F , hence, in the vector form, torque can be defined as ,
D
T
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D
)&
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Hence, torque of a force F , with respect to some point O is defined as the cross product of the position vector )& )& of the point of application of F with respect to O and the force F .
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Unit of torque is N – m and it has the dimensions ML2T-2. You may not that it has the same dimensions as that of work. Torque is also called the moment of force. Some more illustrations are shown in figure 7.13 to decide the direction of torque: Figure 7.13: Torque on a ring about its center due to a force in the plane of the ring and acting tangential.
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Example : 4. www.NetBadi.com A thin uniform rod of mass m and length l is supported by a fixed support about which it can rotate freely, as shown in figure 7.14 (a). Find the torque on the rod about point O when the rod is horizontal.
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Solution: There are two forces acting on the rod: 1. The weight of the rod, mg, in the vertically downward direction; 2. the reaction force from the hinge, R. About the point O, the reaction force from the hinge does not produce any torque because its line of action passes through that point. Only weight of the rod produces torque about O.
weight of the rod u
Perpendicular dis tan ce of O
From the line of action of weight
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The torque of the weight of the rod about 0 is pointing perpendicularly inside the page, as suggested in figure 7.14 (c). Perpendicularly inside direction of torque means it tends to produce rotation in the clockwise sense in the given plane, as shown in figure 7.14(d). When you will turn the fingers of your right hand along the sense of rotation, thumb would point along the direction of the torque. But, at this juncture, I must tell you that in simple & )& cases you need not to first find the direction of torque by finding the direction of r u F and then decide the sense of rotation. You can know it directly by just observing the direction of force and its point of application only. For an example, in figure 7.14(d) it is quite obvious that mg would try to rotate the rod about the point 0 in the clockwise direction only. Now let us find the magnitude of the torque of the weight of the rod. If it is * , then,
l 2
[from figure 7.14 (b)]
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mgu
mg
l 2
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From the previous discussion it is obvious that the net torque on the rod about O is equal to the torque due to its weight only. Example 5: Three tangential forces, each of magnitude F, are applied on a disc of radius R. Find the net torque on the disc due to these three forces about the centre of the disc O.
Solution: A close look of the situation would let you know that each force tends to rotate the disc in the same direction (clockwise direction), as shown in figure 7.15(b). 3
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& )&
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If you use r u F to decide the direction of the torque of the each force, then, also you would get the same result. Each of the three forces provide torque perpendicularly inward, i.e., they try to rotate the disc in the clockwise direction. Magnitude of the torque due to each force is FR (force x perpendicular distance of 0 from the line of action of force), hence, net torque about 0 due to all the forces is 3FR. ROTATIONAL INERTIA: If you want a body to remain static then not only the net force on it should be zero but the net torque on it (about any stationary point) should also be zero. Hence, a necessary condition for a body not to rotate is that the resultant torque about any point be zero. Although this condition is necessary, it is not a sufficient condition for a body not to rotate. If a body is set rotating about an axis and there is no external torque acting, the body will continue to rotate with constant angular velocity. (This is the same as the case of linear motion. A particle can be at rest and remain at rest only if the resultant force is zero, but this is not a sufficient condition for being at rest If the resultant force is zero, the particle would be moving with constant velocity). Hence, a body maintains its rotational state unless an external unbalanced torque is applied on it.
The physical quantity measuring the tendency of a body to maintain its rotational state is called as rotational inertia or moment of inertia of the body.
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Consider the situation shown in figure 7.16. A force F i is acting on the i th particle of mass mi of a wheel which is pivoted about an axis through its centre and perpendicular to the plane of wheel. The component Fir parallel & to the radius vector of the element, r , has no effect on the rotation of the body. The tangential component Fit, &
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which is perpendicular to r i , does affect the rotation of the body The torque produced by the force .P about the centre of the wheel is given by *i
Fit ri
Using Newton’s second law, the tangential acceleration of the particle is given by Fit
mi ait
mi ri D
[ait
ri D ] .
Where D is the angular acceleration of the particle which is same as that of the whole body. Combining the two expression above, we get *i
mi ri2 D
If we now sum over all the particles in the body, we obtain 6 *i
6 mi ri2 D 4
6 mi ri2 D
[ D is same for all elements ]
………… (7.18) www.NetBadi.com
The quantity 6 *i is the resultant torque acting on the body, which we denote as * net . The sum 6 mi ri2 is the property of the wheel and is defined as the moment of inertia I. There fore,
I
6 mi ri2
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The moment of inertia depends upon the distribution of mass relative to the axis of rotation. For the same body moment of inertial can be deferent for different axes. Moment of inertia has the dimensions ML2 and is usually expressed in kg – m2. For a particle of mass m at a distance r from the axis of rotation, moment of inertia is I = mr2. For a body that is not composed of discrete point masses but is instead a continuous distribution of mass, the summation
I
³r
2
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in I 6 mi ri2 becomes an integration. We imagine the body to be subdivided into infinitesimal elements, each of mass dm, as shown in figure 7.17. Let r be the distance from such an element to the axis of rotation. Then the moment of inertia is obtained from ud m
………. (7.19 c_
replaced by the integral
³
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Where the integral is taken over the whole body. The produced by which the summation 6 of a discrete distribution is for a continuous distribution is the same as that discussed for the center of mass.
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Figure For bodies of irregular shape the integrals may be difficult to evaluate. For bodies of simple geometrical shape the integrals are relatively easy when the axis of symmetry is chosen as the axis of rotation. CALCULATION OF MOMENT OF INERTIA: a. A small ball of mass m at a distance r from the axis: I = mr2
axis
axis fig.7.18
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m
m
w
r
5
I = mr
I
Two small balls, each of mass m, with perpendicular bisector of the line joining the two balls as the axis of rotation: 6 mi ri2
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b.
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c. I
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= mr2 + mr2 = 2 mr2.
Three balls, each of mass m, at the vertices of an equilateral triangle of side length l with the line passing perpendicularly through the centriod of the triangle as the axis of rotation: 6 mi ri2
6 Ii
3mr2
§1 · 3 m ¨ sec 3T ¸ 2 © ¹
l2 4 . 4 3
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2m
ml2 .
An uniformed of mass m and length l with the line passing through its center and perpendicular to its length as the axis of rotation:
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d.
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m r2 m r2 m r2
Using equation 7.19, moment of inertia of the element is dI = dm.x2 §m · ¨ l .dx ¸ © ¹
x2
Therefore, moment of inertia of the rod is I
³d I 6
l / 2
³
l / 2
m 2 x .dx l
m ª l3 º « » l ¬« 3 ¼»
m l
l / 2
l / 2
³
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x 2 .dx
l / 2
m ª 3 ºl / 2 l 3 l ¬ ¼ l / 2
l / 2
m ª l 3 § l 3 ·º « ¨ ¸» l 3 «¬ 8 ¨© 8 ¸¹ »¼
m 2l3 3l 8
m l2 12
An uniformed rod of length l and mass m with the line passing through one of its ends and perpendicular to its length as the axis of rotation:
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ml 2 . 12
I = sum of moments of inertia of all elements about the same about the same axis
³
³
l
³
dm.x 2
0
m 2 x .dx l
m l3 l 3
l
0
ml 2 3
An uniform ring of mass m and radius r with the line passing through its center and perpendicular to its plane as the axis of rotation:
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f.
dI
I = sum of moments of inertia of all the elements of the ring about the chose axis.
³d I
= mr2
³ dm.r
2
³
r 2 dm [ All elements are equidistant from the axis] ª dm «¬
³
mº »¼
NOTE: The above result could also be established in the following way : Consider the point mass m at a distance r from the axis of rotation, as shown in figure 724 (a). In this case the moment of inertia of the mass m about the chosen axis is mr2. Now, if we redistribute this mass and make a ring of radius r about the same axis, as shown in figure 7.24 (b), the entire mass still remains at the same distance 7
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from the axis and hence the moment of inertia is mr2 only. Now, if we stretch the ring along the length of the axis and make a hollow cylinder of radius r, as shown in figure 7.24(c), the entire masswww.NetBadi.com is still at the same 2 distance from the chosen axis ad hence the moment of inertia is still mr only.
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Same approach could be use to find the moments of inertia of uniform rectangular plates using the results for uniform rods, as shown in figure 7.25(a) and 7.25(b).
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We can say that when the mass of a system is redistributed in such away that each element shifts parallel to the axis of rotation (so that its distance from the axis remains the same), the moment of inertia of the system remains unaffected by the redistribution. (g) An uniform disc of radius rand mass m with the line passing through its centre and perpendicular to its plane as the axis of rotation:
The top view of the disc is shown in figure 7.26(b). To calculate the moment of inertia of the disc about the chosen axis, it is divided into a large number of elements. Each element being a ring with the same centre as that of the disc. Such an element is shown in figure 7.26 (c). Mass of the element is dm, radius is x and thickness is dx. The moment of inertia of the element is
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dI = dm. x2
[Using I = mr2 for an uniform ring]
§ m · 2 ¨ 2 .2S x.dx ¸ .x r S © ¹
2m r
2
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mass per unit area º ª « mass of the u » « element » «¬ area of the element »¼
.x 2 .dx
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The moment of inertia of the disc, I = sum of moments of inertia of all elements
³ dI r2
r
³ x .dx 3
0
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2m
2m r4 . r2 4
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m r2 2
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(h) A disc of uniform mass density having mass M and radius R with a concentric circular hole of radius r with the line passing through the centre of the disc and perpendicular to the plane of the disc as the axis of rotation.
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To find the moment of inertia of the given body about its axis of symmetry we follow the same approach as we did in the last case. Divide the body in several elemental rings, calculate the moment of inertia of each ring and then add the moments of inertia of all elemental rings to get the moment of inertia of the whole body. Such an element, having mass dm, is shown in figure 6.27 (c),r being its radius and its radius and it has a thickness dr, then its moment of inertia, dI = dm.r2
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§ · M u 2S r.dr ¸ .r 2 z ¨¨ 2 2 ¸ © S R2 S R1 ¹ 2M R22
R12
r 3 .dr
Therefore, moment of inertia of the given body is I
³ dI 2M R22 R12
R2
³
R1
r 3 .d r
2M R22 R12
u
R24 R14 4
9
M R12 R22
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NOTE: It could also be calculated using the method I remaining I removed
I actual
r = Rcos
r dm R
Reference line
axis
dm = mass per unit area × area of the ring 4 S R2 M 4 S R2
u 2 S r.R d T
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M
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fig.7.28
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Which is exactly what we have done is center of mass. i. A hollow sphere of mass M and radius R with the line coinciding with its diameter as the axis of rotation: To find the moment of inertia of the given hollow sphere, let us divide it into several elemental rings having the same axis as that of the hollow sphere, as shown in figure 2.28. Let the elemental ring shown in figure is at an angular position T from the reference line. If dm be the mass of the ring, then.
u 2 S R cos T .R d T
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M cos T .d T 2
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The moment of inertia of the elemental ring, about the chosen axis is I = sum of moments of inertia of elemental rings
³
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= dI
M R2 2
S / 2
³ S
cos3 T .d T
/2
M R2 4 u 2 3
[Solve the integral on your own.]
2 M R2 . 3
PARALLEL AXIS THEOREM: We can often simplify the calculation of moments of inertia for various bodies by using general theorems relating the moment of inertia about one axis of the body to that about another axis. Steiner’s theorem or parallel axis theorem, relates the moment of inertia about an axis through the centre of mass of a body to that about a 10
second parallel axis. Let Icm be the moment of inertia about an axis through the centre of mass of a body and I www.NetBadi.com be that about a parallel axis a distance h away. The parallel axis theorem states that I = Icm + mh2 ……………. (7.20) where m is the mass of the body, Proof: Consider the situation shown in figure 7.29(a). A body of mass M is shown with its centre of mass coinciding with the origin of the reference frame. The moment of inertia of an element of mass dm at the point (x, y, z), as shown in figure 7.29 (b), about the axis coinciding with the z-axis and hence through the centre of mass of the body is § square of perpendicular · dm. ¨ ¸ © distnace from the z axis ¹
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dI cm
= dm. (x2 + y2) Now, consider another axis parallel to the z-axis and meeting the x-y plane at the point (a, b) as showninfigure729 (b). The moment of inertia of the same element about this axis is
2
2 y b »º ¼
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dm. «ª x a ¬
dI cm
[Using change of reference from method].
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d m x 2 y 2 d m a 2 b 2 2 a.d m.x 2b.dm. y
y
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I
d
z
b
x
a
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I
Therefore, the moment of inertia of the whole body about this axis is I
³d I
w
³d m
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³d I
³
³
³
x 2 y 2 d m a 2 b2 2 a d m.x 2b. d m. y
cm
³ d m .d
2
³
³
2a d m.x 2b d m. y
ªd m x 2 y 2 d Icm º « » « and a 2 b2 d 2 » ¬ ¼
ª dI I cm ; d m m; dm.x cm « Icm m.d 0 « «¬ d m. y m. ycm 0. 2
I
³ ³
³
³
m.xcm
0; º » » »¼
I cm md 2
Applications of parallel axis theorem: Some applications of parallel axis theorem are shown in figure 7.30.
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I cm m h 2
I
ml 2 §l· m¨ ¸ 12 ©2¹
2
I cm mh 2
I
r
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ml 2 ml 2 12 4 2 ml 3
mr 2 mr 2
2mr 2 I
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I
I
I cm mh2
ad
mr 2 mr 2 2 3 2 mr 2
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Fig. 7.30: Applications of parallel axis theorem
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PERPENDICULAR AXES THEOREM: The plane figure theorem or perpendicular axis theorem relates the moments of inertia about two perpendicular axes in a plane figure to the moment of inertia about a third axis perpendicular to the figure. If x, y, z, are perpendicular axes for a figure which lies in the x-y plane, the moment of in the inertia about the z-axis equals the sum of the moments of inertia about the x and y axes. z
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y
Therefore, we have, Iz = Ix + Iy
x dm r
y x
o
fig.7.31
………….. (7.12)
Before applying this theorem you must make that the body is in x – y plane only and the third axis (z – axis in this case) must pass perpendicular through the intersection points of x and y axis. Proof: The moment of inertia of the chosen element in figure 7.31, about the x – axis is dIx = dm.x2 12
The moment of inertia of the element about the y – axis is dIx = dm. y2 The moment of inertia of the element about the z – axis is dIz = dm.r2 = dm. (x2 + y2) = dm. x2 + dm. y2 = dIx. dIy. Therefore, the moment of inertia of the whole planar body about the z – axis is
³ dI
z
³ dI ³ dI x
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Iz
y
Ix I y Iz
Ix I y
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Applications of perpendicular axis theorem: Some applications of perpendicular axis theorem are shown in figure 7.32:
Fig. 7.32 (b): Finding moment of inertia of a disc about its diameter.
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Fig.7.32 (a): Finding moment of inertia of a ring about its diameter
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Fig 7.32 (c): Finding moment of inertia of a rectangular plate about an axis passing through its axis and perpendicular www.NetBadi.com to its plane. NOTE: For a square plate, put a = b. The following table contains the expression for moments of inertia of uniform bodies of various shapes:
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Figures
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Example 6:
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Find the moment of inertia of a uniform ring of mass m and radius r about an axis, AA’ touching the ring tangentially and lying in the plane of the ring only, as shown in figure 7.33 (a).
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Solution: The axis AA’ is parallel to the axis passing through the center of mass of the ring and lying in the plane of the ring only, as shown in figure 7.33 (b). Using parallel axis theorem, we get,
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I
I cm mh 2
mr 2 mr 2 4
5 2 mr 4
Example 7: A disc of radius R/2 has been removed from a disc of radius R and mass M, as shown in figure 7.34. Find the moment of inertia of the remaining body about an axis which is perpendicular to the plane of the body and touches it tangentially as shown in figure (axis is perpendicular to the plane of the body).
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Solution: We have, Iinitial I removed 2º ª3 2º ª3 « 2 minital R » « 2 mremoved R / 2 » ¬ ¼ ¬ ¼
ª « mremoved ¬«
M
SR
2
u S R/2
2
º M /4 » ¼»
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3 3 M R2 M R2 2 2 4 4
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I remaining
ad
45 MR 2 . 32
Example 8: Find the moment of inertia of a third rod AB of length l about an axis perpendicular to the length of the rod and passing through its ends A. Mass per unit length for the rod at a distance r from the end A is given as O0 D r where O0 and D are positive constants.
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O r
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l
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Solution: Let us divide the rod in to several elements so that each element can be considered as a point mass. Such an element of mass dm and length dr is shown in figure 7.35 (b) at a distance r from the axis of rotation. The moment of inertia of the element dI = dm. r2
O.dr .r (O0 D r ).dr .r 2
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O0 r 2 .dr D r 3 .dr
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Therefore, the moment of inertia of the rod is I
³d I l
l
³
³
O0 r 2 .d r D r 3 .d r 0
0
O0 l 3
D l4 4
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3
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Example 9: Find the moment of inertia of a uniform rod of mass m and length l about an axis passing through one of the ends of the rod and making an angle T with the length of the rod, as shown in figure 7.36(a).
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d m. r sin T
dI
2
§m · ¨ l dr ¸ r sin T © ¹
2
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Solution: The moment of inertia of the element shown in figure 7.36 (b) is
ªr sin T is the dis tan ce of º «the element from the axis » ¬ ¼
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m sin 2 T 2 r dr. l
w
r sin
dr
w
r
axis
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fig 7.36(b)
Therefore, the moment of inertia of the rod is I
³d I
m sin 2 T l
m,
dm
l
³ r .d r 2
[ T is same for all elements]
0
m sin 2 T l 3 . l 3
17
ml 2 sin 2 T . 3
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ALTERNATE METHOD: Consider the situation shown in figure 7.36 (c). The rod AB can be made from the rod A’B’ by displacing its each element parallel to the axis of rotation and by different distances. B m,
A
A'
B' sin
axis
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fig. 7.36 (c)
Therefore, the moment of inertia of the rod AB. = the moment of inertia of the rod A’B’
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3
2
[As the distribution of mass in A’B’ is uniform] ml 2 sin 2 T 3
et B
m l sin T
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stretch
NOTE: Whenever you use this approach of displacement of mass parallel to the axis must be careful. Analyze the situation shown in figure 7.37. A uniform cone is compressed along the length parallel to its axis to get a disc. Hence we cannot say that the moment of inertia of the cone about its axis is
M R2 because the disc we have 2
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got is not a uniform one. If we have the moment of inertia of this non uniform disc then of course that will be equal to the moment of inertia of the cone about the axis of symmetry.
Example 10: Find the moment of inertia of the body shown in figure about the axis passing through the point 0 and perpendicular to the plane of the body. The shown body has a mass M and it is one - fourth part of a uniform disc of radius R.
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R2 2
…………… (ii)
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4M .
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I0
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Solution: If we combine four such bodies to form the complete disc, then, moment of inertia of the inertia of the complete disc is I0 = 4I …………… (i) Where I is the moment of inertia of each of the four parts of the disc. Again, mass of the disc is 4M and radius is R, therefore,
From equations (i) and (ii), we have, 4M
R2 2
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4I
MR 2 2
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I
= moment of inertia of the given body.
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ALTERNATE METHOD: The distribution of mass with respect to the given axis of rotation in the given body has the same fashion as that
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in a uniform disc, hence, the moment of inertia of the given body is
MR 2 . 2
RADIUS OF GYRATION: The radius of gyration, k, of a body about an axis is the distance from the axis where the whole mass of the body can be assumed to be concentrated so that the moment of inertia about that axis remains the same. If k be the radius of gyration of the body shown in figure 7.39, then the entire mass can be assumed to be concentrated at a distance k from the axis. As the moment of inertia is same in both the situations, we have, I = Mk2 k
M /I
…….. (7.22)
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K
M, I
M
axis
axis
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MASS MOMENT OF INERTIA: Single Answer Correct
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2. A wheel has mass of the 1 kg, having 50 spokes each of mass 5g. The radius of the wheel is 40 cm. Find the MOI.
fig. 10.21
Solution: (c)
I
b. 1.73 kgm2 d. 2.73 kgm2.
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a. 0.273 kgm2 c. 0.173 kgm2
m r 2 50
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1u 0.4
2
ml 2 3
2
3 0.173 kgm 2
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0.16 1.083
50 5 u 103 .4
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3. The radius of gyration of a disc about a tangent parallel to one of the diameters is a.
r
2
Solution: (b) I
b.
3 r 3
3 3 m r 2 about the tangent m r2 4 4
c.
2 3
mk 2 or k
r
d. 3 r 2
4. A bout which axis MOI in the given triangular lamina is maximum a. AB b. BC c. AC
20
3 r. 2
d. BL.
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5. The radius of gyration for a right circular cone is 3 r 5
b.
5 r 3
Solution: (c) MOI of cone is
k
3 mr 2 10
3 r 10
d.
mk 2
3 r 10
10 r. 3
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c.
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Solution (b) MOI is 6 mi ri2 . About BC masses are spread far away than any other axis.
a.
ml 2 4
b. ml 2
c.
3 ml 2 4
d.
2 ml 2 . 4
I1 I 2 I3
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Solution: (d) I
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6. Three identical rods each of mass m and length l form an equilateral triangle. MOI about one of the sides is
2 ml 2 . 3
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ml 2 ml 2 0 3 3
7. The MOI of a system of four rods each of length l and mass m about the axis shown is
21
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a.
2 2 ml 3
b. 2 ml 2
c. 3 ml 2
d.
8 2 ml . 3
§l l · 2 ml 2 Solution: (d) Consider a square lamina, then 4m ¨¨ 12 12 ¸¸ about COM = . 3 © ¹
§ l · 2 ml 2 4m¨ ¸ 3 © 2¹
2
8 2 ml . 3
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Apply perpendicular axis theorem
2
om
2
(b)
Fig.10.24
et B
8. The surface density of a circular disc of radius a depends on the distance as U r = A + Br. The MOI about the line perpendicular to the plane of disc is §A
·
2a
§A
4 a. S a ¨ 2 5 B ¸ © ¹
Ba ·
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§A
3 c. 2 S a ¨ 2 5 ¸ © ¹
Solution: (a) dm 2 S r dr U
³
a
d mr2
S A a2
0
d. none of these.
A Br 2 S r d r
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2B ·
4 b. S a ¨ 2 5 ¸ © ¹
2
2 S B a5 . 5
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9. One quarter of the disc of mass m is removed. The new MOI is (Radius is r)
3 mr2 2
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a.
Solution: (c) New mass m '
b.
mr2 2
c.
3 mr2 8
3 m. 4
22
d. none of these.
1 m'r2 2
I
3 m r2. 8
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1.
Three rings each of mass m and radius r are so placed that they touch each other. Find the MOI about the axis as shown in the figure. 10.20 (a)
a.
5mr2
b.
d.
7 m r2 . 2
I1 I 2 I3
I2
I3
mr 2 2
3 2 mr 2
I1 I 2 I 3
om
I1
? I
c. 7mr2
7 2 mr . 2
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Solution: (d) I
5 m r2 2
10. The moment of inertia of a ring an axis passing through its center and lying in the plane of the ring is b.
1 MR 2 2
c.
1 MR 2 4
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a. MR2
d. None of these.
11. Let 1 be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the
et B
centre of the plate and makes an angle T with AB. The moment of inertia of the plate about the axis CD is then equal to: b. I sin2 T
a. I
c. I cos2 T
d. I cos2 ( T /2).
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12. About which of the following axes, the moment of inertia of a thin circular disc is minimum? a. Through center perpendicular to the surface. b. Through center parallel to the surface. c. Tangential and perpendicular to the surface. d. Tangential and parallel to the surface.
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13. The diameter of a flywheel increases by 1%. What will be percentage increase in moment of inertia about axis of symmetry? a. 2% b. 4% c. 1% d. 0.5%.
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14. Figure shows a uniform solid block of mass M and edge lengths a, b and c. Its moment of inertia about an axis through one edge and perpendicular (as shown) to the large face of the block is:
23
a.
M 2 a b2 3
b.
M 2 a b2 4
c.
7K 2 a b2 12
d.
M 2 a b2 . 12
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15. Four holes of radius R are cut from a thin square plate of side 4R and mass M. The moment of inertia of the remaining portion about z – axis is: S 12
§4 S · 2 b. ¨ 3 4 ¸ MR © ¹
MR 2
§ 8 10 S c. ¨ 3 16 ©
§4 S · 2 d. ¨ 3 6 ¸ MR . © ¹
· 2 ¸ MR ¹
om
a.
l 2
b. l
3 2
c.
l
2
d. l 3 .
et B
a.
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16. Three identical rods, each of mass m and length l are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plate of the triangle is
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17. A symmetrical lamina of mass M consists of a square shape with a semicircular section over each of the square as shown. The moment of inertia of the lamina about an axis 1.6 Ma2. The moment of inertia about the tangent AB in the plane of the lamina is
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a. 4.8 Ma2 b. 2.6 Ma2 c. 1.8 Ma2 d. 7.2 Ma2. One or More than one answer(s) correct: 18. The moment of inertia of a body depends on a. mass of the body b. angular velocity of the body c. axis of rotation of the body d. distribution of mass in the body. 19. A ring, a solid sphere and a disc have the same mass and radius. Then, which of the following is/are true? a. I ring ! I disc
b. I disc ! I sphere
c. I sphere ! I disc
d. I ring ! I sphere .
20. A closed tube partly filled with oil is lying in a horizontal plane. Which of the following is/are true. 24
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a. When rotated about the perpendicular bisector its inertia decreases www.NetBadi.com b. When rotated about the perpendicular bisector its inertia increases c. When rotated about the end its inertia increases d. When rotated about the end its inertia decreases. 21. A circular ring has a moment of inertia I about an axis passing through perpendicular to the plane. The moment of inertia can be decreased by a. Choosing any other parallel axis b. Decreasing its radius while keeping mass constant c. Decreasing the mass while keeping radius constant d. Decreasing its angular velocity. 22. Consider the following homogeneous rectangular lamina, where AB = 2 BC then, b. I yy ' ! I zz '
c. I yy ' ! I xx '
d. All of these.
23. A rod length L is composed of a uniform length
L L of wood whose mass is Mw and a uniform length 2 2
et B
of brass whose mass is Mb.
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a. I zz ' ! I xx '
1 M w L2 . 12
The moment of inertia of rod about an axis passing through the center and perpendicular to its is
b.
The moment of inertia of rod about an axis as specified in (A) is
c.
The moment of inertia of rod about an axis passing through the wood end perpendicular to the rod is
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a.
1 M w M b L2 . 2
1 M w M b L2 . 3
The moment of inertia of rod about an axis as perpendicular in (C) is 1 §L· Mw ¨ ¸ 12 ©2¹
2
IG1
1 §L· Mb ¨ ¸ 12 ©2¹
2
IG2
1 M w 7 M b L2 . 12
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d.
I1 = Moment of inertia of wooden part about specified axis shown in figure 1 2
I1
1 §L· §L· Mb ¨ ¸ M w ¨ ¸ 12 ©2¹ ©4¹
2
I1
I1
§L· I G1 M w ¨ ¸ ©4¹
1 M w L2 12 25
2
Similarly, I2
I1
1 M w L2 12
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1 M b L2 12
It I is the total moment of inertia of the combined system about axis shown in figure 1, then Further, I 2
2
{OPTION (B)}
1 9 §L· M b ¨ ¸ M b L2 12 © 2 ¹ 16
I
I
7 M b L2 12
I
I1 I 2
I
1 M w 7M b L2 {option (d)} 12
om
1 M w M b L2 12
I
et B
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2
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I = I1 + I2
§ 3L · I G2 M b ¨ ¸ © 4 ¹
24. The moment of inertia of a solid cylinder of mass M, length 2R and radius R about an axis passing through the center of mass and perpendicular to the axis of the cylinder is I1 and about an axis passing through one end of the cylinder and perpendicular to the axis of cylinder is I2. I2
b. I 2 ! I1
MR 2
I1
1 1 MR 2 M 4R 2 2 12
I1
1 1 MR 2 MR 2 2 3
I2
1 1 MR 2 M 4 R 2 2 3
1 1 MR 2 Ml 2 2 12
I2
1 4 MR 2 MR 2 2 3
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w
I1
c. I 1
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.N
a. I 2 I1
Also,
9 12
d. I1 I 2
MR 2 .
5 MR 2 6
11 MR 2 6
PASSAGE – 1 : The moment of inertia of a particle of mass m about an axis distant r from it is given by mr2. For a body, it is given by I
6 Mi ri2 where mi, ri are the mass and distance from the axis of rotation of the ith particle. The theorem of
parallel axis states that the moment of inertia about an axis is equal to the moment of inertia about a parallel axis passing through the C. G. of the body plus the product MR2 where M is the mass of the body and R is the distance between the axis of rotation and the C.G. of the body. 26
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Questions on Passage – 1 1.
The moment of inertia of a circular disc of radius R and mass M about one of its diameters is I. The moment of inertia about the axis XX’ will be.
a.
I
2. a. 3.
A solid sphere and a hollow sphere both have same mass and same radii. Which of them has a larger moment of inertia about one of the diameters? Solid b. Hollow c. Both have same monument of inertia. If Ix, Iy, Iz denote the moment of inertia of a particle about X – Y and Z – axis, which of the following is true?
a.
Iz
c.
Iz
4.
Tow bodies of moment of inertia I and 2I are subjected to torques 2L and L. The ratio of their angular acceleration will be 1:2 b. 2 : 1 c. 1 : 4 d. 4 : 1.
d. I
c. 2I
MR 2 . 4
b. I z ! I y ! I x
ad
I x2 I y2 Ix I y
d. I z I x I y
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I Match The Following: Body
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om
b. I MR 2
0
et B
a.
1 MR 2 2
Radius of gyration. (p)
b. Solid cylinder of radius R and length L
(q)
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a. Solid cylinder of radius R about its axis
2 R 5 R 2
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about its central diameter c. Annular cylinder of radius R and outer
(r)
radius L about cylinder axis
d. Solid sphere of diameter 2R about its
(s)
diameter
27
R 2 L2 2
R 2 L2 4 12
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II Match The Following: Moment of inertia.
a. A ring about its axis
(p)
MR 2 2
b. A uniform circular disc about its axis
(q)
2 MR 2 5
(r)
7 MR 2 5
d. A solid sphere about any tangent
(s)
MR 2.
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c. A solid sphere about any diameter
om
System
12. B
3. C
Match the following: c – s, c – q,
14. A 15. C 20. B,C 21. B, C
4 D.
d – p. d – r.
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I, a – q, b – r, II a – s, b – p,
13. A 19. A, B, D
et B
10. B 11. A 17. A 18. A, C, D Passage 1. B 2. B
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Answers:
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Page no 39 and 40 type
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16. C 22. A, C.
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