ch-10

Chapter 10 Rotational Motion r O

P

Reference line

(a)

P r O

u

s Reference line

(b)

F I G U R E 10.1

A compact disc rotating about a fixed axis through O perpendicular to the plane of the figure. (a) To define angular position for the disc, a fixed reference line is chosen. A particle at P is located at a radial distance r from the rotation axi s at O. (b) As the disc rotates, point P moves through an arc length s on a circular path of radius r. The angular position of P is .

y

,t f

r

,ti

θf θi O

F I G U R E 10.2

x

A particle on a rotating rigid object moves from to along the arc of a circle. In the time interval t tf t i , the radial line of length r sweeps out an angle f i.

ω

ω

FIGURE

10.3 The orange disk rotates in the directions indicated. The right-hand rule determines the direction of the angular velocity vector.

y v P s

r u O

x

Figure 10.4 As a rigid object rotates about the fixed axis through O, the point has a tangential velocity : v that is always tangent to the circular path of radius r.

y at P

a ar

O

F I G U R E 10.5

x

As a rigid object rotates about a fixed axis through O, a particle at point P experiences a tangential component at and a radial component a r of translational acceleration. The total translational acceleration of this : a at : ar , where particle is : : a c ˆr. ar

z axis

v vi mi ri

O

F I G U R E 10.6

A rigid object rotating about the z axis with angular speed . The kinetic energy of the particle of mass mi is 12 m i v i 2. The kinetic energy of the rigid object is called its rotational kinetic energy.

T A B L E 10.2

Moments of Inertia of Homogeneous Rigid Objects With Different Geometries Hoop or thin cylindrical shell I CM = MR 2

R

Solid cylinder or disk I CM = 1 MR 2 2

R

Hollow cylinder I CM = 1 M(R 12 + R 22) 2

R1

Rectangular plate I CM = 1 M(a 2 + b 2) 12 b a

Long thin rod with rotation axis through center I CM = 1 ML 2 12

Long thin rod with rotation axis through end

L

I = 1 ML 2 3

Solid sphere I CM = 2 MR 2 5

Thin spherical shell I CM = 2 MR 2 3 R

L

R

R2

y

m b

m

M

M a

a

x

M b

a

b O

m

a

b

M

m (b)

F I G U R E 10.7

(Example 10.3) Four spheres form an unusual baton. (a) The baton is rotated about the y axis. (b) The baton is rotated about the z axis.

(a)

z dr r R L

F I G U R E 10.8

(Example 10.4) The geometry for calculating the moment of inertia about the central axis of a uniform solid cylinder.

L

Pivot Mg

F I G U R E 10.9

(Example 10.5) A uniform rod rotates freely under the influence of gravity around a pivot at the left end.

F1 F sin φ d1

F

r

O

φ O

d2

r

F cos φ Line of action

φ

d F2

Figure 10.11 :

The force F 1 tends to rotate the object counterclockwise about an : axis through O, and F 2 tends to rotate the object clockwise.

F I G U R E 10.10

:

A force F is applied to a wrench in an effort to loosen a bolt. The force has a greater rotating tendency about O as F increases and as the moment arm d increases. The component F sin tends to rotate the system about O.

z

τ = r

F

y

O r P x

φ F

Figure 10.12 The torque vector : lies in a direction perpendicular to the plane formed by the position vector : r and the applied : force vector F .

Right-hand rule C = A × B

A

θ B

–C = B × A

F I G U R E 10.13

:

:

The vector product A B is a third vector C having a magnitude AB sin equal to the area of : the parallelogram shown. The vector C is perpendicular to : : the plane formed by A and B, and its direction is determined by the right-hand rule. :

y

T1

R1 R2 O

x

z T2

F I G U R E 10.14

(Example 10.6) A solid cylinder pivoted about the z axis : through O. The moment arm of T1 is : R 1, and the moment arm of T2 is R 2.

F

F

O

(a)

F 2r r

2F O

(b)

F I G U R E 10.15

(a) The two forces acting on the object are equal in magnitude and opposite in direction. Because they also act along the same line of action, the net torque is zero and the object is in equilibrium. (b) Another situation in which two forces act on an object to produce zero net torque about O (but not zero net force).

53.0° 8.00 m (a) R

T

θ

53.0° 200 N 600 N (b)

R sin θ

T sin 53.0°

R cos θ

T cos 53.0° 200 N

2.00 m 600 N 4.00 m

(c)

F I G U R E 10.16

(Example 10.8) (a) A uniform beam supported by a cable. A man walks out on the beam. (b) The free-body diagram for the beam – man system. (c) The free-body diagram with forces resolved into horizontal and vertical components.

P

n

θ θ O (a)

F I G U R E 10.17

mg g

fs (b)

(Example 10.9) (a) A uniform ladder at rest, leaning against a frictionless wall. (b) The free-body diagram for the ladder.

M, R

m1

+

+

m2 (a)

F

T1

T2 +

m1

+

m2

m 1g T1

T2 Mg

m 2g

(b)

F I G U R E 10.18

(Example 10.10) (a) An Atwood machine with a massive pulley. The pulley is modeled as a disk. (b) Free-body diagrams for the two hanging objects and the pulley.

F

φ ds dθ

P

r

O

F I G U R E 10.19

A rigid object rotates about an axis through O under the action of an external : force F applied at P.

M

O

R T

T

m

mg

F I G U R E 10.20

(Example 10.11) An object hangs from a cord wrapped around a wheel. The tension in the cord produces a torque about the axle passing through O.

z L = r × p

O r

m

y

p

φ x

Figure 10.21 :

The angular momentum L of a particle of mass m and linear momentum : p located at the position : : : r is given by L : r p . The value : of L depends on the origin about which it is measured and is a vector perpendicular to both : p. r and :

(© Stuart Franklin/Getty Images)

F I G U R E 10.22

Angular momentum is conserved as Russian figure skater Evgeni Plushenko performs during the 2004 World Figure Skating Championships. When his arms and legs are close to his body, his moment of inertia is small and his angular speed is large. To slow down for the finish of his spin, he moves his arms and legs outward, increasing his moment of inertia.

O

R

vi m

F I G U R E 10.24

(Example 10.13) When the string is pulled downward, the speed of the puck changes.

(David Malin, Anglo-Australian Observatory)

F

F I G U R E 10.23

The Crab Nebula, in the constellation Taurus. This nebula is the remnant of a supernova explosion, which was seen on Earth in the year A.D. 1054. It is located some 6 300 lightyears away and is approximately 6 lightyears in diameter, still expanding outward.

z L

CM

n

r

Mg

(a) y

O

x

∆L Li

Lf

(b)

F I G U R E 10.25 Precessional motion of a top spinning about its symmetry axis. (a) The only external forces acting on the top are the normal force : n and the gravitational force M : g . The : direction of the angular momentum L is along the axis of symmetry. The righthand rule indicates that : : : : r F r M: g is in the xy : plane. (b). The direction of L is : parallel to that of in part : : : (a). That L f L i L indicates that the top precesses about the z axis.

Light sources at the center and rim of a rolling cylinder illustrate the different paths these points take. The center moves in a straight line (green line), whereas a point on the rim moves in the path of a cycloid (red curve).

(Courtesy of Henry Leap and Jim Lehman)

F I G U R E 10.26

P′

2 v CM

vCM

CM

P

F I G U R E 10.27

All points on a rolling object move in a direction perpendicular to a line through the instantaneous point of contact P. The center of the object moves with a velocity : vCM, whereas the point P moves with a velocity 2: vCM.

M R

h

ω

x

θ

vCM

Figure 10.28 (Example 10.15) A round object rolling down an incline. Mechanical energy of the object –surface – Earth system is conserved if no slipping occurs and there is no rolling resistance.

Sprocket Chain

Crank

Figure P10.11

y 4.00 kg

y = 3.00 m

x

O 2.00 kg

3.00 kg

y = –2.00 m y = –4.00 m

Figure P10.14

m

(John Lawrence/Stone/ Getty Images)

Figure P10.15

Figure P10.16 Problem 10.16

R

I

m1 2h m2

Figure P10.17

Figure P10.19

2.00 m

20.0° 37.0°

20.0°

100 N

Figure P10.20 10.0 N

30.0°

a O

12.0 N

b 9.00 N

Figure P10.21 2.00 m

Fg1

Fg 2

Figure P10.26

d

2 m1

m2

P

O

CG x

Figure P10.27

F

30.0 cm

30.0°

Single point of contact

5.00 cm

Figure P10.29

θ

d

2L

Figure P10.31

A 1.00 m B 2.00 m

(3 000 kg)g 10 000 kg 6.00 m

Figure P10.32

Figure P10.35

Figure P10.36 2.00 m/s2 T1

T2

15.0 kg m1

m 2 20.0 kg

37.0°

Figure P10.37

Figure P10.44 Problems 10.44 and 10.50.

I2 ωi

ωf

1

Before

After

Figure P10.45 ωi

ωf

(a)

(b)

Figure P10.48

1.50 m/s

(a)

(b)

Figure P10.49 North Star

(a)

(b)

(NASA)

Cone of precession

Figure P10.52 (a) At present, the spin axis of the Earth points toward the North Star. (b) Torque on the spinning Earth will cause it to precess, so the spin axis will no longer be pointing in this direction in the future.

Figure P10.55