ME3112 - Free Vibration - Solution

Question 1 Assume small oscillation, the cylinder rotates about C. Based on this, the following relations can be defined 𝑥𝑂 = 𝑟𝜃, and 𝑥𝐴 = 2𝑟𝜃 where 𝑥𝑂 and 𝑥𝐴 are the cylinder displacement at point O and A from its equilibrium position. And at equilibrium position, the moment equation about the contact point gives 𝑘𝛿2𝑟 − 𝑚𝑔𝑟𝑠𝑖𝑛(15𝑜 ) = 0 The spring force at the positive displaced position is 𝐹𝑆 = 𝑘(𝑥𝐴 + 𝛿) = 2𝑘𝑟𝜃 + 𝑘𝛿 Now, from FBD of the cylinder

 𝐼𝜃̈ o

15

C

C mg

Taking moment about C gives the equation of motion of the system 2𝑟𝐹𝑆 − 𝑚𝑔𝑟𝑠𝑖𝑛(15𝑜 ) = 𝐼𝜃̈ + 𝑚𝑟𝑥̈ 𝑂 ⟹ 2𝑟(2𝑘𝑟𝜃 + 𝑘𝛿) − 𝑚𝑔𝑟𝑠𝑖𝑛(15𝑜 ) = 𝐼𝜃̈ + 𝑚𝑟𝑥̈ 𝑂 ⟹ 4𝑟 2 𝑘𝜃 + [𝑘𝛿2𝑟 − 𝑚𝑔𝑟𝑠𝑖𝑛(15𝑜 )] = 𝐼𝜃̈ + 𝑚𝑟𝑥̈ 𝑂 𝑥𝑂 1 𝑥̈ 𝑂 ⟹ 4𝑟 2 𝑘 ( ) = ( 𝑚𝑟 2 ) ( ) + 𝑚𝑟𝑥̈ 𝑂 , 𝑟 2 𝑟 3 ⟹ 𝑚𝑥̈ 𝑂 + (4𝑘)𝑥𝑂 = 0, or 2 8𝑘 ⟹ 𝑥̈ 𝑂 + ( ) 𝑥 = 0, 𝑜𝑟 3𝑚 𝑂

1 or 4𝑟 2 𝑘𝜃 = ( 𝑚𝑟 2 ) 𝜃̈ + 𝑚𝑟 2 𝜃̈ 2 3 𝑚𝜃̈ + (4𝑘)𝜃 = 0 2 8𝑘 𝜃̈ + ( )𝑥 = 0 3𝑚 𝑂

Hence, the natural frequency of this system is 𝜔𝑛 = √(

8𝑘 8 200 )=√ ( ) = 7.303 𝑟𝑎𝑑/𝑠 3𝑚 3 10

(a) And its period is 𝜏𝑛 =

2𝜋 2𝜋 = = 𝟎. 𝟖𝟔𝟎 𝑠 𝜔𝑛 7.303

(b) Given that it is displaced by 5 cm, and then set into vibration, which is given by 𝑥𝑂 = 0.05cos(𝜔𝑛 𝑡) ⟹ 𝑥̈ 𝑂 = −0.05𝜔𝑛 2 cos(𝑡) Hence, the maximum acceleration during the vibration is (𝑥̈ 𝑂 )𝑚𝑎𝑥 = 0.05𝜔𝑛 2 = 0.05(7.303)2 = 𝟐. 𝟔𝟕 𝑚𝑠 −2

*Note: In this example, you can again choose to ignore the effects of the initial spring force and the weight of the cylinder (because they counter-balanced each other during the oscillation). This condition, however, cannot be generalized to all cases, such as in Question 2. So if you are not sure whether this condition is satisfied or not, please follow the above approach.

Question 2 At equilibrium, the FBD is

Taking moment about O, gives

𝐿 𝑚𝑔 ( 𝑠𝑖𝑛𝛼1 ) = 𝑘𝛿0 𝐿𝑐𝑜𝑠𝛼3 2

Now, suppose the rod oscillates by a small angular displacement 𝜃, the FBD becomes

Again, taking moment about O, the EOM of the rod is 𝐿 1 𝐿 2 −𝑚𝑔 ( 𝑠𝑖𝑛(𝛼1 + 𝜃)) + (𝑘𝛿0 𝑐𝑜𝑠𝛼3 − 𝑘𝐿𝜃)𝐿 = ( 𝑚𝐿2 + 𝑚 ( ) ) 𝜃̈ 2 12 2 𝐿 1 ⇒ −𝑚𝑔 (𝑠𝑖𝑛𝛼1 𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠𝛼1 𝑠𝑖𝑛𝜃) + (𝑘𝛿0 𝐿𝑐𝑜𝑠𝛼3 − 𝑘𝐿2 𝜃) = 𝑚𝐿2 𝜃̈ 2 3 Using small angle assumption, 𝐿 1 ⇒ −𝑚𝑔 (𝑠𝑖𝑛𝛼1 + 𝑐𝑜𝑠𝛼1 𝜃) + (𝑘𝛿0 𝐿𝑐𝑜𝑠𝛼3 − 𝑘𝐿2 𝜃) = 𝑚𝐿2 𝜃̈ 2 3 1 𝐿 𝐿 ⇒ 𝑚𝐿2 𝜃̈ + [𝑘𝐿2 + 𝑚𝑔 𝑐𝑜𝑠𝛼1 ] 𝜃 + [𝑚𝑔 𝑠𝑖𝑛𝛼1 − 𝑘𝛿0 𝐿𝑐𝑜𝑠𝛼3 ] = 0 3 2 2 1 𝐿 ⇒ 𝑚𝐿2 𝜃̈ + [𝑘𝐿2 + 𝑚𝑔 𝑐𝑜𝑠𝛼1 ] 𝜃 = 0 3 2 Hence, the natural angular frequency is 𝐿 𝑘𝐿2 + 𝑚𝑔 2 𝑐𝑜𝑠𝛼1 𝜔𝑛 = √ 1 2 3 𝑚𝐿

Question 3 Position 1: Suppose collar A is offset from its equilibrium position by xA . At this position, the system has maximum potential energy (due to spring), but zero kinetic energy. Position 2: And during oscillation, the system will attain zero potential energy when collar A is at the equilibrium position, which is also when it has the maximum kinetic energy. At position 1, the following displacement diagrams can be drawn



From the geometry of the system, one observes that 𝑥𝐴 = 𝑙𝜃𝑐𝑜𝑠𝛽 = (𝑙𝑐𝑜𝑠𝛽)𝜃 And for SHM, 𝜃 = 𝛩𝑠𝑖𝑛(𝜔𝑛 𝑡 + 𝜙) ⇒ (𝑥𝐴 )𝑚𝑎𝑥 = (𝑙𝑐𝑜𝑠𝛽)𝛩 Hence, the elastic potential energy is 1 1 2 𝑉1 = 𝑘(𝑥𝐴 )𝑚𝑎𝑥 = 𝑘(𝑙𝑐𝑜𝑠𝛽)2 𝛩2 2 2 At position 2, the velocity diagram at that instant can be drawn as shown.

Consider the velocities of collars A and B, 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝑣⃗𝐵/𝐴 ⇒ −𝑣𝐵 𝑗⃑ = 𝑣𝐴 𝑖⃑ + 𝜃̇ 𝑘⃑⃑ × 𝑟⃗𝐵/𝐴 ⇒ 𝑣𝐵 𝑗⃑ = 𝑣𝐴 𝑖⃑ + 𝜃̇ 𝑘⃑⃑ × (−𝑙𝑠𝑖𝑛𝛽𝑖⃑ + 𝑙𝑐𝑜𝑠𝛽𝑗⃑) And matching the velocity components, give ⇒ 𝑣𝐴 = 𝑙𝑐𝑜𝑠𝛽𝜃̇,

and

𝑣𝐵 = 𝑙𝑠𝑖𝑛𝛽𝜃̇

Similarly, the velocity at G is 𝑣⃗𝐺 = 𝑣𝐴 𝑖⃑ + 𝜃̇ 𝑘⃑⃑ × 𝑟⃗𝐺/𝐴 𝑙 𝑙 = (𝑙𝑐𝑜𝑠𝛽𝜃̇ )𝑖⃑ + 𝜃̇ 𝑘⃑⃑ × (− 𝑠𝑖𝑛𝛽𝑖⃑ + 𝑐𝑜𝑠𝛽𝑗⃑) 2 2 𝑙 𝑙 = ( 𝑐𝑜𝑠𝛽𝜃̇ ) 𝑖⃑ − ( 𝑠𝑖𝑛𝛽𝜃̇ ) 𝑗⃑ 2 2 Hence, the kinetic energy at this instant is 1 1 1 1 𝑇2 = 𝑚|𝑣⃗𝐺 |2 + 𝐼𝐺 𝜃̇ 2 + 𝑚𝑐 |𝑣⃗𝐴 |2 + 𝑚𝑐 |𝑣⃗𝐵 |2 2 2 2 2 1 𝑙 2 1 1 1 1 2 2 = 𝑚 ( 𝜃̇ ) + ( 𝑚𝑙 2 ) 𝜃̇ 2 + 𝑚𝑐 (𝑙𝑐𝑜𝑠𝛽𝜃̇ ) + 𝑚𝑐 (𝑙𝑠𝑖𝑛𝛽𝜃̇ ) 2 2 2 12 2 2 1 1 = ( 𝑚 + 𝑚𝑐 ) 𝑙 2 𝜃̇ 2 2 3 1 1 2 3

= ( 𝑚 + 𝑚𝑐 ) 𝑙 2 (𝜔𝑛 𝛩)2

(since 𝜃 = 𝛩𝑠𝑖𝑛(𝜔𝑛 𝑡 + 𝜙))

Finally, applying the principle of conservation of energy, gives 𝑉1 = 𝑇2 1 1 1 ⇒ 𝑘(𝑙𝑐𝑜𝑠𝛽)2 𝛩2 = ( 𝑚 + 𝑚𝑐 ) 𝑙 2 (𝜔𝑛 𝛩)2 2 2 3 𝒌𝒄𝒐𝒔𝟐 𝜷 ⇒ 𝜔𝑛 = √ 𝟏 (𝟑 𝒎 + 𝒎𝒄 )

Question 4 The FBD of this system when it is subjected to small oscillation is

Taking moment about A, we get

And for small angle rotation, this is reduced to

Substituting the numeric, gives

Hence, the EOM for this system when expressed in the equivalent 1D spring-mass-damper system is

And its critical damping would be given by

And hence the damping ratio is 𝜁=

𝑐𝑒𝑞 0.09 = = 𝟎. 𝟎𝟒𝟑𝟏 𝑐𝑐𝑟 2.087