Question

There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s. You carefully put a ring with moment of inertia Ir = 0.0353 kgm2 , on the plate, center to center of plate and they rotate with same angular speed later. Ignore all the friction other than the one between the plate and the ring. Please (a) calculate their common angular speed; (b) the kinetic energy loss during this process.

Answer 1

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Step-by-step explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J