A solid, vertical cylinder with a mass of 8.90 kg and a radius of 2.5 m rotates with an angular speed of 3.54?

Question

rad/s about a fixed vertical axis through its center. A 0.349 kg piece of putty is dropped vertically at a point 0.810 m from the cylinder's center of rotation and sticks to the cylinder. What is the final angular speed of the system? Answer in units of rad/s.

Answer 1

This is a typical problem involving conservation of the angular momentum. Since there is no external torques on the system cylinder-putty, the angular momentum remains the same:
I1 x w1 = I2 x w2
A moment of inertia of the cylinder is I1 = 1/2 x m x R2 = 1/2 x 8.90 kg x (2.5 m)2 = 27.8125 kg m2.
A piece of putty can be treated as a point particle. A particle itself has got moment of inertia m x r2 = 0.349 kg x (0.810 m)2 = 0.229 kg m2.
A new moment of inertia is the sum (because the axis of rotation remains the same): I2 = 28.0412 kg m2.
A new angular speed is w2 = I1 x w1 / I2 = 3.51 rad/s.

Answer 2

The rotational kinetic power is given via a million/2 the made of the 2nd of inertia and the sq. of the angular velocity. Ek = I?² 2nd of inertia of a solid cylinder: I = ½ mr² = 0.5*3*0.a million² = 0.0.5 angular velocity: ? = 3.5 rad/s Ek =0.0.5*3.5² = 0.183 J