Question

A solid, horizontal cylinder of mass 10.6 kg and radius 1.00 m rotates with an angular speed of 8.00 rad/s about a fixed vertical axis through its center. A 0.250-kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

Answer 1

Answer:

The final angular speed is 7.71 rad/s

Explanation:

Given

Cylinder mass, M = 10.6 kg

Cylinder radius, R = 1.00 m

Angular speed, w = 8.00 rad/s.

Mass of putty, m = 0.250-kg

Radius, r = 0.900 m

First, we set up an expression for the initial and final angular momentum of the system.

The moment of inertia of the cylinder is given as I = ½MR²

While the moment of inertia if the putty is mr².

Initial Momentum of the system = Initial momentum of the cylinder =

Li = Iw --- Substitute ½MR² for I

Li = ½MR²w

By

Substituton

Li = ½ * 10.6 * 1² * 8

Li = 42.4kgm²/s

Calculating the final momentum of the system.

First we calculate the final momentum of the cylinder

Li = Iw --- Substitute ½MR² for I

Li = ½MR²wf where wf = final angular speed

By

Substituton

Li = ½ * 10.6 * 1² wf

Li = 5.3w kgm²/s

Then we calculate the final momentum of the putty

Final Momentum of the putty =

L2 = Iwf --- Substitute mr² for I;

L2 = mr²wf --- By Substituton

L2 = 0.25 * 0.9² * wf

L2 = 0.2025wf kgm²/s

Final momentum = Li + L2

Lf = (5.3wf + 0.2025wf) kgm²/s

Lf = 5.5025wf kgm²/s

By conservation of momentum

Li = Lf

Where Li = 42.4kgm²/s and Lf = 5.5025wf kgm²/s

So, we have

5.5025wf kgm²/s = 42.4kgm²/s --- make wf the subject of formula

wf = 42.4/5.5025

wf = 7.71 rad/s

Hence, the final angular speed is 7.71 rad/s