Question

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertical axis through its center. A 0.8 kg piece of putty is dropped vertically onto the cylinder at a point 0.300 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

Answer 1

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$