Question

A 1 kg mass of clay, moving with a velocity of 10$\frac{m}{s}$, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. Assuming that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately

• A. Zero
• B. $\frac{1}{3}rad/s$
  
• C. $\sqrt{\frac{10}{3}}rad/s$
• D. $\frac{10}{3}rad/s$

Answer 1

The correct option is B $\frac{1}{3}rad/s$


In this question we need to neglect the mass of clay w.r.t. mass of wheel.

We can't conserve linear momentum as friction is there so we will conserve angular momentum about the point of the wheel which is in contact with the surface. (because torque about A of all forces is zero)

Taking both clav and wheel as a system.

$L_i=L_f$ (about A)

$L_i=mvr=1\times 10\times 1=10$

$L_f=m{v}_{cm}f+I\omega$

$=20\times r\omega \times r+\frac{m{r}^2}{2}\omega$

$=20\times 1\times \omega \times 1+\frac{20}{2}\times 1^2\times \omega$

$L_i=L_f$

$10=20\omega +10\omega$

$\Rightarrow 10=30\omega$

or $\omega =\frac{10}{30}=\frac{1}{3}rad/s$