Question

What is the angular momentum of a 3.2-kg uniform cylindrical grinding wheel of radius 27 cm when rotating at 1500 rpm ?

Answer 1

Angular momentum is given by

$L = I w$

here we have

I = moment of inertia of wheel

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}*3.2*0.27^2$

$I = 0.117 kg*m^2$

angular speed is given as 1500 rpm

$w = 2 \pi f$

$w = 2 *\pi * \frac{1500}{60}$

$w = 157 rad/s$

now angular momentum is given by

$L = 0.117 * 157 = 18.4 kg*m^2* rad/s$