Question

What is the angular momentum of a 3-kg uni- form cylindrical grinding wheel of radius 0.2 m when rotating at 1500 rpm (Rotational Inertia of a cylinder is mR^2/2).

Answer 1

Answer:

$L = 9.42 kg m^2/s$

Explanation:

Angular speed of the cylinder is given as

$f = 1500 rpm$

$f = 1500 round/60 s$

$\omega = 2\pi f$

$\omega = 2\pi(25) = 50 \pi$

now moment of inertia of the cylinder is given as

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

$I = \frac{1}{2}(3)(0.2)^2$

$I = 0.06 kg m^2$

now we have

$L = I\omega$

$L = (0.06)(50\pi)$

$L = 9.42 kg m^2/s$