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

`L = 9.42 kg m^2/s`

Step-by-step 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 = (1)/(2)mR^2`

`I = (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`