Question

A 2.73 kg cylindrical grinding wheel with a radius of 31 cm rotates at 1416 rpm. What is its angular momentum? Answers:

Answer 1

Answer:

The angular momentum of the wheel is  $19.45\ kg-m^2/s$.

Explanation:

It is given that,

Mass of the wheel, m = 2.73 kg

Radius of the wheel, r = 31 cm = 0.31 m

Angular speed of the wheel, $\omega=1416\ rpm= 148.28\ rad/s$

We need to find its angular momentum. It is given by :

$L=I\omega$

I is the moment of inertia of the wheel, $I=\dfrac{mr^2}{2}$

$L=\dfrac{mr^2}{2}\times \omega$

$L=\dfrac{2.73\times (0.31)^2}{2}\times 148.28$

$L = 19.45\ kg-m^2/s$

So, the angular momentum of the wheel is  $19.45\ kg-m^2/s$. Hence, this is the required solution.