Question

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

27.8 kg*m^2/s
18.8 kg*m^2/s
22.8 kg*m^2/s
17.5 kg*m^2/s

Answer 1

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

Step-by-step 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=(mr^2)/(2)`

`L=(mr^2)/(2)* \omega`

`L=(2.73* (0.31)^2)/(2)* 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.