Question

A sanding disk with rotational inertia 8.2 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 14 N·m about the central axis of the disk. About that axis and with torque applied for 54 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

a)
`L = 0.756\,(kg\cdot m^(2))/(s)`, b)
`\omega = 92.195\,(rad)/(s)`

Step-by-step explanation:

b) Final angular speed is obtained by using the concept of Angular Momentum and Impact Theorem:

`I_(g)\cdot \omega_(o) + T \cdot \Delta t = I_(g)\cdot \omega`

`\omega = \omega_(o) +(T\cdot \Delta t)/(I_(g))`

`\omega = 0\,(rad)/(s) + ((14\,N\cdot m)\cdot (54* 10^(-3)\,s))/(8.2* 10^(-3)\kg\cdot m^(2))`

`\omega = 92.195\,(rad)/(s)`

a) The angular momentum of the disk is:

`L = (8.2* 10^(-3)\,kg\cdot m^(2))\cdot (92.195\,(rad)/(s) )`

`L = 0.756\,(kg\cdot m^(2))/(s)`