Question

Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c

Answer 1

(a)
`50.4 N\cdot m`

The torque exerted on the solid disk is given by

`\tau=Frsin \theta`

where

F is the magnitude of the force

r is the radius of the disk

`\theta` is the angle between F and r

Here we have

F = 180 N

r = 0.280 m

`\theta=90^(\circ)` (because the force is applied tangentially to the disk)

So the torque is

`\tau = (180 N)(0.280 m)(sin 90^(\circ))=50.4 N\cdot m`

(b)
`17.2 rad/s^2`

First of all, we need to calculate the moment of inertia of the disk, which is given by

`I=(1)/(2)mr^2`

where

m = 75.0 kg is the mass of the disk

r = 0.280 m is the radius

Substituting,

`I=(1)/(2)(75.0)(0.280)^2=2.94 kg m^2`

And the angular acceleration can be found by using the equivalent of Newtons' second law for rotational motions:

`\tau = I \alpha`

where

`\tau = 50.4 N \cdot m` is the torque exerted

I is the moment of inertia

`\alpha` is the angular acceleration

Solving for
`\alpha`, we find:

`\alpha = (\tau)/(I)=(50.4 N \cdot m)/(2.94 kg m^2)=17.2 rad/s^2`