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

$I=\frac{1}{2}mr^2$

where

m = 75.0 kg is the mass of the disk

r = 0.280 m is the radius

Substituting,

$I=\frac{1}{2}(75.0)(0.280)^2=2.94 kg m^2$

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:

$\alpha = \frac{\tau}{I}=\frac{50.4 N \cdot m}{2.94 kg m^2}=17.2 rad/s^2$