Question

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

Answer 1

Explanation:

It is given that,

Force on grindstone, F = 180 N

Radius of grindstone, r = 0.28 m

Mass of grindstone, m = 75 kg

We need to find the angular acceleration of the grindstone. In rotational motion, the relation between the torque and angular acceleration is given by :

$\tau=I\times \alpha$

$\alpha=\dfrac{\tau}{I}$

I is the moment of inertia of solid disk, $I=\dfrac{1}{2}mr^2$

$\tau$ is the torque exerted, $\tau=F\times r$

$\alpha=\dfrac{F\times r}{\dfrac{1}{2}mr^2}$

$\alpha=\dfrac{2F}{mr}$

$\alpha=\dfrac{2\times 180\ N}{75\ kg\times 0.28\ m}$

$\alpha =17.14\ rad/s^2$

So, the angular acceleration of the disk is $17.14\ rad/s^2$. Hence, this is the required solution.