Question

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (

Answer 1

Answer:

Explanation:

Given mass of grindstone $m=90\ kg$

radius of stone $r=0.34\ m$

angular speed of disc $\omega =90\ rpm$

Steel Axle applying a force of $F=20\ N$

coefficient of kinetic friction $\mu =0.2$

Frictional Torque applied by steel is given by

$\tau=r\times f_r$

$\tau =r\times \mu F$

where $f_r$=frictional force

$\tau =r\times \mu \times F$

$\tau =0.34\times 0.2\times 20$

$\tau =1.36\ N-m$

Torque is also given by

$\tau =I\cdot \alpha$

where $\alpha$=angular acceleration

I=moment of Inertia

$\tau =0.5Mr^2\times \alpha$

$0.5Mr^2\times \alpha =1.36\ N-m$

$0.5\times 90\times 0.34^2\times \alpha =1.36$

$\alpha =0.261\ rad/s^2$