Question

You have a grindstone (a disk) that is 98.0 kg, has a 0.335-m radius, and is turning at 100 rpm, and you press a steel axe against it with a radial force of 23.6 N. Assuming the kinetic coefficient of friction between steel and stone is 0.192, calculate the angular acceleration of the grindstone.

Answer 1

Answer:

$a=0.276$

Explanation:

From the question we are told that:

Mass $m=98.kg$

Radius $r=0.335$

Angular velocity $\omega=100rpm$

Radial force of $F_r=23.6 N.$

Kinetic coefficient of friction  $\mu=0.192$

Generally the equation for Kinetic Force is mathematically given by

 $F_k=\mu.F_r$

 $F_k=0.192*23.6$

 $F_k=4.5312$

Generally the equation for Torque on Center is mathematically given by

 $Ia=f_k*r$

Where

 $I=\frac{Mr^2}{2}$

Therefore

 $a=\frac{2f_k}{Mr}$

 $a=\frac{2*4.5312}{98*0.335}$

 $a=0.276$

Therefore Angular acceleration of the grindstone is

$a=0.276$