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

`a=0.276`

Step-by-step 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=(Mr^2)/(2)`

Therefore

`a=(2f_k)/(Mr)`

`a=(2*4.5312)/(98*0.335)`

`a=0.276`

Therefore Angular acceleration of the grindstone is

`a=0.276`