Question

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torque brings the grindstone to rest in 10 s. What is the number of revolutions of the grindstone after the power is shut off?

Answer 1

`\theta=50\ revolution`

Step-by-step explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone,
`\omega_i=600\ rpm=62.83\ rad/s`

Finally it shuts off,
`\omega_f=0`

Time taken, t = 10 s

Let
`\alpha` is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

`\alpha =(\omega_f-\omega_i)/(t)`

`\alpha =(0-62.83)/(10)`

`\alpha =-6.283\ rad/s^2`

Let
`\theta` is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

`\omega_f^2-\omega_i^2=2\alpha \theta`

`\theta=(\omega_f^2-\omega_i^2)/(2\alpha )`

`\theta=(-62.83^2)/(2* -6.283)`

`\theta=314.15\ radian`

`\theta=49.99\ revolution`

or

`\theta=50\ revolution`

So, the number of revolutions of the grindstone after the power is shut off is 50.