Question

A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 rev/min whent ehmotor is turned off. Neglecting the frictionat the axle, calculate the force that must be applied tangenticaly to the wheel to bring it to rest in 10 revolutions.

Answer 1

Answer:

Given in the question :-

Rotation of the wheel = 100 revolution / min or  \omega = 5/3 revolution/second.

Now ,  \omega' = 0  

Total rev.  \theta  = 10 revolution

Now we have to find  \alpha  = ?

We know the formula,

\omega' =\omega ^2 - 2 \alpha \theta  

0 = (10/6)² - 2 × α × 10

0 = 100/36 - 20α

α = 5/36 rev/second²

or α = 2 π × (5/36)

α = 10π / 36 radian/second² .

Now, we have to moment of inertia of the wheel.

I = mr^2 /2  

I = 1/2 (0.2² × 10)

I = 0.2 kg.m²

Since the force is F, Now torque

F = F.r

0.2 F

Now torque = I α

0.20 / (10 π /36)

= 2 π /36

0.2 F = 2 π /36

F = 2 π/7.2

F = 0.87 N