Question

A solid flywheel of mass 20 kg and radius 100 mm revolves at 600 revolutions per minute. If the coefficient of friction is 0.1 and to stop the flywheel in 3.14 s, the force to be applied against is approximately:

• A. 30 N
• B. 20 N
• C. 300 N
• D. 200 N

Hint:

When solving problems involving rotational motion, use the relationship between torque, moment of inertia, and angular acceleration. Also, consider the role of friction in providing the necessary force.

Answer 1

The Correct Option is D

Given: 
Mass of the flywheel, \( m = 20 \, {kg} \)
Radius, \( r = 100 \, {mm} = 0.1 \, {m} \)
Angular velocity, \( \omega = 600 \, {rpm} = \frac{600 \times 2\pi}{60} = 62.83 \, {rad/s} \)
Time to stop, \( t = 3.14 \, {s} \) 
Coefficient of friction, \( \mu = 0.1 \) 
Step 1: Calculate the angular deceleration (\( \alpha \)) The flywheel comes to rest in 3.14 s, so the angular deceleration is: \[ \alpha = \frac{\Delta \omega}{t} = \frac{0 - 62.83}{3.14} = -20 \, {rad/s}^2 \] The magnitude of angular deceleration is \( 20 \, {rad/s}^2 \). 
Step 2: Calculate the torque (\( \tau \)) The torque required to stop the flywheel is: \[ \tau = I \alpha \] where \( I \) is the moment of inertia of the flywheel. For a solid disk: \[ I = \frac{1}{2} m r^2 = \frac{1}{2} \times 20 \times (0.1)^2 = 0.1 \, {kg m}^2 \] Thus, the torque is: \[ \tau = 0.1 \times 20 = 2 \, {Nm} \] Step 3: Calculate the frictional force (\( F \)) The torque is also given by: \[ \tau = F \cdot r \] Solving for \( F \): \[ F = \frac{\tau}{r} = \frac{2}{0.1} = 20 \, {N} \] However, considering the coefficient of friction \( \mu = 0.1 \), the normal force \( N \) required to produce this frictional force is: \[ F = \mu N \implies N = \frac{F}{\mu} = \frac{20}{0.1} = 200 \, {N} \] Thus, the force to be applied against the flywheel is 200 N.
Final Answer: 200 N