# Rotational Dynamics

A grinding wheel in the shape of a uniform disk of radius 16 cm is rotating at 900 revolutions per minute. The total mass of the wheel is 2.1 kg, which is uniformly distributed. A piece of steel is pressed in the normal direction against the wheel with a force of 9 Newtons. The coefficient of friction between steel and wheel is .21. How long will it take the wheel to stop?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

A grinding wheel in the shape of a uniform disk of radius 16 cm is rotating at 900 revolutions per minute. The total mass of the wheel is 2.1 kg, which is uniformly distributed. A piece of steel is pressed in the normal direction against the wheel with a force of 9 Newtons. The coefficient of friction between steel and wheel is .21. How long will it take the wheel to stop?

Solution:

R = 0.16m Frictional force f

M = 2.1 kg

N = 9 N

Moment of inertia of the wheel = I = Â½ MR2 = Â½ x 2.1 x 0.162 = 0.027 kg.m2

Frictional force f = Î¼N = 0.21 x 9 = 1.89 N

Stopping torque acting on the wheel due to frictional force = Ï„ = fR = - 1.89x0.16 = - 0.3 Nm (-ve sign implies direction of force opposes motion)

As per Newton's second law of motion as applied to rotational motion:

Torque (Ï„) = Moment of inertia (I) x Angular acceleration (Î±)

Substituting values: - 0.3 = 0.027Î±

Î± = - 0.3/0.027 = - 11.11 rad/s2

Initial frequency of rotation of the wheel = f = 900 rpm or 900/60 = 15 rps

Initial angular velocity Ï‰0 = 2ÐŸf = 2x3.14x15 = 94.2 rad/sec

Final angular velocity = Ï‰ = 0

Using the equation Ï‰ = Ï‰0 + Î±t we get: 0 = 94.2 - 11.11t

t = 94.2/11.11 = 8.5 sec

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