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# Rotational Dynamics

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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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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 = ½ MR2 = ½ x 2.1 x 0.162 = 0.027 kg.m2

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

Stopping torque acting on the wheel due to frictional force = τ = 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 (τ) = Moment of inertia (I) x Angular acceleration (α)

Substituting values: - 0.3 = 0.027α

α = - 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 ω0 = 2Пf = 2x3.14x15 = 94.2 rad/sec

Final angular velocity = ω = 0

Using the equation ω = ω0 + αt we get: 0 = 94.2 - 11.11t

t = 94.2/11.11 = 8.5 sec

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 6, 2022, 1:28 pm ad1c9bdddf>
https://brainmass.com/physics/torques/rotational-dynamics-grinding-wheel-396780