The flywheel of an old steam engine is a solid homogeneous metal disk of mass M

Question

The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 112 kg and radius R = 80 cm. The engine rotates the wheel at 520 rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force F = 125 N. If the coefficient of kinetic friction between the pad and the flywheel is k = 0.2.

How many revolutions does the flywheel make before coming to rest?

How long does it take for the flywheel to come to rest?

Calculate the work done by the torque during this time.

Explanation / Answer

a) The force F = 125 N applied to the flywheel "produces" the friction force that will eventually stop the flywheel. The friction force F(f) = (0.2)(125N) = 25 N. The torque on the wheel is:
T = F(f)*r = (25N)(0.8m) = 20 Nm.
The deceleration of the flywheel = (applied net torque) / (moment of inertia). The moment of inertia of a solid disk I = (1/2)mr^2 = (0.5)(112)(0.8^2) = 35.84 kgm^2, therefore the deceleration =
(20Nm) / (35.84 kgm^2) = 0.558 rad/s^2.
To calculate the number of revolutions we will use:
(final angular speed)^2 = (initial angular speed)^2 + 2*(angular acceleration)(angle of rotation). The same equation with numbers: 0 = (54.45 rad/s)^2 + 2*(-0.558rad/s^2)(theta). Solving for theta we have:
theta = 2656.63 radians. Converting to revolutions we have the answer: 422.82 revolutions

b) angular acceleration = (change in angular speed) /(time). Solving for time we have t =54.45/.558 = 97.58 s

c) the work done by the torque = the change in the rotational kinetic energy = 0.5Iw^2 =0.5*0.5mR^2w^2

W = 0.25 * 112* 0.8^2 * 54.45^2 = 53129.26 J = 53.13 kJ

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