An electric motor turns a flywheel through a drive belt that joins a pulley on t

Question

An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel as shown in the figure below. The flywheel is a solid disk with a mass of 57.0 kg and a radius R = 0.625 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. The tension Tu in the upper (taut) segment of the belt is 158 N, and the flywheel has a clockwise angular acceleration of 1.67 rad/s2. Find the tension in the lower (slack) segment of the belt.

Explanation / Answer

torque about pulley = torque about flywheel Tensile force x radius of pulley = moment of inertia of flywheel x angular acceleration (T1 -T2) x radius= 0.5 x mass of flywheel x square of radius of flywheel x angular acceleration (T1-154) 0.23 = 0.5 x 52.5 x 0.625^2 x 1.67 T1 =79.5N Note: why the torque about the pulley and flywheel would be the same? answer: Torque is taken to be applied around some center of rotation, and the center of rotation of the pulley and flywheel are one and the same. Torque is the circular motion analog of force. Moment of inertia is the circular motion analog of mass. You don't expect an applied force to increase because it's applied to a more massive object; a force is whatever values it's specified to be. Similarly for torque Now it's certainly true that it takes more force to produce a given acceleration if the mass is larger, but changing the mass doesn't automatically change the force; the force is applied by some outside agency that is separately specified. The same holds for circular motion and torque. In the present problem the pulley is taken to be bonded to the flywheel and has negligible mass. So the two together are taken to have the same moment of inertia as the flywheel alone.

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