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. The flywheel is a solid disk with a mass of 70.5 kg and a diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. If the tension in the upper (taut) segment of the belt is 165 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.

The tension on the slack segment of the belt is 65 N

Explanation:

The given parameters are;

The mass of the solid disk = 70.5 kg

The diameter of the flywheel disk, D = 1.25 m

The radius of the pulley = 0.230 m

If the tension in the upper segment of the belt = 165 N

The angular acceleration of the flywheel = 1.67 rad/s²

The moment of inertia of a circular disc, I = 1/2·M·r²

The radius of the flywheel, r = D/2 = (1.25 m)/2 = 0.625 m

The moment of inertia of the flywheel, I = 1/2 × 70.5 kg × (0.625 m)² = 13.7695312 m²·kg

The moment of inertia, I = 13.7695312 m²·kg

Torque, τ = Moment of Inertia, I × Angular acceleration, α

Torque, τ = I × α

τ = 13.7695312 m²·kg × 1.67 rad/s² = 22.9951171 N·m ≈ 23 N·m

The torque applied from the pulley side is given as follows;

Let represent the tension on the slack segment of the belts, we have;

23 N·m = (165 N - ) × 0.230 m

∴ 165 N - = 23 N·m/0.230 m = 100 N

= 165 N - 100 N = 65 N

The tension on the slack segment of the belt, = 65 N

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