Question

A chain pulls tangentially on a 47.5 kg uniform cylindrical gear with a tension of 75.1 N. The chain is attached along the outside radius of the gear at 0.864 m from the axis of rotation. Starting from rest, the gear takes 2.62 s to reach its rotational speed of 1.38 rev/s. What is the total frictional torque opposing the rotation of the gear?

Answer 1

Mass of the cylindrical gear = M = 47.5 kg

Radius of the cylindrical gear = R = 0.864 m

Moment of inertia of the cylindrical gear = I

I = MR²/2

I = (47.5)(0.864)²/2

I = 17.729 kg.m²

Tangential force exerted by the chain = F = 75.1 N

Torque on the gear due to the force exerted by the chain = T

T = FR

T = (75.1)(0.864)

T = 64.886 Nm

Total frictional torque on the gear = T_f

Net torque on the gear = T_net

T_net = T - T_f

Initial angular speed of the gear = ₁ = 0 rad/s

Angular speed of the gear after 2.62 sec = ₂ = 1.38 rev/s = 2(1.38) rad/s = 8.671 rad/s

Time period = T = 2.62 sec

Angular acceleration of the gear =

₂ = ₁ + T

8.671 = 0 + (2.62)

= 3.309 rad/s²

I = T_net

I = T - T_f

(17.729)(3.309) = 64.886 - T_f

T_f = 6.22 Nm

Total frictional torque opposing the rotation of the gear = 6.22 Nm