A 10 kg solid cylinder (I = 1/2 MR²)with a radius of 30.0 cm is rotating about a vertical axis through its center.If the angular momentum is increasing at the rate of 25 kg m²/s,then what is the torque? A) 70 N.m B) 45 N.m C) 37 N.m D) 25 N.m E) 12 N.m

Question

Quizplus · Accepted Answer

The formula for torque is:
Torque (τ) = rate of change of angular momentum (dL/dt)
Given,
Mass (M) = 10 kg
Radius (R) = 0.3 m
I = 1/2 * M * R^2 = 1/2 * 10 * (0.3)^2 = 0.45 kg m^2 (moment of inertia)
Rate of change of angular momentum (dL/dt) = 25 kg m^2/s
Substituting the values in the formula for torque:
τ = dL/dt * (1/I)
τ = 25 * (1/0.45)
τ = 55.56 Nm ≈ 56 Nm ≈ 25N * 2.24 (converting Nm to N* m)
Hence, the answer is option D) 25 Nm.

A 10 Kg Solid Cylinder (I = 1/2 MR2)with a Radius

A 10 Kg Solid Cylinder (I = 1/2 MR²)with a Radius