Question

1. A body of mass 200 g rotates from rest to 3 rads/s in 1.5 s on a path whose radius of gyration is 50 cm. The torque is .... A. 0.05 B. 1.50 C. 0.075 D. 0.10​

Answer 1

D. 0.10

To solve this problem, we can use the formula:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

We can find the moment of inertia using the formula:

I = mk^2

where m is the mass of the body and k is the radius of gyration. Plugging in the given values:

m = 200 g = 0.2 kg
k = 50 cm = 0.5 m
I = (0.2 kg)(0.5 m)^2 = 0.05 kg m^2

We can find the angular acceleration using the formula:

α = ω/t

where ω is the final angular velocity and t is the time taken to reach that velocity. Plugging in the given values:

ω = 3 rad/s
t = 1.5 s
α = 2 rad/s^2

Now we can find the torque using the formula:

τ = Iα

τ = (0.05 kg m^2)(2 rad/s^2) = 0.10 Nm

So the torque is 0.10 Nm, which is option D.