Question

A piece of machinery with a moment of inertia of 2.6 kg m^2 rotates with a constant angular speed of 4.0 rad/s when a frictional torque of 0.56 Nm acts on it. What is the net torque acting on the piece?

a) 3.44 Nm
b) 4.56 Nm
c) 2.04 Nm
d) 0.56 Nm

Answer 1

The net torque acting on the piece of machinery is equal to the frictional torque of 0.56 Nm, because the machinery rotates at a constant angular speed, implying that the net torque must be zero.

Step-by-step explanation:

The question asks for the net torque acting on a piece of machinery that has a moment of inertia of 2.6 kg m2 and is rotating with a constant angular speed of 4.0 rad/s under the action of a frictional torque of 0.56 Nm. To find the net torque, we use the concept that if an object rotates at a constant angular speed, the net external torque must be zero (Newton's first law for rotational motion). Since the machinery is rotating with a constant angular speed, the net torque is indeed equal to the frictional torque, but in the opposite direction, making the net torque zero. However, the only torque mentioned in the problem is a frictional torque, so we're looking for the magnitude of this torque as it would be the net torque required to counteract this friction and maintain a constant angular speed.

Therefore, the net torque acting on the piece of machinery is the magnitude of the frictional torque, which is given as 0.56 Nm.