### Video Transcript

The wheel of a train carriage has a
moment of inertia of 28 kilogram meters squared. As the train is increasing in speed
as it leaves the station, the angular acceleration of the wheel is 1.5 radians per
second squared. What is the magnitude of the torque
being applied to the wheel?

Drawing a diagram can help us
visualize the situation. In our diagram, we chose one of the
wheels of our train carriage and labeled it with the information from the
problem. The moment of inertia, 𝐼, is given
as 28 kilogram meters squared. The angular acceleration of the
wheel is given as 1.5 radians per second squared. And we are trying to solve for the
torque applied to the wheel.

We need an equation that relates
these three variables together. We need to remember that Newton’s
second law of motion applied to rotational motion is the net torque, 𝜏 net, is
equal to the moment of inertia of the object, 𝐼, times the angular acceleration of
the object, 𝛼. Looking at our problem, we are
given 𝐼, 𝛼, and solving for 𝜏. Therefore, we do not need to
rearrange our formula to solve for our unknown variable.

Substituting in our values, we have
28 kilogram meters squared for 𝐼 and 1.5 radians per second squared for 𝛼. When we multiply these two numbers
together, we get a 𝜏 of 42 newton meters. The magnitude of the torque being
applied to the wheel of the train carriage is 42 newton meters.