Question

A solid flywheel is rotating counterclockwise at 69.3 rad/s. It has a mass of 4.80 kg
and a diameter of 32.7 cm. A force F applied tangentially to the rim as shown brings
the wheel to rest in 5.29 s. Ignore friction. Calculate:

a) the angular acceleration. (ans: 1.31 rad/s2)
b) the net torque. (ans: 3.36 Nm)
c) the force F. (ans: 10.3 N)

Answer 1

Remember that rotational kinematics are a lot like linear kinematics
\[\cancel{\omega}^0 =\omega_i + \alpha t\]

then
Torque
\[T=I\alpha\]

and
\[T=rF\sin \theta\]
the force is applied tangentially, so theta=90
\[T=rF\]

Answer 2

I'm sure this is a perfectly good explanation. I'm just tired and gonna go to bed. I appreciate your time and effort.

Answer 3

It's sort of thrown out without much explanation, sorry.

For the first part, it closely resembles its linear counterpart

\[v=v_i + at\]
vs
\[ \omega = \omega_i + \alpha t\]
where alpha is the angular acceleration, omega is the final angular velocity (which is zero) and omega i is the initial angular velocity.


For the second part, that's the definition of Torque - the moment of inertia I times he angular acceleration. I of the flywheel (which I think you can treat as a large solid cylinder) is
\[I=\frac{mr^2}{2}\]
, where m and r are the mass and radius of the flywheel, respectively) and alpha is the angular acceleration found in the first part of the problem.

Lastly, again, that's the definition of Torque on an object

\[\textbf T = \textbf r \times \textbf F\]
\[T = rF\sin \theta\]
and for tangential force
\[T=rF\]
and thus
\[F=\frac{T}{r}\]

where r is the radius of the flywheel, and T is the torque found in the second part of the problem.