A wheel with a hanging mass is a translation–rotation system. The hanging mass accelerates downward, while the
wheel rotates because the cord applies a torque to the wheel.
1. Physical setup
A mass \(m\) hangs from a light cord wrapped around a wheel of radius \(r\) and moment of inertia \(I\).
The cord does not slip, so the linear acceleration of the mass and the angular acceleration of the wheel are linked:
\[
a=\alpha r.
\]
Equivalently,
\[
\alpha=\frac{a}{r}.
\]
2. Force equation for the hanging mass
Taking downward as positive for the hanging mass, the forces are weight \(mg\) downward and tension \(T\) upward.
Newton’s second law gives
\[
mg-T=ma.
\]
If the wheel had no rotational inertia, the tension would be nearly zero and \(a\) would approach \(g\). But for a
massive wheel, some of the gravitational effect must rotate the wheel.
3. Torque equation for the wheel
The tension produces a torque about the wheel’s axle:
\[
\tau=Tr.
\]
The rotational form of Newton’s second law is
\[
\tau=I\alpha.
\]
Therefore,
\[
Tr=I\alpha.
\]
Using \(\alpha=a/r\),
\[
Tr=I\frac{a}{r}.
\]
So the tension can be written as
\[
T=\frac{Ia}{r^2}.
\]
4. Solving for acceleration
Substitute \(T=Ia/r^2\) into the mass equation:
\[
mg-\frac{Ia}{r^2}=ma.
\]
Move all terms involving \(a\) to one side:
\[
mg=ma+\frac{Ia}{r^2}.
\]
Factor out \(a\):
\[
mg=a\left(m+\frac{I}{r^2}\right).
\]
Therefore,
\[
a=\frac{mg}{m+I/r^2}.
\]
The angular acceleration is then
\[
\alpha=\frac{a}{r}.
\]
The tension is
\[
T=m(g-a).
\]
5. Effective rotational mass
The term
\[
\frac{I}{r^2}
\]
has units of mass. It acts like an extra inertia term in the denominator:
\[
a=\frac{mg}{m+I/r^2}.
\]
A larger wheel inertia \(I\), or a smaller cord radius \(r\), makes \(I/r^2\) larger. This reduces the acceleration.
6. Common wheel inertia models
| Wheel model |
Moment of inertia |
Effect |
| Direct input |
\(I=\text{given}\) |
Use when the wheel inertia is known from a problem statement or experiment. |
| Solid cylinder / disk |
\(I=\frac12Mr^2\) |
Moderate rotational inertia. |
| Hoop / thin wheel |
\(I=Mr^2\) |
Larger rotational inertia for the same \(M\) and \(r\). |
| Custom factor |
\(I=kMr^2\) |
Useful for non-standard pulleys or flywheels. |
7. Energy view
If the hanging mass falls a distance \(s\) from rest, the loss in gravitational potential energy is
\[
\Delta U=mgs.
\]
This becomes translational kinetic energy of the mass plus rotational kinetic energy of the wheel:
\[
mgs=\frac12mv^2+\frac12I\omega^2.
\]
Because the cord does not slip,
\[
v=\omega r.
\]
The acceleration result from Newton’s laws is consistent with this energy relationship.
8. Worked example
Suppose the wheel has
\[
I=0.45\ \mathrm{kg\,m^2},
\]
radius
\[
r=0.12\ \mathrm{m},
\]
and the hanging mass is
\[
m=3.0\ \mathrm{kg}.
\]
First compute the effective rotational mass:
\[
\frac{I}{r^2}
=
\frac{0.45}{(0.12)^2}.
\]
\[
\frac{I}{r^2}=31.25\ \mathrm{kg}.
\]
Then
\[
a=\frac{mg}{m+I/r^2}.
\]
\[
a=
\frac{(3.0)(9.80665)}{3.0+31.25}.
\]
\[
a\approx0.859\ \mathrm{m/s^2}.
\]
The angular acceleration is
\[
\alpha=\frac{a}{r}.
\]
\[
\alpha=\frac{0.859}{0.12}\approx7.16\ \mathrm{rad/s^2}.
\]
The tension is
\[
T=m(g-a).
\]
\[
T=3.0(9.80665-0.859)\approx26.8\ \mathrm{N}.
\]
Therefore,
\[
\boxed{\alpha\approx7.16\ \mathrm{rad/s^2}},
\qquad
\boxed{a\approx0.859\ \mathrm{m/s^2}},
\qquad
\boxed{T\approx26.8\ \mathrm{N}}.
\]
Key idea: the hanging mass does not fall freely because part of the gravitational pull is used to spin up the
wheel.
