A sphere rolling down an incline without slipping has both translational and rotational motion.
Gravity pulls it down the slope, while static friction supplies the torque needed for rotation.
1. Rolling constraint
Rolling without slipping means the center-of-mass speed and angular speed satisfy
\[
v_{\mathrm{cm}}=\omega R.
\]
Differentiating with respect to time gives
\[
a_{\mathrm{cm}}=\alpha R.
\]
Therefore,
\[
\alpha=\frac{a_{\mathrm{cm}}}{R}.
\]
2. Inertia factor
It is useful to write the sphere’s center-of-mass moment of inertia as
\[
I_{\mathrm{cm}}=kMR^2.
\]
The value of \(k\) depends on how the mass is distributed.
| Object |
Moment of inertia |
\(k\) |
| Solid sphere |
\(I_{\mathrm{cm}}=\frac{2}{5}MR^2\) |
\(k=\frac{2}{5}\) |
| Thin spherical shell |
\(I_{\mathrm{cm}}=\frac{2}{3}MR^2\) |
\(k=\frac{2}{3}\) |
| Custom object |
\(I_{\mathrm{cm}}=kMR^2\) |
Entered by user |
3. Acceleration down the incline
Along the slope, gravity contributes
\[
Mg\sin\theta.
\]
Static friction acts up the incline for a sphere rolling downward from rest. The translational equation is
\[
Mg\sin\theta-f_s=Ma.
\]
The torque equation about the center of mass is
\[
f_sR=I_{\mathrm{cm}}\alpha.
\]
Using \(\alpha=a/R\),
\[
f_sR=I_{\mathrm{cm}}\frac{a}{R}.
\]
\[
f_s=\frac{I_{\mathrm{cm}}}{R^2}a.
\]
Since \(I_{\mathrm{cm}}=kMR^2\),
\[
f_s=kMa.
\]
Substitute this into the translational equation:
\[
Mg\sin\theta-kMa=Ma.
\]
\[
Mg\sin\theta=Ma(1+k).
\]
Therefore,
\[
a_{\mathrm{cm}}=\frac{g\sin\theta}{1+k}.
\]
4. Final speed from distance along the incline
If the sphere travels a distance \(s\) along the incline with constant acceleration,
\[
v_f^2=v_0^2+2a_{\mathrm{cm}}s.
\]
Therefore,
\[
v_f=\sqrt{v_0^2+2a_{\mathrm{cm}}s}.
\]
If it starts from rest,
\[
v_f=\sqrt{2a_{\mathrm{cm}}s}.
\]
5. Final speed from vertical drop
The vertical drop \(h\) and distance along the incline \(s\) are related by
\[
h=s\sin\theta.
\]
Energy conservation gives
\[
Mgh=\frac12Mv^2+\frac12I_{\mathrm{cm}}\omega^2.
\]
Using \(\omega=v/R\) and \(I_{\mathrm{cm}}=kMR^2\),
\[
Mgh=\frac12Mv^2+\frac12(kMR^2)\left(\frac{v}{R}\right)^2.
\]
\[
Mgh=\frac12Mv^2(1+k).
\]
The mass cancels:
\[
gh=\frac12v^2(1+k).
\]
Solving for speed,
\[
v=\sqrt{\frac{2gh}{1+k}}.
\]
6. Static friction requirement
Static friction is needed for rolling without slipping. For a rolling sphere on an incline,
\[
f_s=\frac{k}{1+k}Mg\sin\theta.
\]
The normal force is
\[
N=Mg\cos\theta.
\]
Therefore the minimum coefficient of static friction is
\[
\mu_{s,\min}=\frac{f_s}{N}.
\]
\[
\mu_{s,\min}
=
\frac{k}{1+k}\tan\theta.
\]
If the available \(\mu_s\) is smaller than this value, rolling without slipping is not possible and the sphere will
slip.
7. Energy split
At the bottom of the incline, the translational kinetic energy is
\[
K_{\mathrm{trans}}=\frac12Mv^2.
\]
The rotational kinetic energy is
\[
K_{\mathrm{rot}}=\frac12I_{\mathrm{cm}}\omega^2.
\]
With \(I_{\mathrm{cm}}=kMR^2\) and \(\omega=v/R\),
\[
K_{\mathrm{rot}}=\frac12kMv^2.
\]
The total kinetic energy is
\[
K_{\mathrm{total}}=\frac12Mv^2(1+k).
\]
8. Worked example
A solid sphere rolls from rest down a \(25^\circ\) incline over a distance
\[
s=4.0\ \mathrm{m}.
\]
For a solid sphere,
\[
k=\frac25=0.4.
\]
The acceleration is
\[
a_{\mathrm{cm}}=\frac{g\sin\theta}{1+k}.
\]
\[
a_{\mathrm{cm}}
=
\frac{(9.80665)\sin(25^\circ)}{1+0.4}.
\]
\[
a_{\mathrm{cm}}\approx2.96\ \mathrm{m/s^2}.
\]
The final speed after traveling \(4.0\ \mathrm{m}\) is
\[
v_f=\sqrt{2a_{\mathrm{cm}}s}.
\]
\[
v_f=\sqrt{2(2.96)(4.0)}.
\]
\[
v_f\approx4.87\ \mathrm{m/s}.
\]
The required coefficient of static friction is
\[
\mu_{s,\min}
=
\frac{0.4}{1.4}\tan(25^\circ)
\approx0.133.
\]
Therefore,
\[
\boxed{a_{\mathrm{cm}}\approx2.96\ \mathrm{m/s^2}},
\qquad
\boxed{v_f\approx4.87\ \mathrm{m/s}}.
\]
Key idea: a rolling sphere accelerates more slowly than a sliding frictionless block because some gravitational
energy becomes rotational kinetic energy.
