Rolling with slipping occurs when the center-of-mass speed and angular speed do not satisfy the no-slip condition.
In that case, kinetic friction acts at the contact point and changes both the linear speed and the angular speed
until pure rolling begins.
1. No-slip condition
For pure rolling,
\[
v_{\mathrm{cm}}=\omega R.
\]
When this is not true, the contact point slides relative to the ground. Define the slip speed
\[
u=v_{\mathrm{cm}}-\omega R.
\]
If \(u=0\), the object rolls without slipping. If \(u\neq0\), kinetic friction acts to reduce \(|u|\).
2. Friction direction
The sign of \(u\) determines the direction of kinetic friction:
| Condition |
Physical meaning |
Friction direction |
| \(u>0\) |
The contact point slips forward because \(v_{\mathrm{cm}}>\omega R\). |
Backward |
| \(u<0\) |
The contact point slips backward because \(\omega R>v_{\mathrm{cm}}\). |
Forward |
| \(u=0\) |
No slipping. |
No kinetic friction is needed. |
3. Moment of inertia model
For many rolling objects,
\[
I_{\mathrm{cm}}=kMR^2.
\]
Common values are:
| Object |
\(I_{\mathrm{cm}}\) |
\(k\) |
| Solid cylinder / disk |
\(I_{\mathrm{cm}}=\frac12MR^2\) |
\(k=\frac12\) |
| Solid sphere |
\(I_{\mathrm{cm}}=\frac25MR^2\) |
\(k=\frac25\) |
| Thin spherical shell |
\(I_{\mathrm{cm}}=\frac23MR^2\) |
\(k=\frac23\) |
| Hoop / thin ring |
\(I_{\mathrm{cm}}=MR^2\) |
\(k=1\) |
4. Slipping accelerations
On a horizontal surface, kinetic friction has magnitude
\[
f_k=\mu_kMg.
\]
If \(u>0\), friction acts backward, so
\[
a=-\mu_kg,
\qquad
\alpha=\frac{\mu_kg}{kR}.
\]
If \(u<0\), friction acts forward, so
\[
a=+\mu_kg,
\qquad
\alpha=-\frac{\mu_kg}{kR}.
\]
A compact way to write both cases is
\[
a=-\operatorname{sgn}(u_0)\mu_kg,
\]
\[
\alpha=\operatorname{sgn}(u_0)\frac{\mu_kg}{kR}.
\]
5. Time until pure rolling
During slipping,
\[
v(t)=v_0+at,
\]
\[
\omega(t)=\omega_0+\alpha t.
\]
Pure rolling begins when
\[
v(t)-\omega(t)R=0.
\]
This gives
\[
t_{\mathrm{roll}}
=
\frac{|u_0|}{\mu_kg(1+1/k)}.
\]
where
\[
u_0=v_0-\omega_0R.
\]
6. Slipping distance and energy loss
The slip speed decreases linearly from \(|u_0|\) to \(0\), so the slipping distance at the contact point is
\[
s_{\mathrm{slip}}=\frac12|u_0|t_{\mathrm{roll}}.
\]
The work done by kinetic friction on the combined translation-plus-rotation motion is
\[
W_f=-\mu_kMg\,s_{\mathrm{slip}}.
\]
This work is negative, so mechanical energy decreases. The lost mechanical energy becomes thermal energy at the
contact surfaces.
7. Kinetic energy before and after slipping
At any instant, the total kinetic energy is
\[
K=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2.
\]
Initially,
\[
K_0=\frac12Mv_0^2+\frac12I_{\mathrm{cm}}\omega_0^2.
\]
When pure rolling begins,
\[
K_f=\frac12Mv_f^2+\frac12I_{\mathrm{cm}}\omega_f^2,
\]
with
\[
v_f=\omega_fR.
\]
The mechanical energy lost is
\[
\Delta E=K_0-K_f.
\]
8. Worked example
A solid cylinder has
\[
k=\frac12,
\qquad
R=0.25\ \mathrm{m},
\qquad
v_0=8\ \mathrm{m/s},
\qquad
\omega_0=12\ \mathrm{rad/s}.
\]
The initial slip speed is
\[
u_0=v_0-\omega_0R.
\]
\[
u_0=8-(12)(0.25)=5\ \mathrm{m/s}.
\]
Since \(u_0>0\), friction acts backward. With \(\mu_k=0.25\) and \(g=9.80665\ \mathrm{m/s^2}\),
\[
t_{\mathrm{roll}}
=
\frac{5}{(0.25)(9.80665)(1+1/(1/2))}.
\]
\[
t_{\mathrm{roll}}\approx0.680\ \mathrm{s}.
\]
The final speed is
\[
v_f=v_0-\mu_kg t_{\mathrm{roll}}.
\]
\[
v_f\approx8-(0.25)(9.80665)(0.680)\approx6.33\ \mathrm{m/s}.
\]
The final angular speed is
\[
\omega_f=\omega_0+\frac{\mu_kg}{kR}t_{\mathrm{roll}}.
\]
\[
\omega_f\approx25.3\ \mathrm{rad/s}.
\]
Check:
\[
\omega_fR\approx(25.3)(0.25)\approx6.33\ \mathrm{m/s}.
\]
So pure rolling begins when \(v_f=\omega_fR\).
Key idea: kinetic friction does not simply stop the motion. It adjusts translation and rotation until the contact
point no longer slips.
