# If, for a body rolling on an incline, the friction coefficient isn't enough to allow pure rolling will it still roll?

More specifically that it won't be pure rolling (obviously) but would it still have some rotational motion along with its translational motion? (if yes how would we write their mechanical equations). Will this also apply to a body which is on a horizontal surface?

• Have a look at this: physics.stackexchange.com/questions/217843/… – Gert Nov 14 '15 at 14:46
• – Gert Nov 14 '15 at 15:11
• @Gert I read through both the answers you posted and they were very informative but I'm still not clear on this concept . Could you please elaborate a bit more ? – Ishita Gupta Nov 17 '15 at 6:42

Firstly let's look at the case of a horizontal plane first.

In my answer here I derived that the critical friction coefficient is:

$$\large{\mu_c=\frac{FI}{mg(I+mR^2)}}$$

Now we have three scenarios:

a) No friction at all, $\mu=0$:

Assuming no forces or couples act on the object then Newton tells us that the state of motion remains unchanged, or:

$$v=v_0,$$

and:

$$\omega=\omega_0.$$

Specifically, if the object was rotating it will keep doing so (of course this does not constitute 'rolling'), if it wasn't then it won't start doing so.

b) Lots of friction, $\mu>\mu_c$:

Assuming no forces or couples act on the object then $v$ and $\omega$ will adjust themselves until:

$$v=\omega R$$

c) Small amount of friction $0<\mu<\mu_c$:

The object will both roll and slide.

If for simplicity's sake we assume the object was completely at rest at $t=0$ and a constant accelerating force $F$ acts on its centre of gravity, then the equations of motion are:

$$v=\frac{F-\mu mg}{m}t$$

and:

$$\omega=\mu\frac{mgR}{I}t$$

also:

$$v>\omega R$$.

Secondly, the case of the object on an inclined plane. Here we must bear in mind that on an inclined plane a gravitational force $mg\sin\theta$ always acts on the object.

In my answer here I derived that for that case the critical friction coefficient is:

$$\large{\mu_c=\frac{I}{I+mR^2}\tan\theta}.$$

Again we have three scenarios:

a) No friction at all, $\mu=0$:

Assuming no forces other than gravity are acting on the object and that the object was completely stationary at $t=0$ then the equations of motion are:

$$v=mg\sin\theta t$$

and:

$$\omega=0$$

b) Lots of friction, $\mu>\mu_c$:

Assuming no forces other than gravity are acting on the object and that the object was completely stationary at $t=0$ then the equations of motion are:

$$v=(\sin\theta - \mu_c \cos\theta)gt.$$

$$\omega=(\mu_c \frac{mgR}{I}\cos\theta)t.$$

$$v=\omega R$$

c) Small amount of friction $0<\mu<\mu_c$:

The object will both roll and slide.

If for simplicity's sake we assume the object was completely at rest at $t=0$, then the equations of motion are:

$$v=(\sin\theta - \mu \cos\theta)gt.$$

$$\omega=(\mu \frac{mgR}{I}\cos\theta)t.$$

$$v>\omega R$$
• @IshitaGupta: $v>\omega R$ is always true when there is slipping. Only without slipping is $v=\omega R$. The critical $\mu_c$ needs to be applied when there no slipping. Applying the actual $\mu$ in that case leads to an over-estimate of the friction forces. Only when there's slippage use the true $\mu$. Re. your last point, friction keeps acting if: a) an accelerating force keeps acting or b) equilibrium speed hasn't been achieved yet. – Gert Nov 19 '15 at 17:13
• @IshitaGupta: that for slipping $v>\omega R$ can also be seen from the equations under c): just reduce $\mu$: the $v$ increases and $\omega$ decreases! – Gert Nov 19 '15 at 17:23