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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?

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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:

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

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

In addition:

$$v=\omega R$$

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:

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

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

In addition:

$$v>\omega R$$

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  • $\begingroup$ thank you for you lucid and straightforward answer ,it made things a lot better for me .However I still have 3 doubts regarding your answer : 1. How did you establish that v>w in case 1 (c) from the equations as we have variables (F and radius of gyration for w) in both . 2. In case 2 you began with writing the critical condition for coeff of fric. for rolling but I wanted to ask why we don't also include the critical condition for sliding in this ( i.e coeff>tantheta) because if sliding starts then it won't be toppling $\endgroup$ – Ishita Gupta Nov 19 '15 at 16:30
  • $\begingroup$ 3. In case 2 (b) you used umg in your equations but isn't that the maximum value /limiting value of friction ? which is at work when object is just about the slide .as friction is a self adjusting force couldn't its value be lesser than that as well . Also if friction keeps acting even after rolling then wouldn't it have angular acceleration /linear acceleration ruining the rolling condition ? $\endgroup$ – Ishita Gupta Nov 19 '15 at 16:35
  • $\begingroup$ @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. $\endgroup$ – Gert Nov 19 '15 at 17:13
  • $\begingroup$ @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! $\endgroup$ – Gert Nov 19 '15 at 17:23

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