# For a solid sphere rolling (pure roll) up a slope (with friction) does friction play a role in slowing it down?

A solid sphere rolling up a slope is slowed down - is this only due to gravity, or is it also because of friction? I need to know this, to calculate the final translational and angular velocity of a solid sphere, at the top of the slope, initially rolling (pure roll) with velocity v on a plane surface and then comes onto this slope.. I'll do this using work - energy theorem, but I need to know the total work done on the sphere first.

I think that friction is the only force that can affect the rotational motion of the sphere, as only it exerts a torque on the sphere..

In the case of pure rolling (no slipping or sliding) enough friction is provided so that at all times:

$$v=\omega R$$

where $$R$$ is the radius of the sphere.

Does friction play a part in the energy balance? Yes.

Let $$T$$ be the total energy of the sphere, $$U$$ be the potential energy of the sphere and $$K$$ its kinetic energy, then because the friction force does no work and we assume no other external forces:

$$\Delta T=\Delta U+\Delta K=0$$

when rolling up the incline.

Of course $$\Delta U=mg\Delta h$$ (for smallish $$\Delta h$$).

But for $$\Delta K$$ the situation is slightly more complicated because the sphere is translating and rotating at once, so that:

$$K=K_{trans}+K_{rot}$$

You'll need to find the expression for $$K_{rot}$$ and use $$v=\omega R$$ to make everything about $$v$$.

Without friction, $$\omega$$ would remain constant and thus also $$K_{rot}$$.

• So, friction doesn't do any work on the sphere, while the sphere is rolling up? If I use the work - energy theorem: Work done= ΔKinetic Energy. Will the work done term only have -mgΔh ? – karun mathews Dec 5 '18 at 12:07
• Correct. Only if the friction is insufficient to maintain $v=\omega R$ would the friction force do work, because it 'slides' over the surface. $-mg\Delta h$ is the only work done here. – Gert Dec 5 '18 at 12:26
• Thanks a lot! Another doubt, I think that friction is the only force present - that can produce torque in the sphere, as all the other forces act through the center of mass of the sphere. So if the sphere slows down $while $$in$$pure$$roll$ shouldn't the friction be acting so as to maintain it in pure roll, and so then wouldn't friction do work? – karun mathews Dec 5 '18 at 13:16
• shouldn't the friction be acting so as to maintain it in pure roll, and so then shouldn't friction do work? A common misconception. The friction force does not do any work because it doesn't move wrt the surface. If you hold your finger against a surface against friction but w/o moving the finger, is work performed? No because no displacement! – Gert Dec 5 '18 at 13:20

$$\let\a=\alpha \let\om=\omega \let\tm=\times \def\D#1#2{{d#1\over d#2}} \def\ba{\mathbf a} \def\br{\mathbf r} \def\bv{\mathbf v} \def\bF{\mathbf F} \def\bL{\mathbf L} \def\bM{\mathbf M} \def\bP{\mathbf P} \def\bR{\mathbf R} \def\cM{{\cal M}} \def\frac#1#2{{\textstyle {#1 \over #2}}}$$

I think that friction is the only force that can affect the rotational motion of the sphere, as only it exerts a torque on the sphere..

As you'll see, this statement isn't true, and is also ill-defined. Let's discuss your question not in terms of energy but in terms of forces acting. To do so I find expedient to recall a more general form equation of moments can take.

I'm referring to $$\D\bL t = \bM \tag1$$ where $$\bL$$ is angular momentum, $$\bM$$ resultant moment of external forces (commonly called "torque" in english mechanics literature, but I don't like that term, for reasons I can't pause to explain here).

There are some points I have to clarify, however. The first is that eq. (1) is only meaningful if a reference frame K is assumed. This I will take for granted and never change in the following. The first reason why a frame is needed is to give a meaning to points' velocities. For the present problem K is most conveniently taken as the frame where the slope is stationary.

In order to define a moment of forces we must also choose a reference point. This is not to be confused with the reference frame. It can be a fixed point in K (often the origin of coordinates) but any other point, fixed or moving, is allowed and occasionally useful.

This is why I defined your final paragraph "ill-defined". You didn't say which reference point for moments you were assuming. Very likely you were thinking of ball's c.o.m. But in this case too your statement isn't correct, as the normal reaction of the slope has a (negative) moment wrt to ball's centre. As you will presently see, it's possibile (and convenient) to choose a different reference point, giving rise to a wholly different analysis of force moments.

Assume our system consists of mass points of masses $$m_i$$, position vectors $$\br_i$$ (taken from coordinates origin O), velocities $$\bv_i$$. If the reference point is O, we have $$\bL = \sum m_i\,\br_i \tm \bv_i.$$ If the net external force acting on $$i$$-th point is $$\bF_i$$, then $$\bM= \sum \br_i \tm \bF_i.$$ We have $$\D\bL t = \sum m_i\,(\bv_i \tm \bv_i + \br_i \tm \ba_i) = \sum m_i\,\br_i \tm \ba_i = \sum \br_i \tm \bF_i = \bM$$ and eq. (1) is proven.

Let's now change the reference point to O$$'$$ (position $$\br'$$). (Note che $$\br'$$ may also depend on time.) Define $$\bL' = \sum m_i\,(\br_i - \br') \tm \bv_i$$ $$\bM' = \sum (\br_i - \br') \tm \bF_i$$ and compute \eqalign{ \D{\bL'}t &= \sum m_i\,(\bv_i - \bv') \tm \bv_i + \sum m_i\,(\br_i - \br') \tm \ba_i \cr &= -\sum m_i\,\bv' \tm \bv_i + \sum (\br_i - \br') \tm \bF_i \cr &= -\bv' \tm \bP + \bM' \tag2 \cr} where $$\bP$$ is the system's total momentum.

Eq. (2) simplifies if $$\bv'$$ is parallel to $$\bP$$ i.e. to c.o.m. velocity ($$\bv'=0$$ included): $$\D{\bL'}t = \bM'.\tag3$$ This is the only case we'll have to consider.

Let's now come to our problem. I'll take $$x$$-axis along the slope, positive upwards, $$y$$-axis positive downwards. As a consequence $$z$$-axis is horizontal, parallel to slope plane. Origin O is on the slope. It's assumed that the ball's center is moving in $$(x,y)$$ plane.

As a reference point O$$'$$ for moments the best choice is the contact point between ball and slope. Then O$$'$$ is moving, but its velocity is the same as that of the ball's c.o.m., so that eq. (3) applies. A bonus of this choice for O$$'$$ is that moment of the unknown force the slope exerts on the ball (friction included) vanishes. So as external force we are left with weight alone.

Weight can be reduced to just one force applied to ball's centre. Its moment is parallel to $$z$$. Then if $$\cM$$ is ball's mass, $$r$$ its radius we have $$M'_z = -\cM\,g\,r \sin\a.\tag4$$ As to $$\bL'$$, it's also parallel to $$z$$ and is written $$L'_z = I \om.\tag5$$ $$\om>0$$ if the ball is rolling up. $$I$$ is the moment of inertia wrt to an axis parallel to $$z$$ through O$$'$$. For a homogeneous ball its value is $$I = \frac75\,\cM\,r^2.\tag6$$

From (3--6) we find $$\D \om t = -{5 g \over 7 r}\,\sin\a$$ $$\om(t) = \om(0) - \left(\!{5g \over 7r}\,\sin\a\!\right) t.$$ Since $$v(t) = \om(t)\,r$$, we have also $$v(t) = v(0) - \frac57\,g\,\sin\a\>t.$$

As you can see, there is no trace of friction in the final result. It is required, of course, in order to force pure roll. But it does no work, nor contributes to slowing and halting the ball. (Incidentally, not even mass and radius of the ball have any effect as far as a homogeneous ball is considered.)

• Sorry for taking so much time to respond. Another doubt, because we're taking moment and angular velocity all relative to the contact point, the angular velocity that we find should be relative to that point. Then would the velocity relative to the com of the sphere be different? Thanks for answering. – karun mathews Dec 11 '18 at 1:49
• @karun mathews the angular velocity that we find should be relative to that point. No, there is one angular velocity in a rigid motion. Here is a useful equation. Let P, Q be any two points of a rigid body, $\mathbf v(P)$, $\mathbf v(Q)$ their velocities (both relative to the same frame). Then there is one $\omega$ (vector) such that $\mathbf v(Q) = \mathbf v(P) + \omega\times PQ$. – Elio Fabri Dec 11 '18 at 14:08
• @karun mathews Then would the velocity relative to the com of the sphere be different? I'm not sure to understand what you mean. I clearly stated that I never changed reference frame. Only the reference point for computing moments is changed. All velocities are always referred to one and the same frame. – Elio Fabri Dec 11 '18 at 14:14
• sorry I meant "what would be the $angular$ velocity of the sphere relative to the centre of mass", but now I know that the angular velocity of the sphere is the same with reference to all points. Thanks. – karun mathews Dec 11 '18 at 14:51

The sphere has kinetic energy in 2 forms, translational and rotational (because it is rolling). Because the sphere has inertia and energy stored in its rotation it will actually go higher up the slope than if it didn't. The height it reaches will be where mgh equals kinetic energy of v trans PLUS I times omega squared.

If the slope had no friction the sphere would only go up to a point and stop but would still be rotating.