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In the figure above mass $m_1$(1.5kg) is rotating with uniform speed in a circular path of radius $r$ and mass $m_2$(2kg) is tied to $m_1$ with a thread. The coefficient of friction on the table is stated to be $0.2$. The problem asks us to find the necessary speed for which mass $m_2$ remains at equilibrium.

The solution given was divided into two cases:

Case-1 The frictional force acting along the centrifugal force which gives the equation $T=\frac{mv^2}{r}+f_k$ which gives us an answer of $3.87\sqrt{r}$.

Case-2 The frictional force acting along the tension which gives us the equation $T+f_k=\frac{mv^2}{r}$ which in turn gives the result $3.33\sqrt{r}$ for speed.

Then they asserted that $3.87\sqrt{r}$ is the maximum speed and $3.33\sqrt{r}$ is the minimum speed and hence showed that this speed will be in the range $[3.33\sqrt{r},3.87\sqrt{r}]$.

My confusion arises in this step. First of all, $f_k=0.2\times m_1\times g$ is not the exact frictional force,rather the maximum frictional force acting. So the $f_k$ which they calculated is the maximum. Now the answer for case-2 turns out to be less than that of case-1 but how does it make it the minimum? I mean how are they so sure that any other velocity in case-1 can't be less than that of case-2?Also i couldn't decipher why they took the velocity of case-2 to be the minimum. Why isn't it possible for any other velocity in case-2 to be less than that?

I would like to draw the kind attention of the physics lovers in this issue.

  • $\begingroup$ The two cases are just describing which way the frictional force is pointing. Don't write the frictional force on the same side as the net acceleration, ever. It belongs on the $F$ side of $F=ma$. Now, if $m_1$ is spinning too fast, which way will the friction force be pointing? And if it is spinning too slow, then? $\endgroup$ Aug 24, 2021 at 7:47
  • $\begingroup$ I understand why two cases are being taken but that was not the intention of this post. My question is why the speed should remain in that interval and why are the two speeds achieved the extreme speeds. $\endgroup$
    – madness
    Aug 24, 2021 at 10:48

1 Answer 1


From what i understood, you must be asking why direction of friction is changing and why speed have to remain in that particular range.

Lets start the question freshly. Firstly, $T=m_2g$

Now, we dont know about direction of friction which is acting, that's why we take 2 cases (extreme values) and we get a particular range. Friction has property to always oppose the change in motion. Tension however will always act inwards in our situation.


Lets talk about minimum velocity, at which the $m_1$ has tendency to slip towards the hole in table. That's why friction will act outwards to oppose slipping of $m_1$ body. Note, since $m_1$ is trying to slip (change in motion), so static friction will not act but limiting friction will act (highest value of friction) .So we will get: $$ T-f_k=\frac{mv_{min}^2}{r}$$


Now for maximum velocity, the $m_1$ will have tendency to slip outwards, so friction will oppose this by acting inwards (together with tension). Hence we get: $$ T+f_k=\frac{mv_{max}^2}{r}$$

Now for centrifugal or centripetal, these are frame dependant quanitities, i.e. they change on frame you are working. Its misleading to say that if $f_k$ is acting outwards, then its centrifugal force or vice-versa. By default we always work in ground frame and use centripetal force (we used it here). Centrifugal force is used if we work in the rotating frame. Also working with respect to different frames, must not change our calculations, (unless they are non-inertial i.e. accelerating).

  • $\begingroup$ I am only at UG level and maybe wrong. I have learnt that friction opposes relative motion and hence in this question friction should act in direction opposite to that of velocity vector. But your answer is contradictory. Can you please clear my doubt? $\endgroup$
    – Jay
    Aug 27, 2021 at 10:21
  • $\begingroup$ @Jay since the question is talking about uniform circular motion, they expect you to consider friction along only radial direction. $\endgroup$
    – ACB
    Aug 27, 2021 at 11:41
  • $\begingroup$ @ACB then I must say question is expecting wrong thing. They should explicitly mention : Assume friction only acts in direction to keep body performing UCM OR Assume friction only acts in radial direction. $\endgroup$
    – Jay
    Aug 27, 2021 at 12:33
  • $\begingroup$ My query is why are both of these cases the extreme one?Please have a look at my last paragraph,there i explained my dilemma. $\endgroup$
    – madness
    Aug 28, 2021 at 20:38

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