I came across the following question in one of my reference materials:

enter image description here

Method 1: I found the radius of curvature of the trajectory (for options B, C and D) by using the equation a=v^2/r, and I got options B and C as correct, which is the answer given.

Method 2: I tried drawing a diagram using the instantaneous axis of rotation (since at the moment give, we can consider the entire sphere to be rotated about the bottom most point, as in the diagram I've drawn). This method yields only option D as correct, which does not match with the given answer.

enter image description here

Why is there an inconsistency with these methods?

Moreover, I noticed that the answer given by Method 2 is half of the answer given by Method 1. Is this always the case, for any point? Can we use it as a trick to solve certain MCQ questions?

  • $\begingroup$ When you use a=v^2/ r, what did you take as 'a'? $\endgroup$ Nov 12, 2020 at 17:57
  • $\begingroup$ here it is said that "The radius of curvature equals the distance to the instantaneous axis of rotation (IAOR" so I think second method gave the right one tho it didn't match w/ options $\endgroup$ Nov 12, 2020 at 18:00
  • $\begingroup$ @Buraian for option b, I took acceleration as a, which is v^2/r (since only the tangential velocity needs to be accounted for this). I equated it to (2v)^2/R, where R is the radius of curvature that we need to find, and the 2v stems from the net velocity of point A. $\endgroup$ Nov 12, 2020 at 18:28

2 Answers 2


Statements A) and D) are obviously incorrect. Regarding statements B),C), the method 1 is correct. It works like this: we express acceleration of the mass point A on the circumference of the body twice, because we have two frames of reference:

  1. frame of the rolling body: here any point on the circumference moves with speed $v$ around circle of radius $r$. From kinematics we know that such motion has acceleration $a_1 = v^2/r$.

  2. frame of the laboratory where ground is stationary: here the motion of A isn't circular, but cycloidal, but for a small part of the trajectory, we can approximate the cycloid with osculating circle of radius $R$. Since A is on the top, it has speed $2v$ and its acceleration can be expressed as $$ a_2 = \frac{(2v)^2}{R} $$ where $R$ is unknown radius of the osculating circle.

Since acceleration is the same in all inertial frames, we have $a_1 = a_2$, so we can find $R$: $$ v^2/r = \frac{(2v)^2}{R} $$ $$ R=4r. $$ So the statement B) is true.

EDIT: I was wrong about radius of curvature of B: it turns out the answer in the option C) is correct.

We can do similar analysis for mass point B, except now the acceleration, pointing towards the center of the body, isn't normal to trajectory. Radius of curvature depends only on the normal component of acceleration. Its magnitude is $1/\sqrt{2}$ of the total acceleration, due to fact trajectory has a downwards slope of 45 degrees.

  1. in frame of the body, B moves in circle with speed $v$, so total acceleration magnitude is again $$ a = \frac{v^2}{r} $$
  2. in frame of the lab, B has the same total acceleration, but normal component is $$ a_n = \frac{v^2}{\sqrt{2}r}. $$

B moves in cycloid with speed $\sqrt{2}r$. Radius of curvature $R$ obeys the equation $$ \frac{(\sqrt{2}v)^2}{R} = a_n $$ $$ \frac{(\sqrt{2}v)^2}{R} = \frac{v^2}{\sqrt{2}r} $$

so we get

$$ R = 2\sqrt{2}r. $$ So the statement C) is true.

I don't understand what you mean by Method 2. Instantaneous axis of rotation is relevant only for finding velocities, not accelerations or shape of trajectory. In other words, the trajectory of mass points of the rolling body are not as if the body rotated around O. The idea of instantaneous axis of rotation applies only to velocities.

Method 3 could work, but you have to find the cycloid trajectory of the mass point A, mathematically, find height $y$ of A in terms of rolled distance $x$. This is much more work than Method 1 but it will give the same results.

  • $\begingroup$ According to the answer given, option C is correct. In Method 2, I am NOT finding the shape of the trajectory. For this particular instant, the motion of the ball can be thought of as rotation about that bottom most point, only for this instant. After some time, the bottom most point would be different (and displaced) and we would consider rotation about that point. Since there is both translation of the bottom point, and rotation about it, a cycloid can be generated in this case also. IAOR is defined as a point about which rotation can be considered for an instant,so why doesn't it work here? $\endgroup$ Nov 13, 2020 at 7:23
  • $\begingroup$ Yes O is instantaneous axis of rotation, but how do you use this to find radius of trajectory curvature? $\endgroup$ Nov 13, 2020 at 11:45
  • $\begingroup$ look at the diagram ive attached. if we can consider rotation about O for that instant, why cant we directly say that its radius of curvature is r root2? $\endgroup$ Nov 13, 2020 at 12:04
  • $\begingroup$ Ah, I get what you're doing now. This doesn't work, because the idea about instantaneous rotation works only for instantaneous velocities. It does not work for trajectory - trajectory isn't (not even locally) a circle with the same radius one uses to get velocity of rotation. The proper radius for trajectory curvature is that of osculating circle, not that of instantaneous angular rotation. Yeah, this is confusing. Just remember that instantaneous rotation was devised and works only for velocities. Trajectory and accelerations are not as if the particle actually rotated in circles around it. $\endgroup$ Nov 13, 2020 at 12:20
  • $\begingroup$ Oh alright, thank you! One last thing, is it a mere coincidence that the IAOR method yielded half the actual answer? Or does this work for all cases? $\endgroup$ Nov 14, 2020 at 6:08

The inconsistency between the two methods is because the cylinder has two components to its velocity: it has a rotational velocity and a linear velocity. In your second method you are not accounting for how the point at A moves due to the linear motion - you are effectively assuming that the cylinder is rotating in place.

The second method will give an answer that is half that of the first method - but only for points on the edge of the cylinder! This is because for these points the contribution from the rotational and linear velocity is equal. As you move towards the center of the cylinder, the rotational velocity contribution becomes smaller.

Consider the point at C - what is its rotational velocity? It must be 0 because this point is not rotating about any other point. It is only in linear motion. Thus its radius of curvature is infinite (and method 2 fails).

  • $\begingroup$ IAOR is defined as a point about which the other points can be assumed to rotate, for any particular instant of time. Because of this definition, it shouldn't matter whether point A is moving linearly or not. The net velocity of A is 2V (where V is the velocity of the center of mass) to the right, which can be considered as a tangential velocity (for this instant) for circular motion about O. However, from your last point, it is obvious that my reasoning is wrong. Could you please help me out with this? Why is the definition not consistent? $\endgroup$ Nov 12, 2020 at 18:25
  • $\begingroup$ I think the issue is that this is true for "an instant of time". At this instant, yes, you can take the velocity at point C to be the tangential velocity for circular motion about O. But for you to find the radius of curvature you need to trace out the trajectory in time. When you do this you see that the trajectory for the point at C has no curvature. $\endgroup$
    – Framazu
    Nov 12, 2020 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.