Inconsistency Regarding Instantaneous Axis of Rotation

Question

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

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.

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?

Answer 1

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.

Answer 2

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).