1
$\begingroup$

I found this problem in my book and which by the look of it doesn't makes sense from the mechanical point of view.

The problem is as follows:

The figure from below shows a circular plate rolling over an incline without slipping. The speed on point $B$ is $30\frac{m}{s}$, find the speed given in (meters per second) of $A$ and that of $O$ in the instant shown.

Sketch of the problem

The alternatives given are:

$\begin{array}{ll} 1.&21\frac{m}{s}\,\textrm{and}\,0\,\frac{m}{s}\\ 2.&13\frac{m}{s}\,\textrm{and}\,0\,\frac{m}{s}\\ 3.&15\frac{m}{s}\,\textrm{and}\,10\,\frac{m}{s}\\ 4.&15\frac{m}{s}\,\textrm{and}\,0\,\frac{m}{s}\\ 5.&18\frac{m}{s}\,\textrm{and}\,10\,\frac{m}{s}\\ \end{array}$

Upon inspecting this problem, it doesn't make sense that the plate is rolling in the incline and not going down. Can this happen? By the way the drawing is reproduced as it is in the original source but other than the lines indicating that the plate is insinuating that is going down the incline. Can the later be inferred to be a rolling in the clockwise direction?

If I'm not mistaken the speed on point $O$ must be zero as the components of velocity in the point of contact cancel because the translation and rotation cancel each other. But what makes me confused is that, if the plate is rotating without slippage, wouldn't mean that there is no translation and only rotation? And because of such wouldn't it mean that there will be velocity in point O?

The rest is where I'm lost. If the speed at point B is $30\frac{m}{s}$ wouldn't it be the same for that point $A$?

Can someone help me here? Supposedly the answer is the first option which does seem to check with what I was guessing for the point of contact. I'd appreciate that an answer could teach me on the physical aspect that it is happening here and be detailed the most as possible, because I'd like to know what's going on.

$\endgroup$
  • $\begingroup$ it doesn't make sense that the plate is rolling in the incline and not going down. Where does it say this...? $\endgroup$ – Aaron Stevens Dec 23 '19 at 11:26
0
$\begingroup$

I assume here that speed means velocity in the direction of the slope. Your assumption about point $O$ is correct. It must be stationary. To think about the other points, try changing the frame of reference.

Imagine the plate is rotating but stationary (or better still, is a stationary wheel on a fixed axle) and is sitting on top of a moving belt.

The belt is travelling at speed $-v$. This must also be the speed of the bottom of the plate as it is not slipping on the belt. The top of the plate ($B$) must be travelling in the opposite direction at the same speed ($v$) as it is symmetrical. Clearly, point $A$ isn't moving at all in the horizontal direction so its speed is zero.

To answer your question, just adjust your reference frame again by adding $v$ to everything.

$\endgroup$

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