Edit: To clarify, this is a homework question. But I couldn't care less about getting the answer to this specific question as it's not even assigned yet. It looks interesting and I want to understand the concepts behind it. Not understanding these things bothers me :p

Here is the given question:

The face of block M in the figure below is shaped like a semicircular bowl of radius R. A mass m is placed at the top-left corner of the bowl and then let go. Find the acceleration of block M relative to the surface it is sitting on when m is a distance of 0.8R from the bottom of the bowl. There is no friction between M and m, or between M and the surface on which it sits.

There is a diagram provided:


I define $cos(\theta) = \sqrt{1-0.2^2}$ , as if the ball is 0.8R from the bottom it is 0.2R from the top, then using the Pythagorean Identity to solve for $cos(\theta)$.

It states the answer is $a_{block}=\frac{-mgcos(\theta)}{M}$

My answer was that $a_{block}=\frac{-mgsin(\theta)cos(\theta)}{M}$

I split the $F_g$ vector into its parallel and perpendicular components such that $F_{g\perp}=m_1gsin(\theta)$

The parallel component is irrelevant for this question.


Then finding the horizontal component of $F_N$ is simple. Drawing a triangle it is apparent that:



Since $a=\frac{F}{m}$, I get my equation: $a_{block}=\frac{m_1gsin(\theta)cos(\theta)}{M}$.

Could anyone enlighten me on what my mistake is? I think I'm running into a similar issue in another problem involving a block sliding down a wedge that is free to move as well (literally this exact problem, but instead of a semicircular surface it's simply a wedge).

  • 1
    $\begingroup$ Are you sure about the answer of the textbook? I did the exercise and agree with you. Moreover, the answer of the textbook states that the acceleration is greatest directly when the ball is released (theta = 0) for any value of R. This would imply (if we take the limit R-> infinity) that a bowl with straight and vertical edges would accelerate if a ball is falling down its side. $\endgroup$ – Steven Mathey Sep 13 '14 at 20:37
  • $\begingroup$ The textbook has many errors, perhaps this is one of them. Do you think that my answer here is correct? !Link to image Or perhaps the denominator should be $m+M$..not sure. $\endgroup$ – Mavvie Sep 13 '14 at 20:49
  • $\begingroup$ As I said: I think that you are correct and that the books' answer does not make sense. $\endgroup$ – Steven Mathey Sep 13 '14 at 21:19
  • $\begingroup$ Alright, thanks so much for your help! I agree that acceleration should peak at 45deg not 0. I will talk to a TA or the professor about this. Thanks again! $\endgroup$ – Mavvie Sep 13 '14 at 22:27
  • 2
    $\begingroup$ Here's the solution for the wedge: farside.ph.utexas.edu/teaching/336k/Newtonhtml/node80.html . I had a quick attempt at the bowl and got non-linear equations of motion in the Lagrangian formulation (could easily have made a mistake). You write that $F_N=mg\sin\theta$, by resolving forces on the mass. This implies zero acceleration perpendicular to the bowl surface, but I'm sure that's not the case. Consider, for example, the perpendicular component of the mass's acceleration for the wedge solution given. $\endgroup$ – Michael Sep 13 '14 at 22:32

Do you know about conservation of energy? Do you know about conservation of momentum?

If you do then here's what you do::

Conserve energy for the entire system (Including velocity of the wedge). At the lower point the potential energy of the smaller block changes (potential energy= mgh) here is one relation between the velocity of the block and the wedge.

Conserve momentum for the system horizontally as there is no net external force on the system horizontally. this will give you another relation between the velocities of the wedge and the block. You should be able to find the velocity of the block now (2 equations, 2 variables). Once you find that, you can find centripetal force. The resultant is the answer. I managed to get the same answer as your textbook did. Ask me if you have any doubt in the process.

| cite | improve this answer | |

do it with conservation of energy

first of all find out the potential energyat the top of the semi circular bowl keep it equal with the kinetic energy at lowest point

| cite | improve this answer | |
  • $\begingroup$ I don't think this will work because the mass $m$ isn't at the bottom of the bowl (it's at $0.8R$), you'll have to include the PE at that height (so a $U_1=U_2+K_2$ relation). $\endgroup$ – Kyle Kanos Jun 16 '15 at 12:41
  • $\begingroup$ Nor does this procedure yield an acceleration. It gets you $v(\theta)$. You can proceed from there, but the next step is not entirely obvious to most people and this answer is not particularly helpful without pointing the way. $\endgroup$ – dmckee --- ex-moderator kitten Jun 16 '15 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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