A block of mass m slides down a hemisphere of mass M. What are the accelerations of each mass? Assume friction is negligible.

enter image description here a_M = Acceleration of hemisphere

N_m = Normal force of M onto m

N_M = Normal force of ground onto M

So from the FBD's, I come up with

$$\sum \text{F}_{xm}= mg\sin \theta = m(a_t - a_M \cos \theta)$$

$$\sum \text{F}_{ym} = N_m - mg \cos \theta = -m(a_r + a_M \sin \theta)$$

$$\sum \text{F}_{xM} = -N_m \sin \theta = Ma_M$$

I need another equation, so I tried going the route of work-energy, to find the tangential speed of the block sliding on the hemisphere, but can I look at the energy of the block by itself? I figure I cannot, as it is atop an accelerating body.

However, if I can consider the energy of the block by itself to find the tangential speed, then I can solve for aM,

$$ a_M = gm\sin \theta \frac{2-3\cos \theta}{M-m\sin ^2 \theta} $$

which goes to 0 when M >> m and so then $$a_t = g\sin \theta$$ in that case, which checks out, however Im still a little weary about this.

I'm rather stuck here so any help would be appreciated.

  • $\begingroup$ More Phys.SE posts on blocks sliding down hemispheres. $\endgroup$ – Qmechanic Aug 4 '14 at 21:01
  • $\begingroup$ It this a solid hemisphere or a hollow one? $\endgroup$ – Floris Aug 4 '14 at 21:09
  • $\begingroup$ @Floris I guess the hemisphere I imagine would be solid. Does that really matter? When I thought of this problem, I was thinking more along the lines of the problem where a block slides down on a movable ramp. $\endgroup$ – Plopperzz Aug 5 '14 at 0:02
  • $\begingroup$ @Qmechanic I guess my main question is: In this scenario where both the block and the hemisphere are in motion, can I ignore the kinetic energy of the hemisphere, as well as the work done by the normal force from the mass on the hemisphere in finding the energy of the mass? $\endgroup$ – Plopperzz Aug 5 '14 at 0:09
  • $\begingroup$ The reason that the hollow/solid question matters is that as the sphere moves, the mass is doing work on the sphere, leaving less energy for itself. Thus the mass and moment of inertia of the large sphere do matter. Tricky problem. $\endgroup$ – Floris Aug 5 '14 at 1:19

I did not check thoroughly your free body diagrams,but they look correct. One comment though, I do not think that using polar separation of the accelerations is particularly useful for this problem since the obvious origin for such system is accelerating, as you well indicate.

I think that your missing equations are just your geometrical constrains. You have not used them. I noticed that we are using different convention for the coordinates. Mine is just a static Cartesian coordinate system with the origin in the center of the hemishpere at the beginning, but it does not move with it. I am assuming we can treat this problem in two dimensions. Assuming that the block has not lost contact with the hemisphere, you have $$(x_m-x_M)^2 + y_m^2=R^2,$$ which is an equation that is valid at all times, therefore it relates the three accelerations and velocities in a complicated manner that appears explicitly by deriving two times. And the fifth equation is just $$\tan\theta=\frac{y_m}{x_m-x_M}.$$ therefore you have the normal, the three dinamical variables, and a fourth dependandant dynamical variable introduced just to ease the notation, $\theta$. The FBD equations in my case are \begin{align} m \ddot{y}_m &= N\cos\theta -mg\\ m \ddot{x}_m &= N \sin\theta\\ M \ddot{x}_M &= -N cos\theta \end{align} Five equations for five unknowns, and a very ugly solution that I do not think exists in closed form.

You asked about energy. Indeed, you can use this approach since there is no friction. Actually, the energy approach is useful even when the objects are no longer in contact. Lets assume that the block start moving in the position you depict in your picture, that is $y_m(t=0)=h$ and $x_m(t=0)=\sqrt{R^2-h^2}$, and $x_M(t=0)=0$, with all velocities equal to zero at $t=0$. Therefore, conservation of energy dictates $$\frac{1}{2}M\dot{x}_M^2+\frac{1}{2}m(\dot{x}_m^2+\dot{y}_m^2)+mgy_m=mgh.$$

  • $\begingroup$ I'm not quite clear on where you got your first equation? $\endgroup$ – Plopperzz Aug 5 '14 at 3:53
  • $\begingroup$ @Plopperzz I 've just edited and explain it better. We are using different notation, my $x$ and $y is fixed. $\endgroup$ – Enredanrestos Aug 5 '14 at 13:18

We can solve this problem easily by using only the law of conservation of momentum and the law of conservation of energy.

It is obvious that in given case only the horizontal component of the momentum of the system of the block and the hemisphere is conserved (i am using OP notations):


Now, an equation of conservation of energy:


Let's introduce the ratio


Then the equations look simpler


$$\eta v^2_M+v^2_m=2gR(1-\cos\theta)\qquad(2)$$

Now, the physical part of the problem is completed. We are left with pure mathematics.

From the last two equations we get an expression for $v_m$:


where $\gamma=\frac{\eta}{(1+\eta)^2}$ for simplicity.

Now, the acceleration of the block $a_m=\dot{v}_m$ can be obtained by differentiating (3) by time, remembering that $\theta=\theta(t)$ and noting that $\dot{\theta}=\frac{v_m}{R}$.

I think, the easiest way is the method of logarithmic differentiation of (3).

The acceleration of the hemisphere $a_M=\dot{v}_M$ can be obtained by differentiating (1) by time:



Since $\dot{v}_m$ and $v_m$ have already been calculated we are done.

  • $\begingroup$ That's not as bad as I had thought it would be. Do you feel this is the type of question they may ask in a first year physics course? $\endgroup$ – Plopperzz Aug 6 '14 at 17:52
  • $\begingroup$ This looks simpler indeed compared to what I did. A small comment though: strictly speaking, the energy equation assumes an initial condition for which the only dynamical solution is static. You need to be bear this in mind or you will get stuck again, literally. $\endgroup$ – Enredanrestos Aug 6 '14 at 19:54
  • $\begingroup$ @Plopperzz I think not. This problem is more an analytical mechanics than a physics problem, i think. $\endgroup$ – Martin Gales Aug 7 '14 at 14:12
  • $\begingroup$ @Enredanrestos A good point! To avoid this issue we can take instead of initial condition $\theta=0$ an initial condition $\theta=\theta_0$ where $\theta_0$ is different from zero, no matter how little. $\endgroup$ – Martin Gales Aug 7 '14 at 14:26

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