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Block Sliding Down Hemisphere

  1. Aug 4, 2014 #1
    1. The problem statement, all variables and given/known data

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

    (See attatchment)

    2. Relevant equations

    [itex] a_M [/itex] = Acceleration of hemisphere

    [itex]N_m [/itex] = Normal force of M onto m

    [itex]N_M [/itex] = Normal force of ground onto M

    [itex]\sum \text{F}_x = ma[/itex]

    3. The attempt at a solution

    So from the FBD's, I come up with

    [itex]\sum \text{F}_{xm}= mg\sin \theta = m(a_t - a_M \cos \theta)[/itex]

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

    [itex]\sum \text{F}_{xM} = -N_m \sin \theta = Ma_M[/itex]

    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.

    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 [itex]a_t = g\sin \theta[/itex] 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.

    Attached Files:

    Last edited: Aug 4, 2014
  2. jcsd
  3. Aug 4, 2014 #2


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    Gold Member

    I'm not so sure the units check out in the numerator of that fraction for your a_M solution.
  4. Aug 4, 2014 #3
    Maybe you saw the first one I posted - I corrected it a little while ago- I made some little arithmetic error.

    $$a_M = gm\sin \theta \frac{2-3\cos \theta}{M-m\sin ^2 \theta}$$ this one checks out unit-wise.
  5. Aug 4, 2014 #4
    So after doing a little more work, I'm inclined to believe that the answer is correct. I went and solved for the height at which the block flies off of the hemisphere and ended up with,

    $$-mg\cos \theta _\text{max} = -m(a_r + a_M \sin \theta _\text{max})$$which when substituting in the ratios gives me

    $$0 = \frac{M(2R-3h)}{R(M-mR)+mh^2}$$ so [itex] h = \frac{2}{3}R [/itex] as expected. Further more, when M >> m, it also reduces to [itex] h = \frac{2}{3} R [/itex].

    Another way of doing it, I guess would be to just notice that when the block flies off, [itex] a_M = 0 [/itex] which, by solving for cosine theta, gives h = 2R/3 as well
    Last edited: Aug 4, 2014
  6. Aug 5, 2014 #5
    Do you mind explaining how you got this relation ?

    Presuming the relation obtained is correct , how did you deduce that when M >> m , a_M → 0 ?
    Last edited: Aug 5, 2014
  7. Aug 5, 2014 #6
    This is the result obtained when the hemisphere is stationary . Why do you believe this should be the result when the hemisphere is accelerating ?

    I think it should leave the hemisphere at an angle less than the one obtained when the hemisphere is stationary .

    Again,I haven't been able to prove it as of now .
  8. Aug 5, 2014 #7
    I believe that if it were the case where the acceleration function is monotonically increasing then yes the block would fly off at a height greater than 2r/3, however in this case I believe the acceleration is first increasing then decreasing, so as the block leaves the hemisphere the acceleration of the hemisphere is 0 and so it still must leave at a height of 2r/3.

    However, I could very well be misunderstanding certain concepts which cause the block to fly off earlier, but the fact that as M gets very large everything reduces to the case where the hemisphere is always stationary causes me to feel somewhat more "confident" in my solution (at east that I got it close).
    Last edited: Aug 5, 2014
  9. Aug 5, 2014 #8
    Here is why I said I was "weary" about this:

    I originally wanted to throw in an equation involving work, but then I thought I might be able to only consider the energy of the block sliding on top of the hemisphere and not the energy of the entire system, since there is no work done one the block by any force besides gravity, I just used conservation of energy: $$mgR(1-\cos \theta) = \frac{1}{2}mv^2 $$
  10. Aug 5, 2014 #9
    This is incorrect .You cannot use conservation of energy in an accelerated frame without taking into account inertial forces (pseudo force ) . If you wish to conserve energy then consider a pseudo force ma acting on the block towards right . You need to take into account work done by the pseudo force on the block.
  11. Aug 5, 2014 #10
    "...but can I look at the energy of the block by itself? I figure I cannot, as it is atop an accelerating body..."

    Ya, I just did it to see what would happen and then went with it.:redface:
  12. Aug 5, 2014 #11
    If you are conserving energy, you must be conserving the total energy of the system. Including that of the hemisphere. I would also advise that you use the stationary "table" frame for all the quantities.
  13. Aug 5, 2014 #12
    Here is a reply I got on physics.SE: http://physics.stackexchange.com/questions/129476/block-sliding-down-hemisphere

    Now, I understand all the equations, however I'm not too sure about the first one, [itex] (x_m -x_M)^2 + {y_m}^2 = R^2 [/itex]. Does this one really hold true as the hemisphere moves to the left?

    When dealing with energy, would I not consider the work the normal force from m onto M that causes the motion of the hemisphere to move?
    Last edited: Aug 5, 2014
  14. Aug 6, 2014 #13
    As the author of that equation stated, the variables are in a fixed frame (that "table" frame as I suggested earlier). The equation is a statement of the fact that the block remains on the circle of radius ##R## with center at ##(x_M, 0)##, which obviously takes care of the hemisphere's possible motion along the X axis.

    Regarding the work of the normal force, keep in mind that there are two normal forces, one acting on the block and another on the hemisphere, which are numerically equal but opposite in direction; the works of these two forces cancel each other, so you can simply ignore them.
  15. Aug 6, 2014 #14

    Ya, not realising why the circle relation was true was a pretty big brain fart. It hit me pretty hard out of the blue why it's true. Thanks for the advice; I will take another crack at it. 5 equations, 5 unkowns, so I'll see what I come up with, although the author of that reply on SE speculated that a closed form solution doesn't exist.
  16. Aug 6, 2014 #15
    Personally, I would use conservation of total energy and conservation of the horizontal component of total momentum.
  17. Aug 6, 2014 #16
    Using the energy/momentum approach, I have been able to reduce the problem to a block sliding on a stationary semiellipse. Interesting.
  18. Aug 6, 2014 #17
    Neat, but now that you mention it, it isn't too terribly surprising, as the arc traced out by the sliding block would trace an ellipse. I actually started working on that problem the other day before getting side tracked by the movable hemisphere problem.
  19. Aug 6, 2014 #18
    Would you consider an instantaneous radial acceleration for an ellipse to be [itex] \frac{v^2}{\sqrt{a^2\cos ^2 \theta + b^2 \sin ^2 \theta}} [/itex]?
    Last edited: Aug 6, 2014
  20. Aug 6, 2014 #19
    If "radial" means "from the center of the ellipse", then none of that would be correct. The radius of curvature approach will give you acceleration normal to the curve, which, for an ellipse, is different from radial (except at four points). More importantly, why do you need radial acceleration? And why do you need acceleration to begin with?

    To avoid confusion, the ellipse in the reduced problem is different from the ellipse that the block actually traces in the non-reduced problem.
  21. Aug 6, 2014 #20
    Oh I was just thinking about the angle at which it would fall off sliding on an ellipse, but the acceleration I used was definitely wrong, because when I set a = b in my solution to get a circle I get a very incorrect answer. Thanks for clearing that up.
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