My Math Forum Two-sphere pendulum
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 June 13th, 2017, 06:03 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 371 Thanks: 26 Math Focus: Number theory Two-sphere pendulum A ball of diameter D, inside a spherical shell of radius D, both with adequate friction at initial contact between equators, act as a pendulum with what period? Last edited by Loren; June 13th, 2017 at 06:08 PM.
 June 14th, 2017, 03:38 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,119 Thanks: 710 Math Focus: Physics, mathematical modelling, numerical and computational solutions Um... same period as a regular pendulum with a mass equal to the total mass of the ball and shell? Not sure...
 June 14th, 2017, 05:29 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra A diagram that illustrates how it acts as a pendulum might be useful.
June 14th, 2017, 06:40 AM   #4
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Quote:
 Originally Posted by Benit13 Um... same period as a regular pendulum with a mass equal to the total mass of the ball and shell? Not sure...
I think the shell is static, making its mass irrelevant. The question indicates that the ball rolls rather than slips, but without losing energy. Google finds several worked examples of this type of "pendulum", but they don't allow for the large angle of oscillation specified in this problem.

June 14th, 2017, 07:35 AM   #5
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Quote:
 Originally Posted by skipjack I think the shell is static, making its mass irrelevant. The question indicates that the ball rolls rather than slips, but without losing energy. Google finds several worked examples of this type of "pendulum", but they don't allow for the large angle of oscillation specified in this problem.
Okay... so the ball with diameter D is rolling around in the inside of a spherical shell with diameter 2D? I guess the period of any oscillations (i.e. rolling up and down the sides of the spherical shell given some initial position away from the bottom) can be determined by building an equation of motion (using Newton's laws) and solving. With friction and the whole rolling/slipping situation, you'll probably get something similar to a damped SHO.

 June 14th, 2017, 08:15 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,168 Thanks: 1640 Assuming that the shell is hemispherical, one can reasonably assume that the ball starts as illustrated below. Pendulum.PNG
 June 14th, 2017, 03:02 PM #7 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 371 Thanks: 26 Math Focus: Number theory Would a two-dimensional, frictionless pendulum simplify the OP (now no slipping)? The ball would initially be released where the equators meet. Thanks for the diagram, skipjack.
 June 15th, 2017, 12:24 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,168 Thanks: 1640 Do you mean that the ball doesn't "roll" at all?
 June 15th, 2017, 08:30 AM #9 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 371 Thanks: 26 Math Focus: Number theory I submit two "extremes"; the ball sliding without friction, or the ball rolling with just enough friction to cause it not to slide.

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