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420chan is Getting Overhauled - Changelog/Bug Report/Request Thread (Updated July 26)

Math problem

- Fri, 26 Sep 2014 14:29:27 EST vUNowU0Q No.54441
File: 1411756167954.png -(17766B / 17.35KB, 640x558) Thumbnail displayed, click image for full size. Math problem
Anyone want to help me out with this problem? I'm new to astronomy/physics and I haven't the faintest.
Henry Draper - Tue, 30 Sep 2014 12:05:13 EST YF1E3C5p No.54451 Reply
The ball sits on top of the semi-circle forever because no lateral force has been applied to it
Giuseppe Piazzi - Tue, 30 Sep 2014 13:54:13 EST W19G6Cz7 No.54453 Reply
This. If the mass is "dropped directly on top" it will stay there.
Roger Penrose - Tue, 30 Sep 2014 20:18:02 EST ruPFqd4c No.54456 Reply
The answer is 0.730 radians.
Roger Penrose - Tue, 30 Sep 2014 20:36:39 EST ruPFqd4c No.54457 Reply
I'm not gonna write out the calculations, but to elaborate a little further, here are the steps involved:

  1. Use conservation of energy to find the velocity of the mass along the circle.
  2. Use the velocity to find the centripetal force the mass experiences as a result of rolling (actually sliding) around the circle. This force should be a function of theta.
  3. Find the radially oriented component of the normal force as a function of theta.
  4. Set the forces in steps 2 and 3 equal to each other and solve for theta.
Tycho Brahe - Fri, 26 Jun 2015 03:11:47 EST HotOC/GQ No.55454 Reply
90°, its just going to bounce when it hits the sphere.
Edward Pickering - Sun, 05 Jul 2015 07:31:07 EST sVia2s64 No.55471 Reply
actually it won't, because of uncertainty principle you can't make it stand perfectly still perfectly at the top
Paul Goldsmith - Fri, 24 Jul 2015 21:11:43 EST euFuFwSC No.55558 Reply

> A mass is dropped directly on top of a half circle. As it rolls of the side...

Top lel
Edwin Salpeter - Mon, 31 Aug 2015 01:06:20 EST OMGzRHpD No.55646 Reply
1440997580737.jpg -(163342B / 159.51KB, 483x572) Thumbnail displayed, click image for full size.
is mass M a sphere or a cylinder? that changes the moment of inertia and the angular momentum of mass M. you said no friction but it's rolling so I'm lead to believe we are operating under the rolling-without-slipping regime (translational speed 'v' = radius of the mass M 'r' x 'omega', the angular velocity).

set up a force diagram for the mass at location X. the weight force 'mg' points straight down, we'll split into components in a moment. The normal force 'N' of the hemisphere on mass M points radially outward (in this case north-east). We'll be solving for the condition where N --> 0 (i.e., it juuust loses contact with the hemisphere)

Edwin Salpeter - Mon, 31 Aug 2015 01:26:04 EST OMGzRHpD No.55647 Reply
now the weight force 'mg' can be split into components. one component points radially inward from X with value mgcos(theta). the other component points in the direction of mass M's velocity vector, tangent to the hemisphere or south-east, with value mgsin(theta). the first component balances with the normal force N and provides the centripetal acceleration for mass M. the second component does the accelerating of mass M down the hemisphere.

As >>54457 said, now do the conservation of energy. The translational kinetic energy 1/2mv^2 plus the rotational kinetic energy 1/2Iomega^2 will equal mg(R-y) where y is the height of location X. if you want it dimensionless (R=1) then R-y = 1-sin(theta). make the v = r x omega substitution and solve for v.

the inward radial componenet mgcos(theta) provides the centripetal acceleration v^2/R. So mgcos(theta) = mv^2/R ... too distracted to finish the rest but solve for the max v that the inward radial component can still provide a = v^2/R, then you can solve for the value of cos(theta)

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