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A sphere of radius $R$ is in contact with a wedge. The point of contact is $\frac{R}{5}$ from the ground as shown in figure. Wedge is moving with velocity $20~\text{m/s}$ towards left then the velocity of the sphere at this instant will be

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I have attempted a solution there , however , unable to make any progress . Kindly advice if that is the right path to the solution , if something is there that I am not considering.

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Minor comment: The link (v1) looks prone to future link rot. –  Qmechanic Apr 23 '13 at 11:23
@nonagon google.com.au/search?q=link+rot –  Michael Brown Apr 23 '13 at 12:49

1 Answer 1

up vote 2 down vote accepted

The wedge is tangent to the sphere. Using that it touches at height R/5 you can easily work out the slope of the wedge. The velocity of the sphere follows directly (20m/s times slope).

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how ? can you please elaborate ? –  nonagon Apr 23 '13 at 13:03
Assume for the moment that R=5 (it doesn't matter for the slope), so the circle is given by x^2 + y^2 = 25. Because height=R/5 we are looking for the slope of the tangent line through the point with y=-4, which is (-3,-4). Now the slope is given by -x/y = -3/4. Now, given a velocity of 20m/s to the left, we know that the sphere will have a velocity of 3/4 * 20m/s = 15m/s. I hope that's clear. –  Doukus Apr 23 '13 at 13:23
What physical law are you applying here ? –  nonagon Apr 23 '13 at 13:35
Nothing but the fact that the wedge won't move inside the sphere. –  Doukus Apr 23 '13 at 13:43
Let's say the height of the intersection of the wedge and the circle at a given time is h, and the wedge moves over axis x. We know the slope of the sphere is 3/4, so dh/dx=3/4, dh=3/4*dx. The velocity of the wedge is given by dx/dt=20m/s. The velocity of the sphere is given by dh/dt. We know dh/dt = 3/4*dx/dt = 3/4*20m/s = 15m/s. –  Doukus Apr 23 '13 at 18:46

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