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Artist 52 [7]
3 years ago
15

A ball of mass m is released at the top edge of a frictionless half-pipe with a radius of curvature r. At the bottom of the pipe

, the ball is moving with a speed v. What is the normal force acting on the ball at this point?
Physics
1 answer:
Art [367]3 years ago
3 0

Answer:

mg +\frac{mv^{2}}{r}

Explanation:

At the bottom of the pipe :

N = Normal force acting in upward direction

m  = mass of the ball

W  = weight of the ball acting in downward direction = mg

v  = speed of the ball at the bottom

r  = radius of curvature

Force equation for the motion of ball is given as

N - W = \frac{mv^{2}}{r}

N - mg = \frac{mv^{2}}{r}

N = mg +\frac{mv^{2}}{r}

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