11276 L Take a Test YouTube Course Home tem 7 A Sm omework #11 Problem 7.71 Prob

Question

11276 L Take a Test YouTube Course Home tem 7 A Sm omework #11 Problem 7.71 Problem 7.71 Part A A small block with mass 0.0525 kg slides in a vertical circle of radius 0.0720 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normall force the track exerts on the block has magnitude 3.70 N What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path? Express your answer with the appropriate units. 1.642 N Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining Co

Explanation / Answer

At the bottom of the circle the normal force acts up and it balances the downward weight and downward centripetal force.

N = W + CF
3.70 = 0.0525(9.8) + 0.0525v^2/0.0720
v = 2.09 at the bottom of the circle
From conservation of energy to find v at the top of the circle
0.5mv^2 + mgh = 0.5m(2.09)^2
0.5v^2 + (9.8)(0.144) = 2.18
v = 1.24 m/s at the top of the circle

Normal force at the top is downward, the weight is downward and the centripetal force is upward
N + W = mv^2/r
N + 0.0525(9.8) = 0.0525(1.24)^2/0.072
N = 0.61 N downward

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