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CP A small block with mass 0.0500 kg slides in a vertical
Chapter 9, Problem 77P(choose chapter or problem)
CP A small block with mass 0.0500 kg slides in a vertical circle of radius R = 0.800 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 normal force the track exerts on the block has magnitude 3.40 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?
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QUESTION:
CP A small block with mass 0.0500 kg slides in a vertical circle of radius R = 0.800 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 normal force the track exerts on the block has magnitude 3.40 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?
ANSWER:Solution 77P Introduction We have to calculate the centrifugal force at the bottom and from that we can calculate the kinetic energy of the particle at the bottom. From the kinetic energy at the bottom we can calculate the kinetic energy at the top. And from the kinetic energy at the top we can calculate the centrifugal force at the top. Then from the centrifugal force and the weight we can calculate the normal reaction. Step 1 The weight of the block is W = mg = (0.0500 kg)(9.8 m/s ) = 0.49 N Now at the bottom of the circular loop the centrifugal force and the weight act in the same direction and hence the normal force is the addition of the both centrifugal force and the weight. So the centrifugal force at the bottom is F = N W = 3.40 0.49 = 2.91 N cb b Now we know that the centrifugal force can be written as mv b F cb = r So we have mv 2 b = 2.91 N r