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2 years ago

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A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

a flat straight section at the same height as the center of the loop (19.2 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

1 answer:

dalvyx [7]2 years ago

4 0

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

$F_{C} = F_{G}$

$\frac{mv^{2}}{R} = mg$

where

v = velocity

g = acceleration due to gravity

$v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s$

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See attached figure.

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--------------------

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