Physics
22.04.2023 00:28
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A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 7.00 revolutions every 10.0 seconds. What is the magnitude of the acceleration of the ball?
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aaqa3735
aaqa3735
4,7(83 marks)

This problem can be divided in to 2 parts

First part in which stone moves up and second part in which stone moves down.

Let the time for travel in upper direction be t_1 and time for travel in down direction be t_2

We have v = u+at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

  Up direction

               0 = 10 - 9.81*t_1

                t_1 = 1.02 seconds

We also have s=ut+\frac{1}{2} at^2, s is the distance traveled.

                             s = 10*1.02-\frac{1}{2} *9.81*1.02^2 = 5.09 m

Now considering the downward motion.

                  displacement = 5.09+65=70.09 m

                 s=ut+\frac{1}{2} at^2

                 70.09 = 0*t+0.5*9.81*t_2^2\\ \\ t_2=3.78seconds

a) So total time taken = 3.78+1.02 = 4.8 seconds

b) In case of downward motion v = u+at

          u=0, a = g, t = 3.78 seconds

  So, Velocity on reaching ground = 0+9.81*3.78 = 37.08 m/s

c) Total distance traveled = 5.09+70.09 = 75.18 m

d)Total displacement is height of cliff, which is equal to 65 m

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