Question

A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and forth like a pendulum. When the ball is at its lowest point and the rope is vertical, the tension in the rope is three times the weight of the ball, so T=3mg.
1. What is the speed of the ball as it swings through this point? Express your answer with the appropriate units.
2. What is the speed of the ball if T=mg at this point, where the rope is vertical? Express your answer with the appropriate units.

Answer 1

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

$a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g$

The speed at this point would then be

$v^2 = a_c r = 2gr = 2*9.8*17 = 333.2$

$v = \sqrt{333.2} = 18.25 m/s$

2. Similarly, if T = mg, then the centripetal acceleration must be

$a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0$

As the ball has no centripetal acceleration, its speed must also be 0 as well.