Question

A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 29.0-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.

Required:
a. What is the speed of the ball as it swings through this point? Express your answer with the appropriate units.
b. 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:

  a) 23.8 m/s

  b) 0

Explanation:

a) The tension in the rope serves to balance the force of gravity and make the ball deviate from a straight-line path. The latter acceleration is ...

  a = (T -mg)/m = 2g = v²/r . . . . . m is mass

  v = √(2gr) = √(2·9.8 m/s²·29 m) = √(568.4 m²/s²)

  v ≈ 23.8 m/s . . . . approximate speed of the ball at the low point

__

b) Using the same analysis, the acceleration required to make the ball deviate from its horizontal path is ...

  a = (T -mg)/m = (mg -mg)/m = 0

and the velocity is ...

  v = √(ar) = 0 . . . . . speed of the ball where the rope is vertical

Answer 2

a. The speed of the ball is 23.84 m/s

b. The speed of the ball is 0 m/s

a.

The speed of the ball is 23.84 m/s

At this point, the net force on the rope equals the centripetal force.

So, T - W = ma where

• T = tension in rope = 3mg,
• W = weight of ball = mg and
• g = acceleration due to gravity = 9.8 m/s²,
• m = mass of ball and
• a = acceleration of ball

3mg - mg = ma

2mg = ma

a = 2g

So, the centripetal acceleration a = v²/r where

• v = speed of ball and
• r = length of rope = 29.0 m

So, a = 2g

v²/r = 2g

Making v subject of the formula, we have

v² = 2gr

v = √(2gr)

Speed of the ball

Substituting the values of the variables into the equation, we have

v = √(2gr)

v = √(2 × 9.8 m/s² × 29.0 m)

v = √(568.4 m²/s²)

v = 23.84 m/s

The speed of the ball is 23.84 m/s

b.

The speed of the ball is 0 m/s

Since the net force T - W = ma where

• T = tension in rope = mg,
• W = weight of ball = mg
• g = acceleration due to gravity = 9.8 m/s²,
• m = mass of ball and
• a = acceleration of ball

T - W = ma

mg - mg = ma

0 = ma

a = 0

So, the centripetal acceleration a = v²/r where

• v = speed of ball and
• r = length of rope = 29.0 m

So, a = 0

The speed of the ball

v²/r = 0

v² = 0

v = √0

v = 0 m/s

The speed of the ball is 0 m/s

Learn more about speed of ball here:

https://brainly.com/question/14687626