Question

A heavy ball with a weight of 130 N is hung from the ceiling of a lecture hall on a 4.0-{\rm m}-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?

Answer 1

Answer: T = 228.3N

Explanation: The weight of the ball will be equal to the product of its mass and acceleration due to gravity.

weight, w = mg 

Then we can get the mass.

130 = mX10 

m = 13 kg 

Now let's consider all the forces at the lowest point.

The sum of the forces will be equal to centripetal force because it swings.

Therefore:

T - mg = mv^2 / r

T = 130 + (13 X 5.5^2)/4.0 = 130 + 98.3 = 228.3 N

Answer 2

The tension on the rope as the ball passes through the lowest point is 230.35 N.

The given parameters;

• weight of the ball, W = 130 N
• length of the rope, r = 4 m
• speed of the ball, v = 5.5 m/s

The tension on the rope as the ball passes through the lowest point of the circular motion is calculated as follows;

$T - mg = \frac{mv^2}{r} \\\\T = mg + \frac{mv^2}{r}$

where;

• m is the mass of the ball

The mass of the ball is calculated by applying Newton's law;

W = mg

$m = \frac{W}{g} \\\\m = \frac{130}{9.8} \\\\m = 13.27 \ kg$

The tension on the rope is calculated as;

$T = 130 \ + \ \frac{13.27 \times 5.5^2}{4} \\\\T = 230.35 \ N$

Thus, the tension on the rope as the ball passes through the lowest point is 230.35 N.

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