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

Answer:

T = 228.3N

Explanation:

We are given:

w = 130N

$v = 5.5m/s^2$

r = 4.0m

We take g = 10

Therefore:

w = m×g

130 = m × 10

m = 130/10

m = 10

At the lowest point,we use the equation:

$T - mg = mv^2 / r$;

Since we are asked to find tension T, we now have:

$T = mg + (mv^2)/r$

Therefore, substituting figures in the equation, we have:

$T = 130+(13 * 5.5^2)/4$;

T = 228.3N

The tension in the rope is 228.3N