Circular Motion (top/Bottom)

Question

A 38-g ball at the end of a string is swung in a vertical circle with a radius of 30 cm. The tangential velocity is 200.0 cm/s. Find the tension in the string (hint: sometimes gravity helps keep the ball going in a circle, in that it points towards the center of the circle, causing the tension to be less, and other times, gravity points away from the center of the circle, causing the string's tension to be greater - sketch a free body diagram for both top and bottom!):

Answer 1

To find the tension in the string, we need to analyze the forces acting on the ball when it is at the top and bottom of its vertical circular path.

At the top of the circle, the tension in the string (T) points downwards, and the weight of the ball (mg) points downwards as well. The net force acting on the ball is the centripetal force (Fc), which points towards the center of the circle. Therefore, we have:

T - mg = Fc

The centripetal force is given by:

Fc = mv^2 / r

where m is the mass of the ball, v is its tangential velocity, and r is the radius of the circle.

Substituting the given values, we get:

Fc = (0.038 kg) * (200.0 cm/s)^2 / 0.30 m ≈ 506.7 N

Therefore, at the top of the circle, we have:

T - mg = Fc T - (0.038 kg) * (9.81 m/s^2) = 506.7 N T ≈ 511.4 N

At the bottom of the circle, the tension in the string (T) points upwards, and the weight of the ball (mg) points downwards as well. The net force acting on the ball is again the centripetal force (Fc), which points towards the center of the circle. Therefore, we have:

T + mg = Fc

Substituting the given values, we get:

Fc = (0.038 kg) * (200.0 cm/s)^2 / 0.30 m ≈ 506.7 N

Therefore, at the bottom of the circle, we have:

T + mg = Fc T + (0.038 kg) * (9.81 m/s^2) = 506.7 N T ≈ 469.4 N

Therefore, the tension in the string at the top of the circle is approximately 511.4 N, and at the bottom of the circle, it is approximately 469.4 N.