Question

A 30.0-g ball at the end of a string is swung in a vertical circle with a radius of 25.0 cm. The tangential velocity is 200.0 cm/s. Find the tension in the string: (a) at the top of the circle, (b) at the bottom of the circle, and (c) at a distance of 12.5 cm from the center of the circle (r=12.5cm).

(a) 0.34 N, (b) 0.46 N, (c) 0.25 N
(a) 0.25 N, (b) 0.34 N, (c) 0.46 N
(a) 0.46 N, (b) 0.34 N, (c) 0.25 N
(a) 0.25 N, (b) 0.46 N, (c) 0.34 N

Answer 1

The tension in the string at the top of the circle is 0.24 N, at the bottom of the circle is 0.77 N, and at a distance of 12.5 cm from the center of the circle is 0.96 N.

Step-by-step explanation:

(a) at the top of the circle:

To find the tension at the top of the circle, we need to consider the forces acting on the ball. At the top of the circle, the tension in the string provides the centripetal force to keep the ball moving in a circle. The tension is equal to the centripetal force:

Tension = centripetal force = mass x radial acceleration

T = m x (v^2 / r)

T = (0.03 kg) x (2.0 m/s)^2 / 0.25 m = 0.24 N

So the tension at the top of the circle is 0.24 N.

(b) at the bottom of the circle:

At the bottom of the circle, the tension also provides the centripetal force. However, there is an additional force acting on the ball at the bottom - the force of gravity. The tension needs to be strong enough to provide both the centripetal force and counteract the force of gravity:

Tension = centripetal force + force of gravity

T = m x (v^2 / r) + m x g

T = (0.03 kg) x (2.0 m/s)^2 / 0.25 m + (0.03 kg) x (9.8 m/s^2)

T = 0.48 N + 0.29 N = 0.77 N

So the tension at the bottom of the circle is 0.77 N.

(c) at a distance of 12.5 cm from the center of the circle:

To find the tension at a distance of 12.5 cm from the center of the circle, we can use the same formula for tension:

Tension = m x (v^2 / r)

T = (0.03 kg) x (2.0 m/s)^2 / 0.125 m = 0.96 N

So the tension at a distance of 12.5 cm from the center of the circle is 0.96 N.