Question

A physics student whirls a rubber stopper around in a horizontal circle. Suddenly the string breaks. What is the speed of the rubber stopper the moment the string breaks if the radius of the circle is 0.75 m, the mass of the stopper is 45 grams, the angle θ in the diagram is 30 degrees, and the counter weight is 100 grams? Use g = 10 m/s².

Answer 1

The speed of the rubber stopper the moment the string breaks is 0 m/s.

Step-by-step explanation:

The speed of the rubber stopper the moment the string breaks can be found using the concept of centripetal force. The centripetal force acting on the stopper is provided by the tension in the string. When the string breaks, this force is no longer present, and the stopper continues to move in a tangential direction.

Using the formula for centripetal force, we can equate it to the tension in the string:

Fc = Tension in the string = m * v^2 / r

Where m is the mass of the stopper, v is its velocity, and r is the radius of the circle.

Plugging in the given values, we have:

Tension in the string = (0.045 kg) * (v^2) /(0.75 m)

Since the string breaks, the tension becomes zero, so:

0 = (0.045 kg) * (v^2) /(0.75 m)

Solving for v, we find that:

v^2 = 0

Therefore, the speed of the rubber stopper the moment the string breaks is 0 m/s.