Question

1. a 0.15 - kg rubber stopper is attached to the end of a 1.00 - m string and is swung in a circle.

a. if the rubber stopper makes 0.85 revolutions/sec. what is the force the string exerts on the stopper?
b. if the rubber stopper is swung 2.3 m above the ground and released, how far from the point it was released will it fall to the ground?

Explanation:

Step1: Calculate angular velocity

Angular velocity \(\omega = 2\pi f\), where \(f = 0.85\space rev/s\).
\(\omega=2\pi\times0.85\approx 5.34\space rad/s\)

Step2: Calculate centripetal force

Centripetal force \(F = m\omega^{2}r\), \(m = 0.15\space kg\), \(r = 1.00\space m\)
\(F=0.15\times(5.34)^{2}\times1.00\)
\(F = 0.15\times28.5156\approx4.28\space N\)

Step3: Calculate time of fall (for part b)

Using \(h=\frac{1}{2}gt^{2}\), \(h = 2.3\space m\), \(g = 9.8\space m/s^{2}\)
\(t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times2.3}{9.8}}\approx\sqrt{0.4694}\approx0.685\space s\)

Step4: Calculate linear velocity (for part b)

Linear velocity \(v=\omega r\), \(\omega = 5.34\space rad/s\), \(r = 1.00\space m\)
\(v = 5.34\times1.00 = 5.34\space m/s\)

Step5: Calculate horizontal distance (for part b)

Horizontal distance \(x = vt\), \(v = 5.34\space m/s\), \(t = 0.685\space s\)
\(x=5.34\times0.685\approx3.66\space m\)

Answer:

a. The force the string exerts on the stopper is approximately \(4.28\space N\)
b. The rubber stopper will fall approximately \(3.66\space m\) from the point it was released.