Question

a 12 g object is attached to a 0.65 m string. the object is then swung in a horizontal circle, completing one revolution in 0.85 s. what is the magnitude of the centripetal force on the object? (round to two significant figures.)
0.12 n
0.36 n
0.14 n
0.43 n

Explanation:

Step1: Calculate the speed of the object

The circumference of the circle is \(C = 2\pi r\), where \(r = 0.65\space m\). So \(C=2\pi\times0.65\approx4.084\space m\).
The speed \(v=\frac{C}{T}\), with \(T = 0.85\space s\). Then \(v=\frac{4.084}{0.85}\approx4.805\space m/s\).

Step2: Calculate the centripetal force

The mass \(m = 12\space g=0.012\space kg\).
The centripetal force formula is \(F_c=\frac{mv^{2}}{r}\).
Substitute \(m = 0.012\space kg\), \(v\approx4.805\space m/s\), \(r = 0.65\space m\) into the formula:
\(F_c=\frac{0.012\times(4.805)^{2}}{0.65}\).
First calculate \((4.805)^{2}\approx23.09\).
Then \(0.012\times23.09 = 0.277\).
Finally \(F_c=\frac{0.277}{0.65}\approx0.43\space N\).

Answer:

\(0.43\space N\) (corresponding to the fourth option)