Question

1. a 2.2 kg object is whirled in a vertical circle whose radius is 1.0 m. if the frequency is 1.03 hz, what is the tension in the string (assume uniform speed)

a) at the top?
b) at the bottom?

1. a 915 kg car goes over a hill of circular arc. if the radius of the curve is 43 m, how fast can the car travel without leaving the road at the top of the arc?

Explanation:

Step1: Calculate the speed \(v\)

The formula for speed \(v = 2\pi r f\), where \(r = 1.0\space m\) and \(f=1.03\space Hz\).
\(v = 2\pi\times1.0\times1.03\approx 6.47\space m/s\)

Step2: Analyze forces at the top

At the top of the vertical circle, the net force \(F_{net}=T + mg=\frac{mv^{2}}{r}\) (centripetal force). So, \(T=\frac{mv^{2}}{r}-mg\).
Substitute \(m = 2.2\space kg\), \(v\approx6.47\space m/s\), \(r = 1.0\space m\) and \(g = 9.8\space m/s^{2}\)
\(T=\frac{2.2\times(6.47)^{2}}{1.0}-2.2\times9.8\)
\(T = 2.2\times41.86 - 21.56\)
\(T=92.1 - 21.56=70.54\space N\)

Step3: Analyze forces at the bottom

At the bottom of the vertical circle, the net force \(F_{net}=T - mg=\frac{mv^{2}}{r}\) (centripetal force). So, \(T=\frac{mv^{2}}{r}+mg\)
Substitute \(m = 2.2\space kg\), \(v\approx6.47\space m/s\), \(r = 1.0\space m\) and \(g = 9.8\space m/s^{2}\)
\(T=\frac{2.2\times(6.47)^{2}}{1.0}+2.2\times9.8\)
\(T=92.1+21.56 = 113.66\space N\)

Answer:

a) \(70.54\space N\)
b) \(113.66\space N\)