Question

A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 14.6 m . The physics major has a mass of 70.0 kg , and his motorcycle has a mass of 40.0 kg. (a) What minimum speed must he have at the top of thecircle if the tires of the motorcycle are not to lose contact withthe sphere?
m/s
(b) At the bottom of the circle, his speed is twice the valuecalculated in part (a). What is the magnitude of the normal forceexerted on the motorcycle by the sphere at this point?

Answer 1

Answer:

a) v=23.9 m/s

b) R=2646 N

Explanation:

According to Newton's second law the net force at the top is given by

∑F=-m*ac

according to

$a_{c}=\frac{v^2}{r}$

∑F=$-m*\frac{v^2}{r}$

a).

To lose contact this means that R=0 so the final equation is

$R-m*g=-m*\frac{v^2}{r}$

$-m*g=-m*\frac{v^2}{r}$

Solve to v

$v^2=g*r$

$v=\sqrt{g*r}=\sqrt{9.8*14.6}$

$v=11.96\frac{m}{s}$

b).

v is the twice of part a so

$v=2*11.96$

$v=23.9\frac{m}{s}$

$R_{m}-m_{m}*g-m_{p}=m*\frac{v^2}{r}$

Solve to Rm

$R_{m}=m*\frac{v^2}{r}+(m_{m}+m_{p})*g$

$R_{m}=m*\frac{23.9^2}{14.6}+(40+70)*9.8$

$R_{m}=2646 N$