Question

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius 13.0 m. She has mass 70.0 kg, and her motorcycle has mass 40.0 kg. (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

Answer 1

(a) 11.3 m/s

At the top of the circle, there are two forces acting on the girl and the motorbike:

- Force of gravity: (m+M)g, with m=40.0 kg being the mass of the motorbike, M=70.0 kg being the mass of the girl, and g=9.8 m/s^2, acting downward

- The normal force exerted by the track on the motorbike, N, also acting downward

According to Newton's second law, the resultant of these forces must be equal to the centripetal force, so

$(m+M)g+N=(m+M)\frac{v^2}{r}$

where v is the speed and r = 13.0 m is the radius of the circle.

The problem is asking us for the minimum speed she must have at the top of the circle for the motorcycle tires to remain in contact with the sphere: this means, the minimum speed at which she loses contact, so N=0. Replacing this into the equation, we can solve it for v to find the speed:

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(13.0 m)}=11.3 m/s$

(b) 5400 N

At the bottom of the circle, the speed is twice the value calculated in part (a), so

$v=2\cdot 11.3 m/s=22.6 m/s$

The equation for the forces is now

$N-(m+M)g = (m+M)\frac{v^2}{r}$

because now the normal force is toward the centre of the circle, while the weight is in the opposite direction. Solving for N, we find the normal force:

$N=(m+M)g+(m+M)\frac{v^2}{r}=(110 kg)(9.8 m/s^2)+(110 kg)\frac{(22.6 m/s)^2}{13.0m}=5400 N$