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 15.0 m . The physics major has a mass of 74.0 kg , and his motorcycle has a mass of 40.0 kg.

What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere?

At the bottom of the circle his 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

1. The centripetal force on the vertical circle must be equal or greater than gravitational force over the physics major and his motorcycle:

Gravitational force = Centripetal force

(m1+m2) g ≤ (m1+m2) v²/R

where:

m1 = mass of physics major (74.0 kg)
m2 = mass of motorcycle (40.0 kg)
v = velocity
R = radius (15 m)

v ≥ √(g R) = √(9.8 * 15)

= = = = = = = = = =
v ≥ 12.13 (m/s)
= = = = = = = = = =


2. Applying Newton's Second Law of Motion over the vertical:

∑ Fy = 0

N – (m1+m2) v²/R – (m1+m2) g = 0

N = (m1+m2) v²/R + (m1+m2) g

N = (74.0 + 40,0) (2 * 12.13)²/15.0 + (74.0 + 40,0) 9.81

= = = = = = = = = = = =
N = 5591.30 (N)