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 a radius of 15.0 m. She has a mass of 80.0 kg and her motorcycle has mass 30.0 kg. What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere

Answer 1

Answer:

v = 12.1 m/s

Explanation:

• When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
• These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
• In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
• Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
• This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
• At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

       $F_{c} = N + m*g (1)$

• Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

       $F_{c} = m*\frac{v^{2}}{r} (2)$

• Since the normal force takes any value as needed to make (1) equal to (2),  if the speed diminishes, it will be needed less force to keep the equality valid.
• In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
• In this condition, from (1) and (2), we can find the minimum possible value of  the speed that still keeps the motorcycle touching the surface, as follows:
• $v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$