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 11.8
m
. The physics major has a mass of 66.0
kg
, and his motorcycle has a mass of 40.0
kg
.
A)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?
B)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?
g = 9.81 = v^2/R
at bottom m(9.81 + 4v^2/R) using v from part A
A) To determine the minimum speed the physics major must have at the top of the circle for the tires of the motorcycle not to lose contact with the sphere, we can analyze the forces acting on the motorcycle at that point.
At the top of the circle, the only two forces acting on the motorcycle are the gravitational force (mg) and the normal force (N) exerted by the sphere. The net force in the vertical direction should be equal to the centripetal force required to keep the motorcycle moving in a circle.
The gravitational force acts downwards, while the normal force acts radially inward. The net force in the vertical direction is given by:
net force = mg - N
The centripetal force necessary to keep the motorcycle moving in a circle with radius R is given by:
centripetal force = mv^2 / R
Setting the net force equal to the centripetal force, we have:
mg - N = mv^2 / R
Solving for N, we get:
N = mg - mv^2 / R
To prevent the tires from losing contact with the sphere, the normal force must be greater than or equal to zero. This occurs when N = 0.
Setting N = 0 in the equation, we have:
0 = mg - mv^2 / R
Solving for v, we get:
v^2 = gR
Taking the square root of both sides, we get:
v = sqrt(gR)
Substituting the given values, we have:
v = sqrt(9.8 m/s^2 * 11.8 m)
v ≈ sqrt(115.24)
v ≈ 10.74 m/s
Therefore, the minimum speed the physics major must have at the top of the circle is approximately 10.74 m/s.
B) At the bottom of the circle, the speed is twice the value calculated in part A. Using the same reasoning as above, we can find the magnitude of the normal force at this point.
The net force in the vertical direction is still given by:
net force = mg - N
However, since the speed is now twice the value in part A, the centripetal force is four times as large. Therefore, the centripetal force required to keep the motorcycle moving in a circle at the bottom is:
centripetal force = 4mv^2 / R
Setting the net force equal to the centripetal force, we have:
mg - N = 4mv^2 / R
Rearranging the equation and solving for N, we get:
N = mg - 4mv^2 / R
Substituting the given values, we have:
N = (66.0 kg * 9.8 m/s^2) - (4 * 66.0 kg * (2 * 10.74 m/s)^2) / 11.8 m
N ≈ 582.97 N
Therefore, the magnitude of the normal force exerted on the motorcycle by the sphere at the bottom of the circle is approximately 582.97 N.
To solve this problem, we can use the concept of centripetal force. In order to keep the tires of the motorcycle in contact with the sphere, there must be a minimum centripetal force equal to the weight of the motorcycle and the rider.
Let's begin by solving part A:
A) 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?
To find the minimum speed required, we need to equate the centripetal force to the weight of the system.
The centripetal force acting on the motorcycle-rider system at the top of the circle is given by:
Fc = mv^2 / r
Where:
- Fc is the centripetal force
- m is the mass of the motorcycle-rider system
- v is the velocity of the motorcycle-rider system
- r is the radius of the circle
At the top of the circle, the net force acting on the system is provided by gravity. Therefore, we can write:
Fc = mg
Setting these two expressions equal to each other, we get:
mg = mv^2 / r
Now we can solve for v:
v^2 = rg
v = sqrt(rg)
Substituting the given values, we have:
v = sqrt(11.8 m * 9.8 m/s^2)
v ≈ 10.9 m/s
Therefore, the minimum speed the physics major must have at the top of the circle is approximately 10.9 m/s.
Now let's move on to part B:
B) 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?
At the bottom of the circle, the net force acting on the motorcycle-rider system is the sum of the gravitational force and the normal force. The gravitational force remains the same as before, but now there is an additional force acting on the system due to the higher speed.
The centripetal force acting on the motorcycle-rider system at the bottom of the circle is given by:
Fc = mv^2 / r
Where the velocity (v) is twice the value calculated in part A.
Now, we can write the net force equation as:
mg + N = mv^2 / r
Rearranging the equation, we can solve for the normal force (N):
N = mv^2 / r - mg
Substituting the values, we have:
N = (40.0 kg + 66.0 kg) * (2 * 10.9 m/s)^2 / 11.8 m - (40.0 kg + 66.0 kg) * 9.8 m/s^2
N ≈ 1694.5 N - 1042.4 N
N ≈ 652.1 N
Therefore, the magnitude of the normal force exerted on the motorcycle by the sphere at the bottom of the circle is approximately 652.1 N.