Download Centripetal Force Practice problems

Centripetal Force Practice problems
1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the
mass is 4.0 m/s and the radius of the circle is 0.75 m, what is the centripetal force and
centripetal acceleration of the mass?
2. A roller coaster has a vertical loop that has a radius of 15 m. How fast must a car be
going if the riders in the car are to feel weightless at the top of the loop?
3. Assuming no energy is lost to friction in the roller coaster depicted in problem 2,
(A) How fast will it be going when it reaches the the bottom of the loop?
(B) What will the apparent weight of a 40 kg person be at the bottom of the loop?
4. A student twirls a 75 gram weight at the end of a string in a horizontal circle of radius
50 cm. If the speed of the weight is 7.0 m/s, what is the tension in the string? Show
Work
5. The student then twirls the same mass in a vertical circle of the same radius at the
same speed.
(a) What is the tension in the string when the weight is at the lowest point of the circle?
(b) What is the tension in the string when the weight is at the highest point on the circle?
6. A roller coaster has a vertical loop that has a diameter of 40 meters. What is the
minimum height that the roller coaster must start from in order to make it through the
loop without losing contact with the track?
7. A 1000 kg car goes around turn in the road that has a radius of curvature of 50 meters.
The car is traveling at a constant speed of 12 m/s.
(A) Is the car accelerating? Why or why not?
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(B) What is the centripetal force required to keep the car from sliding out as it goes
around the turn. State where this force comes from and calculate its magnitude.
(C) What must the minimum coefficient of friction be between the tires of the car and
the road in order for the car to not slide as it goes around the turn?
1. F= 64 N a= 21 m/s^2
2. 12 m/s
3.
1
2
2
2
m
v  12 m
v top  m
gh
21 m s
b. 1568 N

4. 7.4 N
5. 8.1 N at bottom (mg + 7.4) 6.7 N (mg – 7.4)
6. 100 m
7. It is accelerating because the direction of its velocity is changing
(B) 2880 N. It comes from the frictional force between the tires and the pavement
(C) Ff= µmg
µ= 0.29