- #1

Vigorous

- 33

- 3

- Homework Statement
- A roller coaster of mass M is at the top of the Loop-the-loop of radius R at twice the

minimum speed possible. What force does the track exert on it? What force does it

exert when it is at the bottom of the circle? (Use conservation of energy if needed.)

- Relevant Equations
- ac=mv^2/R

1 At the top of the loop-the-loop, the track cannot pull the roller coaster upward so it acts along with the weight of the roller coaster towards the center

N+Mg=Ma=Mv^2/R

N=M(v^2/R-g)=M(4Rg/R-g)=3Mg

N cannot act upwards, therefore vmin=(Rg)^1/2. Since its stated in the problem that the roller coaster is moving at twice the minimum speed, its speed is 2(Rg)^1/2.

At the bottom of the circle

If we also analyze the forces in the radial direction at the bottom of the circle

N-Mg=Mv^2/R=4Mg N=5Mg

My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.

N+Mg=Ma=Mv^2/R

N=M(v^2/R-g)=M(4Rg/R-g)=3Mg

N cannot act upwards, therefore vmin=(Rg)^1/2. Since its stated in the problem that the roller coaster is moving at twice the minimum speed, its speed is 2(Rg)^1/2.

At the bottom of the circle

If we also analyze the forces in the radial direction at the bottom of the circle

N-Mg=Mv^2/R=4Mg N=5Mg

My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.