AP Physics: Calculating Release Height for Loop-the-Loop

[Twigs]

A small block of mass (m) slides down a frictionless loop-the-loop track. The radius at the bottom of the track is R. The question is at what height above the bottom of the loop should the block be released so that it is on the verge of losing contact with the track at the top of the loop?

(Currently studying potential and kinetic energy)

Thanks for the help.

 

[Justin Lazear]

https://www.physicsforums.com/showthread.php?t=27932

--J

 

[wais]

To calculate the release height for the loop-the-loop, we can use the conservation of energy principle. At the bottom of the loop, the block has only kinetic energy, which is given by KE = 1/2 * m * v^2, where m is the mass of the block and v is its velocity. As the block moves up the loop, its kinetic energy is converted into potential energy, given by PE = m * g * h, where g is the acceleration due to gravity and h is the height above the bottom of the loop.

At the top of the loop, the block will have lost all of its kinetic energy and only have potential energy. This means that the potential energy at the top of the loop must be equal to the kinetic energy at the bottom of the loop. So we can set the two equations equal to each other and solve for h:

1/2 * m * v^2 = m * g * h

Solving for h, we get h = v^2 / (2 * g). This means that the release height should be equal to the square of the velocity divided by twice the acceleration due to gravity.

To find the velocity at the bottom of the loop, we can use the conservation of energy again. At the top of the loop, the block has only potential energy, given by PE = m * g * 2R, where R is the radius of the loop. At the bottom of the loop, the block has both kinetic and potential energy, so we can set the two equations equal to each other and solve for v:

1/2 * m * v^2 + m * g * R = m * g * 2R

Solving for v, we get v = √(3 * g * R).

Now that we have the velocity at the bottom of the loop, we can plug it into our equation for h and get the release height:

h = (√(3 * g * R))^2 / (2 * g) = 3R/2

Therefore, the release height for the block to just barely maintain contact with the loop at the top is 3/2 times the radius of the loop. I hope this helps with your understanding of potential and kinetic energy in the context of loop-the-loop tracks. Keep up the good work in AP Physics!