Loop the Loop

Solutions Below:

A roller-coaster car may be represented by a block of mass 50.0 kg. The car is released from rest at a height h = 45.0 m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 15.0 m at ground level, as shown. As you will learn in the course of this problem, the initial height 45.0 m is great enough so that the car never loses contact with the track.

Part A

Express the kinetic energy numerically, in joules.

Since the radius of the circle is 15.0m, the diameter is 30.0m. Therefore, at the top of the loop, the height of the car will be 30.0m.

Solve for KE by first finding potential energy at the beginning of the track:

U = mgh

U = 50.0(9.8)(45.0)

U = 22050J

Since there is no friction, there will be a free exchange of potential to/from kinetic energy. So if we find the potential energy at the top of the loop and subtract that from our beginning potential energy, we’ll have the kinetic energy:

U_{top} = mgh

U_{top} = 50.0(9.8)(30.0)

U_{top} = 14700J

KE = 22050 – 14700

KE = 7350J

Part B

_{min}at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.

This is a centripetal force question. Solve for the velocity at the top of the loop using the formula:

a_{c} = V^{2}/r

At the top of the loop and minimum velocity, the centripetal force equals the force of gravity (and centripetal acceleration equals acceleration due to gravity). So:

9.8 = V^{2}/r

9.8 = V^{2}/15

147 = V^{2}

V = 12.12m/s

Now solve for KE at the top of the loop:

KE = 1/2mv^{2}

KE = 1/2(50.0)(12.12)^{2}

KE = 3675J

Next, solve for potential energy at the top of the loop:

U = mgh

U = 50.0(9.8)(30.0)

U = 14700J

Add the two energies:

3675 + 14700 = 18375

And now solve for height using the above as beginning potential energy:

U = mgh

18375 = 50(9.8)(h)

h = 37.5m

As a shortcut, the minimum height to make it through a loop in this type of problem is 2.5 times the radius of the loop. That doesn’t consider friction, rotational vs. translational kinetic energy, etc., but for simple problems like this, it works.