Loop the loop. (PHYSICS)?

Question

A cart slides down a frictionless inclined track to a circular loop of radius R = 13 m. In order for the cart to negotiate the loop safely, the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cart's weight. (Note: This is different from the conditions needed to just negotiate the loop.)

a) What must be the minimum speed |vmin| of the cart at the top of the loop?

b) How high h above the top of the loop must the cart be released?

c) When the car is descending vertically in the loop (point (c) in the picture), what is its speed |v|?

d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 20 m. What retarding acceleration |a| is required?

PICTURE FOR PART C
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1201/fall/homework/Ch-07-Work-Eng/loop_the_loop2/2-alt.gif

Answer 1

> the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cart's weight.

That's a clue. At the top of the loop, there will be two forces acting on the cart:
1) cart's weight (mg)
2) normal force of track against cart (must be >= mg)

Since these two forces both push in the same direction (downward), they add up. So:

F_net = weight + normal_force
= mg + mg
= 2mg

That means the cart's acceleration at the top of the loop is:

a = F_net / m
a = (2mg) / m
a = 2g

Now use the formula for centripetal acceleration: "a = v²/r". From this, we can substitute for "a" in the previous equation, to get:

v²/r = 2g

or:

v = sqrt(2gr)

That answers part (a).

For part (b), use conservation of energy. What is the cart's KE + PE when is is at the top of the loop? (Okay, I'll tell you: it's mv²/2 + mg(2r).). You need to start the cart at a high enough spot so that its PE matches that amount.

Part (c): Sorry, I can't see the picture.

Part (d): Figure out the cart's speed at the bottom of the track (hint: use conservation of energy, combined with the answer you figured out in (b)). Then, use d = Δv²/(2a)

Answer 2

An object in motion will move in a straight line at the same speed unless a force acts on it to change its direction or speed. As an example, suppose I kick a soccer ball so that it is going to roll past you very quickly. As it passes you, if you tap it slightly at a right angle to its path, it will turn a little bit in the way you tap it. Similarly, a third person could stand by that path, tap it a little and cause it to turn a little more. If we got enough people, we could all keep tapping it so as to make it go in a circle. As long as each person kept tapping it towards the middle, it would continue going in a circle. But if one person let it go, it would just keep going in a straight line. For a roller coaster, the track is acting like all those soccer players lined up in a circle. It is constantly forcing the cart to go in a circle, even though it would go in a straight line if the track were not there. The force that the track applies to the cart is called the "normal" force because normal means perpendicular, and the force should be perpendicular to the path in order to keep the cart turning as it goes around the circle. Now you may wonder what role does gravity play? It is important in how the track is designed. If you get the cart going very fast, and then angled it upward on a ramp, it would rise to a certain height, but finally the constant pull of gravity would slow it down, and it would start to fall back down the ramp. The top of the loop is designed to be much lower than the height the cart would fly up in the air if there were no track.. So the track is actually having to push against the cart to make it turn in a tighter loop than it would naturally fly if there weren't a track. For most roller coasters, you will be towed up a large hill at the beginning, and then you start coasting. The top of the loop is lower than the height of the first large hill, and this is done to make sure that the cart has enough speed at the bottom of the loop to get through the loop. Hope this helps!

Answer 3

An item at height h has energy mgh where m is the mass and g is 9.81m/s².

The weight of the cart is mg.

The force exerted on the cart by the circular track is mv²/r where v is the speed along the track.

Since the track is frictionless, there is no energy lost, so kinetic energy must equal gravitational potential energy.

Kinetic energy is mv²/2.

That should be enough to solve it.