TBR Physics Work and Energy Review Passage 2

[heylollipop]

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This is the passage on an engineer who designs a new ride for a carnival.

#9. If the cart is barely able to complete the loop, what is speed at the point P, the top of the loop?
- Why is the normal force zero? When can you make the assumption that normal force is zero?
- The solution states that "at the top of the loop both the normal force and the force of gravity point down and the centripetal force must offset the net downward force in order to keep the cart on the tracks." I thought that the centripetal force would point downward/towards the center of the loop. But if all three forces point down, then the cart won't be able to stay on track... I'm a little confused.

#13. Which of the following pictures best represents the direction of the net force F acting on the cart when it is half way up the loop?
- How come when the cart is half way up the loop, the centripetal force is NOT being considered, like in #9?

Thanks in advance!

 

[milski]

I don't have the books, so I cannot comment on everything, only on the second part of #9.

When the cart is at they top, the net force is pointed down. If there was no net force, it would continue straight, parallel to the ground. Since the net force is pointed down, its velocity will turn slightly downwards, following the track.

The sentence about the centripetal force offsetting the sum of the other two is non-sense. The centripetal force is an abstract force, which makes the body follow a circular path. At the top of the loop, the centripetal force is the sum of the normal and the gravity acting on the car. As another example, the centripetal force at the bottom is pointed up and has a magnitude equal to the magnitude of the normal minus the magnitude of gravity.

This about it this way: For a body to follow a circular path at constant speed, at any point of that path, the net force is pointed towards the center of that circle and has constant magnitude. For the cart at the top, gravity and the normal are pointed in the same direction, so only a little contribution from the normal is needed. For the cart at the bottom, gravity and the normal are pointed in opposite directions, so the normal has to have much larger magnitude.

Now, the least net force that you can have at the top is just gravity, since the normal cannot be pointed up. That means that the centripetal force has a magnitude of m.g. To have it stay constant would mean that at the bottom the normal has to be 2.m.g since the magnitude of the centripetal force there would be 2.m.g-m.g=m.g.

 

[heylollipop]

I don't have the books, so I cannot comment on everything, only on the second part of #9.

When the cart is at they top, the net force is pointed down. If there was no net force, it would continue straight, parallel to the ground. Since the net force is pointed down, its velocity will turn slightly downwards, following the track.

The sentence about the centripetal force offsetting the sum of the other two is non-sense. The centripetal force is an abstract force, which makes the body follow a circular path. At the top of the loop, the centripetal force is the sum of the normal and the gravity acting on the car. As another example, the centripetal force at the bottom is pointed up and has a magnitude equal to the magnitude of the normal minus the magnitude of gravity.

This about it this way: For a body to follow a circular path at constant speed, at any point of that path, the net force is pointed towards the center of that circle and has constant magnitude. For the cart at the top, gravity and the normal are pointed in the same direction, so only a little contribution from the normal is needed. For the cart at the bottom, gravity and the normal are pointed in opposite directions, so the normal has to have much larger magnitude.

Now, the least net force that you can have at the top is just gravity, since the normal cannot be pointed up. That means that the centripetal force has a magnitude of m.g. To have it stay constant would mean that at the bottom the normal has to be 2.m.g since the magnitude of the centripetal force there would be 2.m.g-m.g=m.g.

That makes a lot of sense. I think I understood the other question now, applying the same concept. Thank you!