# Whats normal force at top/bottom of loop

• bfusco
In summary, the small mass m slides without friction along the looped track shown in the figure. At the top of the loop, the normal force exerted by the track is mg. At the bottom of the loop, the normal force is mg.(6h+1/r).
bfusco

## Homework Statement

The small mass m sliding without friction along the looped track shown in the figure is to remain on the track at all times, even at the very top of the loop of radius r.

http://session.masteringphysics.com/problemAsset/1057653/6/GIANCOLI.ch08.p090.jpg

A)In terms of the given quantities, determine the minimum release height h.
B)If the actual release height is 3h, calculate the normal force exerted by the track at the bottom of the loop.
C)f the actual release height is 3h, calculate the normal force exerted by the track at the top of the loop.

## The Attempt at a Solution

A)i was easily able to solve part A, i got 2.5r
B)for this i started with energy conservation. i have that the potential energy at the beginning is equal to the kinetic energy at the bottom (3mgh=1/2mv^2, the potential energy being 3mgh because the starting height is 3h). i calculated the velocity to b v^2=6gh, which i then plugged into my centripetal force equation (ƩF=mv^2/r). N-mg=6mgh+mg, solving for N i get N=mg(6h+1/r), which is incorrect, I am instructed that the answer does not depend on height.

welcome to pf!

hi bfusco! welcome to pf!
bfusco said:
i get N=mg(6h+1/r), which is incorrect, I am instructed that the answer does not depend on height.

how did that h survive?

i thought you found what h was?​

tiny-tim said:
hi bfusco! welcome to pf!

how did that h survive?

i thought you found what h was?​

thank you tiny tim,

hm...you make a good point i guess that's what happens when you stare at a board for hours, you miss the little things. lol I am going to give that a try.

hahahahaha it worked...i feel stupid right now. thank you for your help tiny tim

I would first commend the student for their attempt at solving the problem and using energy conservation and the centripetal force equation to find the normal force at the bottom of the loop. However, I would point out that there are a few errors in their calculations that may have led to the incorrect answer.

First, in the energy conservation equation, the potential energy should be 3mgr (instead of 3mgh) since the height is measured from the bottom of the loop, which is at a distance of r from the center. This would lead to a different value for the velocity and subsequently the normal force.

Secondly, in the centripetal force equation, the student has added the weight of the object (mg) to the centripetal force (mv^2/r), which is incorrect. The centripetal force should only include the normal force (N) and the weight of the object (mg).

Finally, I would also point out that the normal force at the bottom of the loop should be equal to the weight of the object (mg) in order for it to remain on the track without slipping. Therefore, the correct equation would be N=mg.

For part C, I would explain that at the top of the loop, the normal force should be equal to the weight of the object (mg) plus the centripetal force (mv^2/r). This is because at the top of the loop, the centripetal force is acting downwards, while the weight of the object is acting downwards as well. Therefore, the normal force would be N=mg+mv^2/r.

## What is normal force at the top of a loop?

The normal force at the top of a loop is the force that a surface exerts on an object in contact with it. In this case, it is the force that the track exerts on the object as it goes through the loop.

## What is normal force at the bottom of a loop?

The normal force at the bottom of a loop is the force that a surface exerts on an object in contact with it. In this case, it is the force that the track exerts on the object as it goes through the loop. However, at the bottom of the loop, this force is directed upwards to counteract the force of gravity.

## How is normal force related to centripetal force?

The normal force and the centripetal force are both necessary for an object to maintain circular motion in a loop. The normal force provides the necessary upward force to counteract the force of gravity and keep the object moving in a circular path, while the centripetal force pulls the object towards the center of the loop.

## What factors affect the normal force at the top and bottom of a loop?

The normal force at the top and bottom of a loop is affected by the speed and mass of the object, the radius of the loop, and the angle of the loop. These factors determine the amount of centripetal force needed to keep the object moving in a circular path, which in turn affects the normal force.

## How can the normal force be calculated at the top and bottom of a loop?

The normal force can be calculated using the equation FN = mv2/r, where FN is the normal force, m is the mass of the object, v is the speed of the object, and r is the radius of the loop. At the top of the loop, the normal force is equal to the weight of the object, while at the bottom of the loop, it is equal to the weight plus the centripetal force.

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