What is the minimum height for an object to complete a loop without friction?

In summary, the conversation discusses the minimum height from which a small mass must be released in order to complete a loop without friction, as well as the height at which it will lose contact with the track. Using the equations for centripetal acceleration, kinetic energy, and gravitational potential energy, the minimum height is found to be 2.5 times the radius of the loop, and the height at which the object loses contact with the track is 5/3 times the radius. The conversation also addresses the assumption that the velocity of the mass is zero when it leaves the track, showing that it is not the case and providing a correct derivation.
  • #1
Ishaan S
19
0
1. Homework Statement
A small mass m slides without friction along the looped apparatus shown in the attached file. (a) If the object is to remain on the track, even at the top of the circle(whose radius is r), from what minimum height hm must it be released(from rest)? (b) If it is released at a height h = 0.80hm, at what height will it lose contact with the track?

2. Homework Equations
ac = v2/r

KE = .5mv2

PEgrav = mgh
3. The Attempt at a Solution
For a:

I knew that for the object to complete the loop, it must be able to complete the extreme case: the top. I knew that the centripetal acceleration must be g. Therefore, from the centripetal acceleration formula, the speed had to be at least v = √(rg).

From the law of the conservation of energy, I equated the initial and final(at the top of the loop) energies:

mghm = 2rmg + .5mv2

I canceled out the m and substituted v from the centripetal acceleration equation and got

ghm = 2rg + .5rg = 2.5rg

Cancel out g

hm = 2.5r

For b:

The initial height .80hm = 2r.

Using the law of the conservation of energy

2rmg = mgh + .5m(0)2.

I substituted 0 for v in the equation above because I thought that the velocity had to be 0 when the object left the track.

Solving for h, I got h = 2r.

I have no idea if this is right. Could someone please tell me if this is right, and if not, give me a nudge in the right direction.

Thanks.
 

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  • #2
Hello, Ishaan.

Your work for part (a) looks very good!

For part (b) your assumption that v = 0 when the mass leaves the track is not correct. Hint: What force on the mass must go to zero at the instant the mass leaves the track?
 
  • #3
Ok.

So I set up an equation.

Fnormal + mgcosθ = mv2/r

where θ is the angle between the normal force and the gravitational force.

I set Fnormal equal to zero because when the object leaves the track, the normal force would be 0.

mgcosθ = mv2/r

I solved for v2.

v2 = rgcosθ

I then set up the conservation of energy equation.

2rmg = mgh + mv2/2.

I substituted rgcosθ for v2 due to my previous derivation. I then canceled mg from both sides.

2r = h + r(cosθ)/2,

Then I set up a diagram that I have attached.

I solved for cosθ using the diagram and substituted it into the above equation.

2r = h + (h-r)/2.

I solved for h and got

h = (5/3)r.

Is this correct?

Thanks.
 

Attachments

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  • #4
That all looks correct. Very nice work!
 

1. What is the physics behind an object sliding through a loop?

The physics behind an object sliding through a loop involves the principles of inertia, centripetal force, and gravity. When the object is released at the top of the loop, it has a certain amount of potential energy. As it slides down, this potential energy is converted into kinetic energy, causing the object to accelerate. In order to stay in the loop, the object must also experience a centripetal force, which is provided by the curvature of the loop. Gravity also plays a role in keeping the object moving in a circular path.

2. How does the shape and size of the loop affect the object's speed and trajectory?

The shape and size of the loop can greatly affect the object's speed and trajectory. A larger loop will require the object to travel a longer distance, resulting in a lower speed. A smaller loop, on the other hand, will allow the object to travel a shorter distance and maintain a higher speed. The shape of the loop can also impact the object's trajectory. A circular loop will result in a smoother and more predictable path, while a non-circular loop may cause the object to deviate from its expected trajectory.

3. What factors determine if an object can successfully complete a loop?

The two main factors that determine if an object can successfully complete a loop are its speed and the size of the loop. If the object does not have enough speed, it will not have enough kinetic energy to maintain its circular path. Additionally, if the loop is too small or the object is too large, the centripetal force may not be strong enough to keep the object in the loop. The object's shape and mass distribution may also play a role in its ability to complete a loop.

4. How does friction affect an object sliding through a loop?

Friction can have a significant impact on an object sliding through a loop. In order for the object to maintain its circular path, the centripetal force must be greater than the force of friction acting on the object. If there is too much friction, the object may slow down or even come to a stop before completing the loop. On the other hand, if there is very little friction, the object may slide too quickly and fly off the loop.

5. Can an object slide through a loop without any external force acting on it?

No, an object cannot slide through a loop without any external force acting on it. In order for the object to move, there must be a force acting upon it, such as gravity or a push from the person releasing the object. Once the object is in motion, it will continue to move until another external force, such as friction or air resistance, slows it down or stops it.

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