Minimum Height for Loop in Frictionless Incline Track

[mrshappy0]

Homework Statement

A mass is placed at the top of a frictionless incline track. The bottom of the track goes into a loop. At what minimum height does the block with mass m have to be released above the ground in order to reach point b (the top of the loop).

Homework Equations

mg(h_min)=mv_b²/2 +mg2R.

The Attempt at a Solution

The above formula shows that the minimum height has to be 2R because v_b=0. I understand that U_gravity= mgh. But I don't understand why mv^2/2 was added on the right side. I am sitting here study my textbook and can't make sense of it.

 

[Doc Al]

The above formula shows that the minimum height has to be 2R because v_b=0.

How fast must the block be moving at the top of the loop in order to maintain contact? (v_b = 0 won't work.)

 

[mrshappy0]

That is from the answer key of a past exam. Maybe I stated it incorrectly. It seems right. If the block is just barely reaching that point that means that it is also stopping there which would make the velocity zero.

 

[Doc Al]

That is from the answer key of a past exam. Maybe I stated it incorrectly. It seems right. If the block is just barely reaching that point that means that it is also stopping there which would make the velocity zero.

There's a minimum speed required at the top (greater than zero) otherwise the block will leave the track before ever reaching the top.

There's nothing wrong with that formula: it's just energy conservation. But if the answer key says that v_b = 0, that's incorrect.

 

[mrshappy0]

Okay, well more importantly is that equation an example of the Mechanical Energy= Kinetic Energy+ potential energy?

 

[Doc Al]

Okay, well more importantly is that equation an example of the Mechanical Energy= Kinetic Energy+ potential energy?

Sure. But if you actually wanted to solve for the minimum height, you'd need to input a minimum value for the kinetic energy term. (You'd solve for that term by applying Newton's 2nd law at the top of the loop.)