How High Must a Cylinder Roll to Loop-the-Loop?

[GemmaN]

In this problem you will consider the motion of a cylinder of radius r_1 that is rolled from a certain height h so that it "loops the loop," that is, rolls around the track with a loop of radius r_2. The cylinder rolls without slipping.

It looks like your standard matchbox car loop.

I need to find the minimum height h that will allow a solid cylinder of mass m and radius r_1 to loop the loop of radius r_2. "Express h in terms of the radius r_2 of the loop."

I am not quite getting a certain portion of this. I know the important part of this problem is when the cylinder is at the top of the loop.
Now, at this point, I know it has a potential energy of 2*r_2*m*g
To stay on the track without slipping, I think I need v = R*omega
At the top part of the track, the cylinder is upside down, and has no normal force... so weight matters?

I may need to use this: K = 1/2Mv_cm^2 + 1/2 I_cm*omega^2

mgh = (2)r_2(mg) + KE? + ?

I am a bit confused at this point. How am I suppose to put this together?

 

[physics girl phd]

You are correct in that the "important part of this problem is when the cylinder is at the top of the loop".

Hint -- draw a free body diagram for the system at that location.

 

[kyle8921]

I can help you with understanding the dynamics of rotational motion in this problem. First, we need to understand that rotational motion involves both linear and angular motion. In this case, the cylinder is rolling without slipping, which means that its linear velocity is equal to the product of its angular velocity and the radius of the cylinder.

To find the minimum height h, we need to consider the energy conservation principle. At the top of the loop, the cylinder has both potential and kinetic energy. The potential energy is due to its height and the kinetic energy is due to its motion. We can express this as:

mgh = (1/2)mv^2 + (1/2)Iω^2

Where m is the mass of the cylinder, g is the acceleration due to gravity, v is the linear velocity, I is the moment of inertia of the cylinder, and ω is the angular velocity.

At the top of the loop, the cylinder is upside down and the normal force is zero, so the only force acting on it is its weight. This weight is balanced by the centripetal force, which is provided by the normal force at the bottom of the loop. This means that the weight of the cylinder is equal to the centripetal force:

mg = mω^2r_2

Substituting this into our energy conservation equation, we get:

mgh = (1/2)mv^2 + (1/2)(mr_1^2 + 1/2mr_1^2)ω^2

Solving for h, we get:

h = (5/2)r_2 + (1/2)r_1

This is the minimum height required for the cylinder to loop the loop of radius r_2. I hope this helps clarify the problem for you. If you have any further questions, please don't hesitate to ask.