Ball rolling without slipping to reach the top of a loop

[mprm86]

I have attached an image of the problem. Which should be the height h (expressed in terms of R) if the ball needs to reach the point Q. The ball will roll without slipping.
In the image, Q is the highest point in the loop. R is the ratio of the loop (think of it as a circle). H is a point on the straight secton of the track, and it is measured respect to the ground.

 

[Chi Meson]

I'll give you a push but you have to do the work yourself.

Consider the minimum kinetic energy the ball must have to just complete this circular loop without falling. This would be the kinetic energy that corresponds to the tangential speed where the normal force from the track just equals zero.

Does this ring any bells?

 

[thepatientmental]

To find the height h in terms of R, we can use the conservation of energy principle. Since the ball is rolling without slipping, the total energy at the highest point Q will be equal to the total energy at point H.

At point Q, the total energy is only potential energy given by mgh, where m is the mass of the ball and g is the acceleration due to gravity.

At point H, the total energy is a combination of potential and kinetic energy. The potential energy is still mgh, but the kinetic energy is given by 1/2mv^2, where v is the velocity of the ball at point H.

Since the ball is rolling without slipping, the velocity at point H can be expressed as v = ωR, where ω is the angular velocity of the ball and R is the radius of the loop.

Now, equating the total energy at points Q and H, we get mgh = mgh + 1/2m(ωR)^2. Simplifying this equation, we get h = R(ω^2/2g).

Since R is the ratio of the loop, we can express it as R = 2πr, where r is the radius of the loop.

Substituting this into our equation for h, we get h = 2πr(ω^2/2g).

Now, we can express ω in terms of the time it takes for the ball to complete one revolution, T, and the radius of the loop, r. We know that ω = 2π/T, so substituting this into our equation for h, we get h = 2πr(2π/T)^2/2g.

Simplifying further, we get h = (4π^2r/T^2)/g.

Therefore, the height h in terms of R is given by the equation h = (4π^2r/T^2)/g, where R = 2πr and T is the time for one revolution.

I hope this helps you solve the problem and find the height h needed for the ball to reach point Q. Remember to always check your units and don't forget to include the acceleration due to gravity, g, in your calculations. Good luck!