Minimum Speed for Hoop to Roll Without Slipping Around a Loop

[physicsss]

A thin hoop of mass m and radius R is rolling without slipping along a horizontal track. It then encounters a loop of radius r. If the hoop travels fast enough along the horizontal portion of the track it is able to go around the loop without losing contact with it at the top. Find the minimum speed the hoop needs to have on the horizontal portion of the track to accomplish this. Express in terms of m, R, r, and g.

 

[Tide]

What exactly have you tried so far?

Meanwhile, here's a couple of hints:

(1) Energy is conserved.

(2) Centripetal acceleration is speed squared divided by radius of circular path.

 

[wais]

To find the minimum speed needed for the hoop to successfully go around the loop without losing contact, we can use the conservation of energy principle. At the top of the loop, the hoop will have both translational and rotational kinetic energy, as well as potential energy due to its height.

First, we can find the potential energy at the top of the loop by considering the height of the hoop above the ground. Since the hoop is rolling without slipping, the height of the hoop at the top of the loop will be equal to the radius of the loop, r. Therefore, the potential energy at the top of the loop is given by mgh = mgr.

Next, we can find the kinetic energy of the hoop at the top of the loop. Since the hoop is rolling without slipping, its translational kinetic energy is given by 1/2mv^2, where v is the speed of the hoop. Additionally, the hoop will also have rotational kinetic energy given by 1/2Iω^2, where I is the moment of inertia of the hoop (for a thin hoop, I = mr^2) and ω is the angular velocity of the hoop. Since the hoop is rolling without slipping, the linear speed of the hoop is equal to its angular speed, v = ωr. Therefore, the total kinetic energy at the top of the loop is given by 1/2mv^2 + 1/2Iω^2 = 1/2mv^2 + 1/2mr^2ω^2 = 1/2mv^2 + 1/2mr^2(v/r)^2 = 1/2mv^2 + 1/2mv^2 = mv^2.

Using the conservation of energy principle, we can equate the potential energy at the top of the loop to the kinetic energy at the top of the loop, and solve for the minimum speed v needed for the hoop to successfully go around the loop without losing contact:

mgr = mv^2

Solving for v, we get:

v = √(gr)

Therefore, the minimum speed the hoop needs to have on the horizontal portion of the track to successfully go around the loop without losing contact is √(gr). This speed is independent of the mass and radius of the hoop, and only depends on the radius of the loop and the acceleration due to gravity, g.