Rolling Object on a Moving Ramp

Question

A solid cylinder of mass m = 2 kg and radius R = 0.1 m is placed at the top of a frictionless ramp of height h = 1.5 m. The ramp itself has a mass M = 5 kg and rests on a frictionless horizontal surface, so it is free to move. The cylinder rolls without slipping down the ramp.

Questions:

(a) What is the speed of the cylinder relative to the ramp when it reaches the bottom?

(b) What is the final speed of the ramp itself (with respect to the ground)?

(c) How much of the cylinder’s potential energy is converted into translational and rotational kinetic energy?

(d) Is mechanical energy conserved in this system? Explain.

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Answer 1

I have to make two assumptions.

Assumption 1:

The bottom of the ramp (touching the ground) is frictionless,

but the surface on which the cylinder is rolling is not.

(Otherwise the cylinder would be slipping.)

Assumption 2:

The problem statement does not mention the shape of the platform.

And indeed the result would be independent of the shape if the platform itself could not move.

Otherwise the result does depend on the angle which the bottom of the ramp forms with the ground;

let this angle be θ (not given).

Let

g = 9.81 m/s² be gravitational acceleration,

I = mR²/2 be the moment of inertia of the disk,

v (to be computed) be the speed (relative to ground) of the cylinder at the bottom of the ramp,

w (to be computed) be the speed (relative to ground) of the ramp,

ω = v/R be the angular velocity of the cylinder.

(d)

The answer the question (d) is "yes".

The reason is the absence of kinetic friction.

While we do need to assume static friction (to prevent the cylinder from slipping),

static friction does not dissipate mechanical energy.

(c)

For the same reason, ALL of the cylinder's potential energy is converted into translational and rotational kinetic energy.

(b)

As a result we can use conservation of mechanical energy.

The potential energy (mgh) at the top gets converted

into translational energy of the cylinder (mv²/2),

into rotational energy of the cylinder (Iω²/2), and

into translational energy of the ramp (Mw²/2).

That is

mgh = mv²/2 + Iω²/2 + Mw²/2 (1)

After substituting I and ω, and simplifying

mgh = 3mv²/4 + Mw²/2 (2)

We can also use conservation of momentum in the horizontal dimension,

because there are no horizontal forces acting on the system.

The horizontal component of v is vcos(θ).

So we can write conservation of momentum as

Mw = mvcos(θ)

or

w = mvcos(θ)/M (3)

Substituting equation (3) into equation (2) you can compute v, and then w.

(Both will depend on the angle θ, which is not given to you.)

(a)

The velocity of the cylinder with respect to the platform is the difference between the velocities of the two.

But that is a VECTOR difference.

We can perform the vector difference by computing the difference in both the horizontal and vertical direction.

In the vertical direction the platform has no velocity,

Therefore the cylinder has relative velocity of vsin(θ).

In the horizontal direction we have the platform moving with speed w,

and the cylinder moving with speed vcos(θ).

Their directions are opposite, therefore the relative velocity of the cylinder is

vcos(θ) + w = vcos(θ) + mvcos(θ)/M = vcos(θ)(1 + m/M).

You are supposed to compute the relative speed, which is the magnitude of relative velocity:

√((vsin(θ))² + (vcos(θ)(1 + m/M))²) = v√(1 + (mcos(θ)/M))²)