Kinetic Energy of two Rolling Objects on a Slope

[interference]

Hi everyone, not a homework problem, just something I was thinking about when I found in a book.

Homework Statement

A hollow Cylinder is rolling down an incline of angle theta. Inside the cylinder is a smaller solid cylinder rolling freely inside it. (Refer to diagram)

Given an x coordinate from the top of the slope which the cylinders were released, and:

Large Cylinder has mass M, moment of inertia I, radius R
Small cylinder has mass m, moment of inertia i, radius r
(p.s. I labeled the small mass m wongly in the diagram!)

The small cylinder is allowed to roll freely in the large cylinder, and we call the angle it makes from vertical phi.

What is the Total Kinetic Energy of the system?

Homework Equations

v=rw
K=(1/2)Iw^2 > where w is angular velocity

The Attempt at a Solution

The Kinetic Energy for the Large Cylinder seems straightforward:

$$K=\frac{1}{2}\left(M\dot{x}^2+\frac{I\dot{x}^2}{R^2}\right)$$
which is the translational KE and the rotational KE.

However, I am not sure about about the smaller cylinder. I believe there are 3 parts to its KE, namely the translational KE, the KE associated with it doing a circular motion on the rim of the large cylinder (I use the parallel axis theorem), and its own rotational KE.

$$K=\frac{1}{2}\left(m\dot{x}^2+\left(i+m\left(R-r\right)^2\right)\dot{\phi}^2+\frac{i\dot{x}^2}{r^2}\right)$$

However, I am unsure about this because the small cylinder is also rotating somewhat around the large cylinder which is itself rolling down the slope.

Any advice or help is appreciated! Thank you!

 

[graphene]

the last term should be multiplied my R^2 / r^2.

we assume there is no slip at the contact between the 2 cyls. so the (translational) velocities of the points of contact between the 2 cyl is the same.

so R*w1 = r*w2

w2 = (R/r)*w1

this brings in the extra factor

the mistake you made was in assuming the same w for both the cylinders

 

[interference]

the last term should be multiplied my R^2 / r^2.

we assume there is no slip at the contact between the 2 cyls. so the (translational) velocities of the points of contact between the 2 cyl is the same.

so R*w1 = r*w2

w2 = (R/r)*w1

this brings in the extra factor

the mistake you made was in assuming the same w for both the cylinders

Hi,

I did know that the angular velocities for each cylinders are different

$$v=R\omega_{M}=r\omega_{m}$$, thus

$$\omega_{m}=\frac{\dot{x}}{rR}R=\frac{\dot{x}}{r}$$

Which I have taken into account already in the last term which is the small cylinder's own rotational kinetic energy about its own axis.

Also, I found a solution which adds another term:

$$m\dot{x}\dot{\phi}\left(R-r\right)\left(\frac{i}{mr^2}+cos\left(\phi+\theta\right)\right)$$

But I do not understand where that comes from. Can anyone please help shed some light?