Rolling 3 objects on an inclined plane

[EnricoHendro]

Homework Statement

Three objects of uniform density—a solid sphere, a solid cylinder, and a hollow cylinder—are placed at the top of an incline. They are all released from rest at the same elevation and roll without slipping. (a) Which object reaches the bottom first? (b) Which reaches it last?

Relevant Equations

Vcm=\sqrt{\frac{2gh}{(M+\frac{I}{MR^{2}})}}

Hello there, I have a question regarding this problem. I have no problem with part A. However, in part B, my solution manual states that the hollow cylinder will reach the bottom last. Why is it? I mean shouldn't the solid cylinder and the hollow one reach the bottom at the same time? you know since they have basically the same moment of inertia, (for the solid one I=\frac{1}{2}MR^{2} and for the hollow one I=\frac{1}{2}M(R1^{2}+R2^{2})).
If the hollow cylinder is the thin loop one, then I understand why it is the last one to reach the bottom. But if we assume that it is a normal hollow cylinder, than shouldn't both solid and hollow cylinders arrive at the same time?

 

[BvU]

you know since they have basically the same moment of inertia

Think again ...

 

[italicus]

Hi BvU

what do you mean by “normal hollow cylinder” ? IF the two cylinders, solid and hollow, have the same mass, and the same external radius, the hollow one has its mass distributed differently, its inertia radius is greater. In the limit of a thin hollow cylinder, the moment of inertia is mr^2 .

 

[kuruman]

Hi BvU

what do you mean by “normal hollow cylinder” ? IF the two cylinders, solid and hollow, have the same mass, and the same external radius, the hollow one has its mass distributed differently, its inertia radius is greater. In the limit of a thin hollow cylinder, the moment of inertia is mr^2 .

Nothing is said about the masses and external radii being equal. The answer to this problem is independent of the radii and masses of the rolling objects.

 

[italicus]

Sorry BvU
The answer was for Enrico...I have to do practice

 

[EnricoHendro]

Hi BvU

what do you mean by “normal hollow cylinder” ? IF the two cylinders, solid and hollow, have the same mass, and the same external radius, the hollow one has its mass distributed differently, its inertia radius is greater. In the limit of a thin hollow cylinder, the moment of inertia is mr^2 .

by normal hollow cylinder I mean the hollow cylinder that is not a thin hollow cylinder. In my textbook, it is said that the hollow cylinder (not the thin one) has a moment of inertia of 0.5M(R1^2+R2^2). Now, the constant in front of the mass times radius is 0.5 (which is the same as the solid cylinder). So that's why I got confused. Because the solution is independent of mass and radii. From part A, because the moment of inertia of the sphere is smaller (2/5 < 1/2), therefore the sphere arrives first.

 

[EnricoHendro]

Think again ...

wait, I think I misread the moment of inertia of the hollow cylinder. the R2 seems to refer to the radius from the axis of rotation to the outer shell of the cylinder. I read it as the radius from the inner shell to the outer shell. So I ended up dividing the R1^2+R2^2 with R1^2+R2^2. I should have divided the R1^2+R2^2 by only R2^2. So yeah, the hollow cylinder should arrive last. My bad.

 

[BvU]

@italicus :

It is good to divide out the mass and come to a radius of inertia:
solid cylinder $\qquad {1\over 2}\sqrt 2\ r$
thin cylinder $\qquad \ \ \ \ r$

 

[italicus]

@italicus :

It is good to divide out the mass and come to a radius of inertia:
solid cylinder $\qquad {1\over 2}\sqrt 2\ r$
thin cylinder $\qquad \ \ \ \ r$

Yes , BvU

I know the concept of radius of inertia. The acceleration of a body rolling down an inclined plane without slipping is given by (sorry but I don’t know latex, I’m learning it, but it’s not easy to me) :

a = gsinα/ (1 + I_c/(mr^2)= gsinα/(1+ (ρ/r)^2 )

so we have :

hollow thin cylinder : ρ = r ⇒ a = ½g*sinα

solid cylinder $\rho = {1\over 2}\sqrt 2\ r$ ⇒ a = 2/3 * g sinα

and so on for other bodies that can roll. The motion down the plane, inclined of α , is uniformly accelerated, the friction force is static and doesn’t make work ( no relative motion between plane and body in the instantaneous center of rotation), so one can apply formulae for rectilinear u.a. motion, and find speed and time to reach the end of the plane. The same can be done by means of conservation of energy, the gravitational field is conservative.
Thank you .