Homework Help: Rotational and translational motion hybrid quetion. How to digest it?

1. Sep 17, 2011

vkash

1. The problem statement, all variables and given/known data

If two circular discs A and B of mass m and 3m and radii R,2R, respectively. are placed from the top of a rough inclined plane, which disc will reach bottom first.

2. The attempt at a solution
NOTE: u is coefficient of friction (both kinetic and static). g is acceleration due to gravity & t is inclination of inclined plane.
friction on disc of radii R is u*g*cos(t) and that in 2R is 3g*u*cos(t). In starting there will not pure rolling but after some time there will pure rolling(may not if inclined is too small). In any case friction force will remain as it in starting(u*g*cos(t) and u*g*cos(t)). so linear acceleration of both the objects will different so their time to reach bottom should different(Heavier should reach late on bottom larger retardation). BUT MY EXPLANATION IS WRONG WHERE SM I WRONG. that's my question to you.

Translational and rotational motion are confusing my mind. Do you have any way to understand rotational in better way.

Last edited: Sep 17, 2011
2. Sep 17, 2011

vela

Staff Emeritus
What does that mean?
That's wrong.
Try applying Newton's laws to the problem to analyze what will happen.

3. Sep 17, 2011

vkash

I have reached to a solution that is. acceleration of both the object does not depend on the mass or radius(seems amazing). The acceleration comes out to be 2*u*g*cos(t). here t is angle of inclination of inclined plane, u is coefficient of friction and well known g is acceleration due to gravity in free fall with respect to inertial frame.

Am i correct?

4. Sep 17, 2011

vela

Staff Emeritus
No. For example, if the surface was horizontal (t=0), there'd be no acceleration, but your result says there would be.

5. Sep 18, 2011

vkash

I think it's answer is in the way i reach to this result. I do this in this way.
watching through the frame in center of the ring/disc/sphere.
T(torque)=f(friction)*r(radius of the object)=I(moment of inertia around COM))*a(angular acceleration disc/ring etc) ------------------------(1)
any object we take it will have a radius of gyration around that point. let me take moment of inertia around the center of mas I=mk2.
so last equation (1) changed to
$a=\frac{f}{mk}$
k has always a relation with r in the terms of natural number so let r=sk (for ring it is 21/2)
$acceleration =\frac{s*friction}{mass}$
for given condition friction /mass = constant (if inclination and coefficient are same)

you say if inclination is zero. in that line of action of friction and gravity are passing through center so no torque no rotation.
I can realize that i represent it in very complex way. Even i am not sure that when i will see it next time will i able to understand it or not.
have i applied concepts in right way or not no matter if i have done little mistake in calculation(that can be cured easily). Answer of the question in my book that all will reach simultaneously and my result agrees with that(it's amazing).
If i am wrong then point out me there.
In last post i do this for disc.
After all i have got answer. I want to say thanks to you vala and DOc Al. Keep helping guys.

Last edited: Sep 18, 2011