Rotation Question. Is there sufficient data/values? Pondered on question 4 sumtime..

brainracked

1. The problem statement, all variables and given/known data
Okay..here's d question:
A sphere, a cylinder and a hoop start from rest and roll down the same incline.
Determine which body reaches the bottom first.

2. Relevant equations
Sphere: I=2/5 mr²
Cylinder: I=1/2 mr²
Hoop: I= mr²
F=ma

3. The attempt at a solution
To find out which of the three reach first I suspect that their acceleration should b found 1st..but..how do I go about it? I also feel as if there is not much data to work on..is tis question complete?
If it is solvable..a lil clue or idea would b much appreciated... =)

tiny-tim

Hi brainracked!

Hint: "roll" means without slipping …

so the instantaneous velocity of the point of contact is zero …

so the speed of the centre of mass must always be cancelled by the rotational speed of the rim (relative to the centre of mass).

That gives you a relationship between velocity and angular velocity

brainracked

Hiyya tiny-tim..we meet again..haha..
So sorry..but..i dont quite understand..the instantaneous velocity of point of contact?
Relationship?
w=v^2/r?

So sorry for the trouble and thank you so much for ur time and help..

tiny-tim

Hi brainracked!

The point of contact of a wheel with the ground is always stationary, just for an instant.

If it wasn't, the wheel would skid.

So if the centre is moving horizontally at speed v, and the wheel is rotating with angular speed ω, the point at the top of the rim has speed v + rω, and the point at the bottom of the rim (the point of contact) has speed v - rω.

So rolling requires v - rω = 0. (that is the "relationship")

This is geometry, not physics.

Now that the geometry has told you the relationship between v and ω, plug that into KE + PE = constant to get the acceleration.

brainracked

Haha..okay..i get it..thanks!

brainracked

Wait wait wait..I ended having two variables..h and v

PE + KE
= mgh +1/2mv^2 + 1/5 mv^2
=gh + 7/10v^2

I dont see how tis can solve the answer..

brainracked

*solve the question

tiny-tim

v = (dh/dt)/sinθ

brainracked

and to get the acceleration?
I'm lost..

LowlyPion

[tex]\Delta PE = \Delta KE = mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2} = \frac{mv^2}{2} + \frac{I}{2}*\frac{v^2}{r^2}[/tex]

Determine this equation for each of the objects and compare them by inspection. All you are asked is which is fastest.

You can always solve V for each.

As to acceleration you were asking about, lest you forget dv/dt is acceleration.

brainracked

the h is still the problem though..i still dont understand what tiny tim means by v=(dh/dt)/sin [tex]\theta[/tex]
I can only manage to simplify the equations down to:
Sphere:
mgh=[tex]\frac{7}{10}[/tex]mv²

Cylinder:
mgh=[tex]\frac{3}{4}[/tex]mv²

Hoop:
mgh=mv²

LowlyPion

For a given drop in height which has the greater V?

Won't the fastest object at the bottom necessarily be the one there first, if acceleration is uniform?

brainracked

For a given drop in height, the hoop has a greater V? It is possible to find the acceleration of each of the objects right?

LowlyPion

You might want to check again which has the greater speed for a given h.

Of course you can figure acceleration if you wish. But if you know that one object is faster at one point of h below the top in a uniformly accelerated field then you know from induction that it is faster at all points on the incline don't you?

LowlyPion

Forgot to add that if it is faster at every point on the incline then it would be fastest to the bottom wouldn't it?

tiny-tim

Hi brainracked!
v = dx/dt.

It's a slope, with angle θ, say …

so x (along the slope) = h/sinθ.

This isn't physics … this is geometry.

brainracked

The equations mentioned at top, are they right?
I assume the sphere has a greater speed for a given h. Followed by the cylinder and the hoop. This is known by looking at the equations? Is that right?

Yea..it would..



I understand that..it's just that I dont understand how there could be values for x and h..