Which Body Reaches the Bottom First?

[brainracked]

Homework Statement

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.

Homework Equations

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

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]

Homework Statement

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.

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 canceled 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 don't 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]

rolling

So sorry..but..i don't quite understand..the instantaneous velocity of point of contact?
Relationship?

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 don't see how tis can solve the answer..

 

[brainracked]

*solve the question

 

[tiny-tim]

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

v = (dh/dt)/sinθ

 

[brainracked]

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

 

[LowlyPion]

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 don't see how tis can solve the answer..

$$\Delta PE = \Delta KE = mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2} = \frac{mv^2}{2} + \frac{I}{2}*\frac{v^2}{r^2}$$

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 don't understand what tiny tim means by v=(dh/dt)/sin $$\theta$$
I can only manage to simplify the equations down to:
Sphere:
mgh=$$\frac{7}{10}$$mv²

Cylinder:
mgh=$$\frac{3}{4}$$mv²

Hoop:
mgh=mv²

 

[LowlyPion]

the h is still the problem though..i still don't understand what tiny tim means by v=(dh/dt)/sin $$\theta$$
I can only manage to simplify the equations down to:
Sphere:
mgh=$$\frac{7}{10}$$mv²

Cylinder:
mgh=$$\frac{3}{4}$$mv²

Hoop:
mgh=mv²

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 which has the greater V?

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

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]

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?

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!

the h is still the problem though..i still don't understand what tiny tim means by v=(dh/dt)/sin $$\theta$$

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]

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?

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?

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?

Yea..it would..

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.

I understand that..it's just that I don't understand how there could be values for x and h..

 

[LowlyPion]

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?

Yes, that is correct.

Here is a demonstration you might enjoy:

http://hyperphysics.phy-astr.gsu.edu/hbase/hoocyl.html#hc1