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

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brainracked
#1
Sep1-08, 05:10 AM
P: 13
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... =)
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tiny-tim
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Sep1-08, 05:34 AM
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Quote Quote by brainracked View Post
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.
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
#3
Sep1-08, 05:50 AM
P: 13
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
#4
Sep1-08, 06:05 AM
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Rotation Question. Is there sufficient data/values? Pondered on question 4 sumtime..

Quote Quote by brainracked View Post
So sorry..but..i dont 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
#5
Sep2-08, 04:11 AM
P: 13
Haha..okay..i get it..thanks!
brainracked
#6
Sep2-08, 09:59 AM
P: 13
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
#7
Sep2-08, 10:01 AM
P: 13
*solve the question
tiny-tim
#8
Sep2-08, 10:33 AM
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Quote Quote by brainracked View Post
Wait wait wait..I ended having two variables..h and v
v = (dh/dt)/sinθ
brainracked
#9
Sep2-08, 10:43 AM
P: 13
and to get the acceleration?
I'm lost..
LowlyPion
#10
Sep2-08, 11:24 AM
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Quote Quote by brainracked View Post
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..
[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
#11
Sep2-08, 12:13 PM
P: 13
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
#12
Sep2-08, 12:17 PM
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Quote Quote by brainracked View Post
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
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
#13
Sep2-08, 01:10 PM
P: 13
Quote Quote by LowlyPion View Post
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
#14
Sep2-08, 01:56 PM
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Quote Quote by brainracked View Post
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
#15
Sep2-08, 02:05 PM
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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
#16
Sep2-08, 03:14 PM
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Hi brainracked!
Quote Quote by brainracked View Post
the h is still the problem though..i still dont understand what tiny tim means by v=(dh/dt)/sin [tex]\theta[/tex]
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
#17
Sep3-08, 12:23 AM
P: 13
Quote Quote by LowlyPion View Post
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?

Quote Quote by LowlyPion View Post
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..


Quote Quote by tiny-tim View Post
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 dont understand how there could be values for x and h..
LowlyPion
#18
Sep3-08, 11:33 AM
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Quote Quote by brainracked View Post
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...oocyl.html#hc1


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