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Dynamic of Ratational Motion

by azizlwl
Tags: dynamic, motion, ratational
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azizlwl
#1
May1-12, 04:57 AM
P: 963
If the ball were rolling uphill, the force of friction would still be
directed uphill as in FJg. 1O.19b. Can you see why'?



I always assume the friction is always in opposite direction of the motion.
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andrien
#2
May1-12, 08:09 AM
P: 1,020
if the ball is rolling down then

a(cm)=(mg sinβ - f)/m and Iα=fr
and since a=rα
(mgsinβ - f)/m=fr^2/I
f is positive from here so when it is rolling down f is upward opposing the motion of centre of mass.
if it is rolling up then
a(cm)=(f - mgsinβ)/m and Iα=-fr
so (f - mgsinβ)/m=-fr^2/I
from here too it is upward.
azizlwl
#3
May2-12, 09:00 PM
P: 963
thank you.

tiny-tim
#4
May4-12, 04:57 AM
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Dynamic of Ratational Motion

hi azizlwl!

we can always take moments about the centre of mass

let's do so the ball in the diagram is accelerating anticlockwise no matter whether it's rolling downhill or uphill

since the only external torque about the centre of mass is the friction, that must also be anticlockwise in both cases!
azizlwl
#5
May4-12, 08:22 AM
P: 963
Quote Quote by tiny-tim View Post
hi azizlwl!

we can always take moments about the centre of mass

let's do so the ball in the diagram is accelerating anticlockwise no matter whether it's rolling downhill or uphill

since the only external torque about the centre of mass is the friction, that must also be anticlockwise in both cases!


My assumption is that going downhill the rotation is anti clockwise and uphill the rotation will be clockwise. So the friction must reversed its direction.

By your explanation, it means we have another inclined plane on the left side, where the ball keep rolling up with same rotational direction from right position of the inclined plane..
tiny-tim
#6
May4-12, 08:36 AM
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Quote Quote by azizlwl View Post
My assumption is that going downhill the rotation is anti clockwise and uphill the rotation will be clockwise. So the friction must reversed its direction.
yes, the rotation will be clockwise, but the angular acceleration will be anti-clockwise

(because this is a freely rolling ball)

so the friction will be in the same direction, whatever the direction of rotation
azizlwl
#7
May4-12, 08:48 AM
P: 963
Quote Quote by tiny-tim View Post
yes, the rotation will be clockwise, but the angular acceleration will be anti-clockwise

(because this is a freely rolling ball)

so the friction will be in the same direction, whatever the direction of rotation
Sorry i didn't read properly.
It's the rotational acceleration that remains the same uphill or downhill although the direction changes.

Analogy of throwing an object upward, direction changes but acceleration remains.

Doing Physics mental experiment is really a tough for me. Thank you.
azizlwl
#8
May7-12, 02:40 AM
P: 963
Problem with rolling.
Can i use different approach.

Assume No friction as in inclined plane. The instantaneous velocity in contact with the plane is downward so the friction upward. Thus produce torque clockwise.

For the problem http://physicsforums.com/showthread.php?t=603360.
Assuming no friction, the instantaneous velocity in contact with the ground is opposite of the force. Thus the friction is in direction of the force.

For uphill, the rotation will be clockwise or uphill thus instantaneous velocity in contact with the plane is downward giving friction in opposite direction that is uphill.
tiny-tim
#9
May7-12, 03:37 AM
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hi azizlwl!
Quote Quote by azizlwl View Post
Assume No friction as in inclined plane. The instantaneous velocity in contact with the plane is downward so the friction upward. Thus produce torque clockwise.

For the problem http://physicsforums.com/showthread.php?t=603360.
Assuming no friction, the instantaneous velocity in contact with the ground is opposite of the force. Thus the friction is in direction of the force.

For uphill, the rotation will be clockwise or uphill thus instantaneous velocity in contact with the plane is downward giving friction in opposite direction that is uphill.
i'm not sure what you're saying, or what force you're assuming

it doesn't look right, because you're concentrating on the velocity of the point of contact, when the primary consideration is the direction of the acceleration (either linear or angular)

by force, are you still referring to gravity, or is this an extra applied force?

for an "external" applied force, such as a rope or gravity, the direction of friction is opposite the acceleration, but for an "internal" force such as from the engine, it is in the same direction as the acceleration
azizlwl
#10
May7-12, 06:36 AM
P: 963
I just try to find out the direction of friction force which to me it's very confusing.
For instance, a cylinder rolling downhill the friction direction is uphill
For cylinder rolling uphill, frictional force direction also uphill. You have given me the acceleration point view.
For a wound cylinder, the friction is in direction of the force pulling the string (as in http://physicsforums.com/showthread.php?t=603360).

Yes I'm concentrating on the velocity of the point of contact so that i will get direction of the friction which is in opposite direction of the motion(i guess it's always true).
tiny-tim
#11
May7-12, 06:53 AM
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Quote Quote by azizlwl View Post
For instance, a cylinder rolling downhill the friction direction is uphill
For cylinder rolling uphill, frictional force direction also uphill. You have given me the acceleration point view.
For a wound cylinder, the friction is in direction of the force pulling the string (as in http://physicsforums.com/showthread.php?t=603360).
but that wound cylinder isn't rolling (on the table), and there's no friction
Quote Quote by azizlwl View Post
The frictional force between table and cylinder is negligible.
azizlwl
#12
May7-12, 07:06 AM
P: 963
Sorry about the wound cylinder question.
It was second question(repeat problem 11.70 if the frictional force between table and cylinder is negligible.)

11.70
The rope shown is wound around a cylinder of mass 4.0 kg. and I=0.020kg.m2, about the cylinder axis. If the cylinder rolls without slipping, what is the linear acceleration of its center of mass? What is the frictional force? Use an axis along the cylinder axis for your computation?

Solution
Choose left and ccw as positive. Write F=ma=20 +f=4a with f being the friction force at the floor. From τ=Iα=(20-f)(0.01)=0.02(s/0.10) =>a=6.7m/s2 and f=6.8

From the solution given frictional force f is direction of the 20N force. Intuitively it should be opposite of the cylinder motion.
So here i have to figure out how to determine the direction of the frictional force.

Thank you
tiny-tim
#13
May7-12, 12:16 PM
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hi azizlwl!
Quote Quote by azizlwl View Post
Choose left and ccw as positive. Write F=ma=20 +f=4a with f being the friction force at the floor. From τ=Iα=(20-f)(0.01)=0.02(s/0.10) =>a=6.7m/s2 and f=6.8
no, you have a decimal point wrong somewhere
20 + f = 4a

20 - f = 20a


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