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 Author Topic: Ball Rolling without slipping in a hill  (Read 27249 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: March 09, 2009, 09:15:46 am » posted from:Taipei,T\'ai-pei,Taiwan

A ball or cylinder rolling (without slipping) down in a down hill slope.
The condition for rolling without slipping is
v=R*ω (angular velocity) or a=R*α (angular acceleration)
where R is the radius of the ball or cylinder.

Assume the friction force is f
1. m*g*sinθ-f=m*a
2. R*f=I*α
3. I=(1/2)m*R2 for cylinder, (2/5)m*R2 for sphere
solve the above equation will give you
f=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere.

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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: March 09, 2009, 10:32:47 am » posted from:Taipei,T\'ai-pei,Taiwan

The following add a spring make it more complicated!

You are welcomed to post your calculation for further discussion.

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lookang
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 « Embed this message Reply #2 on: March 09, 2009, 05:19:09 pm »

A ball or cylinder rolling (without slipping) down in a down hill slope.
The condition for rolling without slipping is
v=R*ω (angular velocity) or a=R*α (angular acceleration)
where R is the radius of the ball or cylinder.

Assume the friction force is f
1. m*g*sinθ-f=m*a
2. R*f=I*α
3. I=(1/2)m*R2 for cylinder, (2/5)m*R2 for sphere
solve the above equation will give you
f=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere.

1. m*g*sinθ-f=m*a
2. R*f=I*α imply R*f = (1/2*m*R^2)*(a/R) imply 2*f/m = a

sub back

m*g*sinθ - f  = m*(2f/m) imply 1/3*m*g*sinθ = f for cylinder
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: March 09, 2009, 08:48:15 pm » posted from:Taipei,T\'ai-pei,Taiwan

The above calculation give the static friction force Fr=(1/3)m*g*sinθ.
And maximum firction force is proportion to normal force Frmax= μ N
Since N=m*g*cosθ.
So μ*m*g*cosθ ≧ (1/3)m*g*sinθ
It means that μ≧(1/3)tanθ
If the angle is increased or μ is not large enough, then the condition for rolling without slipping is not satisified.
It will loss energy due to sliding friction.

So it will be easier to satisfy the condition if static friction coefficient between surface and the rolling object is larger (instead of smaller).

P.S. static friction coefficient is a property between two objects.
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lookang
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 « Embed this message Reply #4 on: March 09, 2009, 10:31:18 pm »

what i was trying to highlight the typo error. LOL

the applet is fine, the text in your first post seems wrong

solve the above equation will give you
f=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere.
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Fu-Kwun Hwang
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 « Embed this message Reply #5 on: March 09, 2009, 11:18:14 pm » posted from:Taipei,T\'ai-pei,Taiwan

I have updated the applet as soon as I found the error this morning
(the net force was shown instead of friction force).
May be your browser still viewing the cached jar file.
You should be able to view the updated simulation.

I was not reading your message carefully. I thought you were talking about previous error in the simulation.
Thank you. It is fixed now!
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