# NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/

## Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on March 09, 2009, 09:15:46 am

 Title: Ball Rolling without slipping in a hill Post by: Fu-Kwun Hwang on March 09, 2009, 09:15:46 am 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 f1. m*g*sinθ-f=m*a2. R*f=I*α3. I=(1/2)m*R2 for cylinder, (2/5)m*R2 for spheresolve the above equation will give youf=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere. Title: Re: Ball Rolling without slipping in a hill Post by: Fu-Kwun Hwang on March 09, 2009, 10:32:47 am The following add a spring make it more complicated!You are welcomed to post your calculation for further discussion. Title: Re: Ball Rolling without slipping in a hill Post by: lookang on March 09, 2009, 05:19:09 pm Quote from: Fu-Kwun Hwang on March 09, 2009, 09:15:46 amA 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 f1. m*g*sinθ-f=m*a2. R*f=I*α3. I=(1/2)m*R2 for cylinder, (2/5)m*R2 for spheresolve the above equation will give youf=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere.1. m*g*sinθ-f=m*a2. R*f=I*α imply R*f = (1/2*m*R^2)*(a/R) imply 2*f/m = asub backm*g*sinθ - f  = m*(2f/m) imply 1/3*m*g*sinθ = f for cylinder ;D Title: Re: Ball Rolling without slipping in a hill Post by: Fu-Kwun Hwang on March 09, 2009, 08:48:15 pm 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 NSince N=m*g*cosθ. So μ*m*g*cosθ ≧ (1/3)m*g*sinθIt means that μ≧(1/3)tanθIf 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. Title: Re: Ball Rolling without slipping in a hill Post by: lookang on March 09, 2009, 10:31:18 pm what i was trying to highlight the typo error. LOLthe applet is fine, the text in your first post seems wrongsolve the above equation will give youf=(1/3)m*g*sinθ for cylinder or f=(2/7)m*g*sinθ for sphere. Title: Re: Ball Rolling without slipping in a hill Post by: Fu-Kwun Hwang on March 09, 2009, 11:18:14 pm 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.Clear your cache , close your broswer and reload it again. 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!