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April 18, 2014

April 18, 2014

Posted by **Karla** on Tuesday, March 27, 2007 at 12:08pm.

You will find the moment of inertia (I) of a hollow sphere of radius R derived here: (Broken Link Removed)

I = (2/3) MR^2

If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.

M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2

= [(1/2) + (1/3)] M V^2

The M's and R's cancel and

gH sin 38 = (5/6) V^2

V = sqrt (1.2 g H sin 38) = sqrt (2 a H)

where a is the acceleration

a = 0.6 g sin 38

Use that acceleration and Newton's second law t compute the actual friction force, F.

M g sin 38 - F = M a

Mg sin 38 - M*0.6 g sin 38 = F

F = 0.4 M g sin 38

To provide this amount of friction, the static coefficient of friction mu,s must equal or exceed a value given by

M g cos 38 * mu,s = 0.4 M g sin 38

mu,s - 0.4 tan 38

Higher values of mu,s will also prevent slipping

The V/R term is the angular rtation velocity.

- xcqd kxuhc -
**xcqd kxuhc**, Sunday, February 1, 2009 at 7:38amwabv eqjd ncok cnwytrg kdovzwpnl yqbhix xtjhf

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