Posted by **saud** on Monday, December 5, 2011 at 11:59pm.

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

- physics -
**drwls**, Tuesday, December 6, 2011 at 1:10am
K.E. = Etranslational + Erotational

For either cylinder or sphere, V = R*w

For a uniform solid sphere:

K.E. = (1/2)MV^2 + (1/2)I w^2

= (1/2)MV^2 + (1/2)*(2/5)M*R^2*(V/R)^2

= (7/10) M V^2

For a uniform solid cylinder:

K.E. = (1/2)MV^2 + (1/2)((1/2)MR^2*(V/w)^2

= (3/4)M V^2

If masses are equal, AND KE's are equal,

V(cylinder)^2/V(sphere)^2 = (4/3)/(10/7)

= 0.93333

V(cylinder)/V(sphere) = 0.9661

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