The question and the hint are both listed--the value entered (3.923) is obviousl

Question

The question and the hint are both listed--the value entered (3.923) is obviously incorrect. Help would be greatly appreciated!!

A cylinder of radius 2.84 cm and a sphere of radius 7.22 cm are rolling without slipping along the same floor. The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the cylinder's angular speed to the sphere's angular speed be? Number eyl w | | 3.923 sph Hint Previous Give Up & View Solution D Check Answer () Next The moments of inertia of a cylinder and a sphere about their respective centers of mass are: cyl cyl sph sph Each object has more than one "type" of kinetic energy. What are the types, and how do you calculate each?

Explanation / Answer

here,

let the mass of the cyclinder and sphere be m

radius of the cyclinder , rc = 2.84 cm

moment of inertia of cyclinder , Ic = 0.5 m rc ^2

Ic = 0.5 * m * 2.84 ^2

Ic = 4.0328 m kg cm^2

radius of the sphere , rs = 7.22 cm

moment of inertia of sphere , Is = 0.4 m * rs^2

Ic = 0.4 * m * 7.22 ^2

Ic = 20.85 m kg cm^2

let the angular speed of cyclinder and sphere be Wcyl and Wsph

let the translational speed of cyclinder and spher be vcyl and vsph

vcyl = rc * wcyl , vsph = rs * wsph

kinetic energy of cyclinder = kinetic energy of cyclinder

0.5 * Is * wsph^2 + 0.5 * m * vsph^2 = 0.5 * Ic* wcyl^2 + 0.5 * m * vcyl^2

20.85 m * wsph^2 + m * (7.22 * wsph)^2 = 4.0328 m * wcyl^2 + m * (2.84 * Wcyl)^2

(Wcyl/Wsph)^2 = 72.98/12.1

Wcyl/Wsph = 2.46

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