A cylinder of mass 12.0 kg rolls without slipping on a horizontal surface. At a

Question

A cylinder of mass 12.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 11.0 m/s.

(a) Determine the translational kinetic energy of its center of mass.

(b) Determine the rotational kinetic energy about its center of mass.

(c) Determine its total energy.

9. + -/1 points SerPSET9 10.P.059.MI A cylinder of mass 12.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 11.0 m/s. (a) Determine the translational kinetic energy of its center of mass (b) Determine the rotational kinetic energy about its center of mass (c) Determine its total energy

Explanation / Answer

Let the velocity of center of mass be v = 11 m/s

mass of cylinder , m = 12 kg

a) Translational Kinetic Energy is KEtrans = (1/2)m*v^2

KEtrans = 0.5*12*11^2 = 6*121= 726 J

b) The Rotational Kinetic Energy is given by KErot = (1/2)*I*W^2

I is Moment of Inertia of cylinder = (1/2)mR^2

W is angular velocity= v/R, R is radius of cylinder.

So, KErot = (1/2)*I*W^2 =(1/2)* (1/2)mR^2* v^2/R^2 = (1/4)m*v^2 = 726/2 = 363 J

KErot = 363 J

c)

Total Kinetic Energy is sum of translational and rotational kinetic energies

KEtrans+ KErot = 726+363 = 1086 J

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