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# A 10-kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a spe?

A 10-kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10 m/s

determine:

(a) the translational kinetic energy of its center of gravity,

(b) the rotational kinetic energy about its center of gravity, and

(c) its total kinetic energy.

### 2 Answers

- 8 years agoFavorite Answer
(a) translational kinetic energy = (1/2)*m*(v^2)

=(1/2)*10*(10^2)

=500J

(b) rotational kinetic energy =(1/2)*I*(w^2) w-angular velocity(omega), I-moment of inertia

=(1/2)*(2/5*m*r^2)(w^2)

=(1/2)*(2/5)*(m)(v^2) v=rw

=200J

(c)total energy = 700J

Source(s): self - cyriaqueLv 44 years ago
Translational KE = a million/2*m*v^2. Plug in. executed. Rotational KE = a million/2*I*omega^2. I for a sturdy cylinder is .5*mr^2. r isn't given. so which you do not have adequate information to end this area of the situation. complete KE = Translational + Rotational.