kinetic energy with a rolling cylinder?

Question

A 12.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 12.0 m/s.
(a) Determine the translational kinetic energy of its center of gravity.

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

(c) Determine its total kinetic energy

Answer 1

The translational energy is equal to 0.5 * mass * (velocity)^2, or

0.5 * 12 * 12^2 = 864 J

The rotational kinetic energy is equal to 0.5 * (moment of inertia) * (angular velocity)^2.

The moment of inertia for a cylinder is 0.5 * (mass)*(radius)^2

The angular velocity is the (velocity)/(radius).

so substituting in gives:

0.5 (0.5 * mass * radius^2) * (velocity/radius)^2

the radius^2 factor will cancel leaving

0.25 * mass * velocity^2 or

0.25 * 12 * 12^2 = 432 J

Total energy is the sum = 432 + 864 = 1296 J.

Answer 2

a) 1/2(12)(12^2) = 864J

b)E=(I(w^2))/2 where I is the moment of inertia = (m(r^2))/2 and w is the rotational speed = v/r

=> E=m(r^2)(v^2)/4(r^2)
=>E=m(v^2)/4 or half the translational energy, = 432J

c)Total energy = 1296J

Answer 3

The axis of rotation is a factor in the time of the middle of the sphere parallel to the floor. you're saying the 2d of inertia s mr^2, yet that assumes all the mass of the sphere is at radius r from the axis of rotation. it isn't any longer. the assorted mass is honestly on the axis of rotation. you will desire to make certain what the 2d of inertia of a hollow sphere is rotating approximately an axis. For a solid sphere, that's 2/5 mr^2. A hollow sphere would be extra. Get out your calculus e book.