Calculating KE in rotational motion

[MinaHany]

Homework Statement

A 20kg solid disk (I=1/2Mr^2) rolls on a horizontal surface at the rate of 4.0m/s
Calculate its total kinetic energy

The Attempt at a Solution

I think that simply equating the KE to (0.5)(m)(v^2) would be a wrong solution because then I would not use the moment of inertia given in the question, although I don't know why it is wrong.


Thank you.

 

[Doc Al]

I think that simply equating the KE to (0.5)(m)(v^2) would be a wrong solution because then I would not use the moment of inertia given in the question, although I don't know why it is wrong.

It's wrong because it ignores the rotational KE. The rolling disk has both translational and rotational KE. What's the formula for rotational KE? Hint: How does translational speed relate to angular speed? (Assume it rolls without slipping.)

 

[lewando]

Use:

KE = KE_{translational} + KE_rotational = (1/2)mv_cm² + (1/2)Iw²

v_cm : velocity at the center of mass[sorry Doc Al, I didn't see you post]

 

[MinaHany]

Thank you Doc AL and lewando..
Your replies made the idea clear for me.

 

[rwisz]

Your thinking is correct. In rotational motion, the moment of inertia must be taken into account in calculating the kinetic energy. The correct formula for rotational kinetic energy is KE = (1/2)(I)(ω^2), where I is the moment of inertia and ω is the angular velocity. In this case, the moment of inertia for a solid disk is given by I = (1/2)(m)(r^2), so plugging in the values from the problem, we get KE = (1/2)(1/2)(20)(4.0)^2 = 160 J. This is the total kinetic energy of the disk in rotational motion.