Formula for total KE of disk rolling across table 8 m/s?

[premedonna89]

Homework Statement

A solid disk (mass = 5 kg, R=0.9 m) is rolling across a table with a translational speed of 8 m/s.

What is the total KE of the disk?

Homework Equations

K = the sum of (1/2 x mass initial x velocity initial^2)

(in my textbook, there are other derivations of this formula using the variable angular velocity instead of velocity, but they yield the same answer)

The Attempt at a Solution

KE = 1/2 x 5 kg x 8^2 = 1/2 x 5 kg x 64 = 160 (this was incorrect)

I also plugged in 80 just for the heck of it, and this was incorrect as well.

In the previous question, I was asked to find rotational KE, and I made the mistake of forgetting the coefficient (1/2) in calculating the moment of inertia. However, this KE formula does not include the moment of inertia, so this does not seem to be a problem here.

 

[premedonna89]

ugh

nevermind...its K trans + K rot = 160 + 80 = 240

 

[kerimek]

The correct formula for the total KE of a rolling disk is given by:

KE = 1/2 * m * (v^2 + ω^2 * R^2)

Where m is the mass of the disk, v is the translational velocity, ω is the angular velocity, and R is the radius of the disk.

In this case, the translational velocity is 8 m/s and the angular velocity can be calculated using the relationship ω = v/R. Therefore, ω = 8/0.9 = 8.89 rad/s.

Plugging these values into the formula, we get:

KE = 1/2 * 5 kg * (8^2 + 8.89^2 * 0.9^2) = 1/2 * 5 kg * (64 + 71.53) = 1/2 * 5 kg * 135.53 = 337.825 J

Therefore, the total KE of the rolling disk is 337.825 J.