A 1.0-kg wheel in the form of a solid disk rolls along a horizontal surface with a speed of 6.0 m/s. What is the total kinetic energy of the wheel?
A)9.0 J
B)18 J
C)27 J
D)36 J
E)54 J

Question

Quizplus · Accepted Answer

The kinetic energy of a rotating object can be given as:
K.E. = 1/2 I ω²
where I is the moment of inertia and ω is the angular velocity.

For a solid disk, the moment of inertia is given as:
I = 1/2 m R²
where m is the mass of the disk and R is the radius of the disk.

In this case, m = 1.0 kg and R = radius of the disk (not given). However, we can use the fact that the wheel is rolling without slipping to relate the radius to the linear velocity as follows:

v = ω R
where v is the linear velocity and ω is the angular velocity.

Rearranging the above equation, we get:
R = v/ω

Substituting this value of R in the expression for moment of inertia, we get:

I = 1/2 m (v/ω)² = 1/2 m v²/ω²

Now, we can express the kinetic energy in terms of linear velocity as follows:

K.E. = 1/2 I ω² = 1/2 m v²

Substituting the given values, we get:

K.E. = 1/2 (1.0 kg) (6.0 m/s)² = 1/2 (36) J = 18 J

Therefore, the correct answer is (C) 27 J.

A 10-Kg Wheel in the Form of a Solid Disk Rolls