A 10.4-g CD with a radius of 5.85 cm rotates with an angular speed of 33.8 rad/s. What is its kinetic energy?

(in J)

I need help with this question.
I thought you would just use K=1/2mv^2 but when I use m=0.0104kg and v=33.8 I get 5.94 which isn't the correct answer....

To find the kinetic energy of a rotating object, you need to use the equation for rotational kinetic energy. The formula for rotational kinetic energy is K = (1/2) * I * ω^2, where K is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

To solve this problem, you need to calculate the moment of inertia of the CD (I) and then use that value along with the given angular speed (ω) to find the kinetic energy (K).

The moment of inertia of a solid disc can be calculated using the formula I = (1/2) * m * r^2, where m is the mass of the CD and r is the radius.

Given:
Mass (m) = 10.4 g = 0.0104 kg
Radius (r) = 5.85 cm = 0.0585 m
Angular speed (ω) = 33.8 rad/s

First, let's calculate the moment of inertia (I):
I = (1/2) * m * r^2
= (1/2) * 0.0104 kg * (0.0585 m)^2

Now, plug in the values and calculate I.

I = (1/2) * 0.0104 kg * (0.0585 m)^2
= 1.92617 x 10^-5 kg·m^2

Now, we can find the kinetic energy (K):
K = (1/2) * I * ω^2
= (1/2) * (1.92617 x 10^-5 kg·m^2) * (33.8 rad/s)^2

Now, plug in the values and calculate K.

K = (1/2) * (1.92617 x 10^-5 kg·m^2) * (33.8 rad/s)^2
≈ 11.1 J

Therefore, the kinetic energy of the rotating CD is approximately 11.1 Joules.