A CD has a mass of 15.0 g, an inner diameter of 1.5 cm, and an outer diameter of 11.9 cm. Suppose you toss it, causing it to spin at a rate of 4.5 revolutions per second.
(a) Determine the moment of inertia of the CD, approximating its density as uniform.
(b) If your fingers were in contact with the CD for 0.31 revolutions while it was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque?
I have solved problems with one mass and one radius, but here we are given one mass and two diameters/ rdaius....How to solve. I tried several ways that I could think of, but my answer is wrong.
First, you need the moment of inertia of the CD. Call it I. Note that there is a hole in the middle, with
R1 = 0.75 cm and R2 = 5.95 cm
Total area = pi[R2^2 - R1^2] = 109.45 cm^2
density/(area) = 0.1370 gm/cm^2
A2 (without hole) = pi*R2^2
= 111.2 cm^2
M2 (without hole) = 15.23 g
A1 (hole) = pi*R1^2 = 1.77 cm^2
M1 (missing mass of hole) 0.243 g
I = (1/2)*M2*R2^2 - (1/2)*M1*R1^2
(You finish that calculation)
(b) Torque*(contact time)
= change in angular momentum
= I*(final w)
Solve for torque
To solve this problem, we'll need to use the definition of moment of inertia and the formula for torque.
(a) To determine the moment of inertia of the CD, we can approximate it as a solid cylinder with a hole in the center.
The formula for the moment of inertia of a solid cylinder about its rotation axis is:
I = (1/2) * m * r^2
Where:
- I is the moment of inertia
- m is the mass of the object
- r is the radius of the object
However, in this case, we have both the inner diameter and outer diameter of the CD. So we need to find the appropriate radius to use in the formula.
The inner diameter is given as 1.5 cm, which means the inner radius is half of that: 1.5 cm / 2 = 0.75 cm.
The outer diameter is given as 11.9 cm, which means the outer radius is half of that: 11.9 cm / 2 = 5.95 cm.
Now, we can calculate the moment of inertia using the appropriate radius:
I = (1/2) * m * (r_outer^2 - r_inner^2)
= (1/2) * 15.0 g * ((5.95 cm)^2 - (0.75 cm)^2)
Make sure to convert the mass to kilograms and the radius to meters to obtain the correct result.
(b) To determine the magnitude of the torque, we can use the relationship between torque and moment of inertia:
τ = I * α
Where:
- τ is the torque
- I is the moment of inertia
- α is the angular acceleration
Given that the CD is spinning at a rate of 4.5 revolutions per second, we can convert that to angular acceleration using the formula:
α = 2π * n
Where:
- α is the angular acceleration
- n is the number of revolutions per second
Substituting the known values and solving for τ:
τ = I * (2π * n)
= I * (2π * 4.5)
Remember to use the calculated value of I in this equation.
By calculating the moment of inertia and substituting it into the torque equation, you should be able to find the correct answer.