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
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.
1 answer