moment of inertia and angular velocity

1. The problem statement, all variables and given/known data

(a) Calculate the moment of inertia I of the disc when it rotates about the pivot as shown in the figure.
(b) If the disc is released from rest, determine the angular speed, ω, of the disc at its lowest point.

2. Relevant equations

a) Id = Icm + md^2
Icm = 1/2*M*R^2
b) a = r*γ, a=g
(ωf)^2 = (ωi)^2 + 2*γ*θ

3. The attempt at a solution

a) Id = Icm + md^2
= 1/2MR^2 + 5(0.3^2)
= 1/2*5*0.3^2 + 5(0.3^2)
= 0.675 kg/m^2

b) a = r*γ, a=g, γ = g/r
θ = ¼*2pi
(ωf)^2 = (ωi)^2 + 2*γ*θ
ωf = sqrt(2*g/r*1/4*2pi) = sqrt(2*9.8/0.3*1/4*2pi = 10.1 rad/s

Am I doing the question correctly?
Thank you very much.

Attached Images
File Type: jpg q5pImid.jpg (26.2 KB)

http://ift.tt/O5Pu8r

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