Moment of inertia and angular velocity

[nomorenomore]

Homework Statement

(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.

Homework 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

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.

 

[voko]

I am not sure what the "relevant equation" for (b) means.

 

[nomorenomore]

I am not sure what the "relevant equation" for (b) means.

b) a = r*γ, a=g
(ωf)^2 = (ωi)^2 + 2*γ*θ

Oh, I'm sorry for this.
I was trying to show that

linear acceleration = radius * angular acceleration,
where linear acceleration = gravitational acceleration, in this case

(final angular velocity)^2 = (initial angular velocity)^2 + 2*angular acceleration*angular displacement

 

[voko]

Don't you think angular acceleration should depend on the moment of inertia?

Besides, why do you think it will be constant?

 

[HalfLostAndSearching]

Your calculations and equations seem correct. However, it would be helpful to provide more context and information about the figure and the disc in order to fully understand and validate your solution. Additionally, it would be beneficial to show the units for each value and to properly label your final answer. Overall, your approach seems reasonable and your solution appears to be correct. Well done!