A uniform circular disc has mass M and diameter AB of length 4a. The disc rotates in a vertical plane about a fixed smooth axis perpendicular to the disc through the point D of AB where AD=a. The disc is released from rest with AB horizontal. (See attached diagram)
(a) Calculate the component parallel to AB of the force on the axis when AB makes an angle θ with the downward vertical.
(b) Calculate the component perpendicular to AB of the force on the axis when AB makes an angle θ with the downward vertical
KE = (1/2)Iω2
I of a disc about axis perpendicular to disc through centre is (1/2)mr2
Transverse component of acc. = r(dω/dt)
Radial component of acc. = rω2
The Attempt at a Solution
I have done (a) okay with no problems and got the correct answer. I used a similar approach for part (b) but got the incorrect answer. For part (b):
The moment of inertia is (1/2)M(2a)2 + Ma2 = 3Ma2
(using the parallel axis theorem).
Using conservation of energy:
(1/2)(3Ma2)ω2 = Mgacosθ
(3/2)aω2 = gcosθ
Differentiating w.r.t. θ gives:
3aω(dω/dθ) = -gsinθ
3aω(dω/dt)(dt/dθ) = -gsinθ
(dω/dt) = -(g/3a)sinθ *
Resolving forces: Mgsinθ - Y = Ma(dω/dt)
Substituting in *: Mgsinθ - Y = -(Mag/3a)sinθ
Y = Mgsinθ + (1/3)Mgsinθ = (4/3)Mgsinθ
The answer is supposed to be (2/3)Mgsinθ. I can't see what I have done wrong. Any help would be greatly appreciated!