Disk falling while rotating round a point

AI Thread Summary
The discussion focuses on the dynamics of a disk falling while rotating around a point on its circumference. The moment of inertia is calculated as I=1/2 mR^2, and the angular velocity is derived using energy conservation, yielding ω^2=4g/3R. The center of mass velocity is determined to be v_cm=2√(gR/3), while the centripetal acceleration is a_cen=4g/3. Tangential acceleration is calculated from torque, resulting in a_tan=2g/3. The calculations and concepts presented appear to be correct, with a minor correction noted regarding terminology.
Karol
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Homework Statement


The disk of mass m and radius R is at position (a), upwards. it is free to rotate round point O on it's circumference.
If falls and reaches state (b), horizontal position.
What are the velocities and accelerations in position (b)

Homework Equations


Moment of inertia of a disk: ##I=\frac{1}{2}mR^2##
Torque of rigid body: ##M=I\alpha##
Energy of a rigid body: ##E=\frac{1}{2}I\omega^2##
Centripetal acceleration: ##a_{cen}=\frac{v^2}{R}=\omega^2 R##

The Attempt at a Solution


The angular velocity ω is from energy conservation:
$$mgR=mR^2\left( \frac{1}{2}+1 \right)\omega^2\; \rightarrow \omega^2=\frac{4g}{3R}$$
COM's velocity is ##v_{cm}=\omega R=\sqrt{\frac{4g}{3R}}R=2\sqrt{\frac{gR}{3}}##
The radial velocity from the circular motion:
$$a_{cen}=\omega^2 R=\frac{4g}{3}$$
Tangential acceleration is from the gravity:
$$M=I\alpha\rightarrow mgR=\frac{3}{2}mR^2\alpha\rightarrow\alpha=\frac{2g}{3R}$$
$$a_{tan}=R\alpha=\frac{2g}{3}$$
 

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One typo... where you wrote radial velocity you meant radial acceleration.
 
Thanks
 
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