Find the linear speed of the particle when system rotates about axis

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The discussion focuses on calculating the potential energy (PE) of a particle in a rotating system using conservation of mechanical energy. The initial potential energy is derived from an integral involving surface density and gravity, while the final potential energy is set to be equal to the negative of the initial potential energy. Participants debate the correctness of the integral and suggest that the center of gravity method may simplify the problem without needing complex integration. Ultimately, the integral is confirmed to be correct, leading to a potential energy expression that includes a pi term, indicating a successful evaluation of the integral. The conversation emphasizes the importance of understanding potential energy in the context of rotational dynamics.
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Homework Statement
A uniform circular disc radius R and mass M with a particle of mass M fixed on edge. The system is free to rotate about chord PQ. It is allowed to fall from vertical. Find the linear speed of the particle when it reaches the lowest point.
Relevant Equations
##P.E.=\int{Fdx}##
Question image:
20231126_230807.jpg

The question should be solved by conservation of mechanical energy.( I assume surface density##\sigma## and acceleration due to gravity##g=const.##)Therefore:
$$PE_i+KE_i=PE_f+KE_f$$
The axis of rotation ##PQ## is line of zero potential. Then
1) ##PE_i=\int Fdy##
Since coordinates of center are ##(0,\frac{R}{4})## I define the integral as:
$$PE_i=\int^{R}_{-R}\left( 2g \sigma \left(y+\frac{R}{4}\right) \sqrt{R^2-y^2}\right)dy +\frac{5MgR}{4}$$ y is with respect to center.

2) $$KE_i=0$$
3) $$PE_f=-PE_i$$
4) $$KE_f=\frac{I\omega^2}{2}=\left( \frac{MR^2}{4}+\frac{MR^2}{16}+\frac{25MR^2}{16} \right)\frac{\omega^2}{2}$$

I thought I had it but then integration struck and I could not do it no more. Moreover the final answer keeps giving a ##pi##(wrong integration on my part I guess) term that does not cancel.

So my questions are :
1) How to find the potential energy of the system?
2) Is there any other method that could give answers faster?
3) Finally is my integral correct? Any hint/help in further solving it?
 
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I looked for the solution online but there is no explanation of how they find the potential energies. Just this:
Screenshot_20231126_235517_Chrome.jpg

Thank you 😊
 
Aurelius120 said:
So my questions are :
1) How to find the potential energy of the system?
2) Is there any other method that could give answers faster?
3) Finally is my integral correct? Any hint/help in further solving it?
(1) Consider that the external force of gravity acts at the center of gravity. That's what the solution that you found did.
(2) See (1) above.
(3) No need to integrate. See (1) above.
 
kuruman said:
kuruman said:
(1) Consider that the external force of gravity acts at the center of gravity. That's what the solution that you found did.
(2) See (1) above.
(3) No need to integrate. See (1) above.
Ok but just for the sake of it, is the integral correct? You know in case it had to be proven that the center of gravity method is correct
 
Aurelius120 said:
Ok but just for the sake of it, is the integral correct?
Finish integrating and if you get ##PE_i=mg\left(R+\dfrac{R}{4}\right)+mg\dfrac{R}{4}##, then it is correct.
 
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Aurelius120 said:
I guess because the particle is at A . ##5R/4## above reference line
Whoops! I was misled by the diagram, which makes it look at least R/2.
 
Have evaluated the integral. It was correct. Adding it for future reference.
$$PE_{i_{ disc}}=\int^{R}_{-R}\left( 2g \sigma \left(y+\frac{R}{4}\right) \sqrt{R^2-y^2}\right)dy$$
$$=\int^{R}_{-R}\left( 2g \sigma

y \sqrt{R^2-y^2}\right)dy +\int^{R}_{-R}\left( 2g \sigma \frac{R}{4}\sqrt{R^2-y^2}\right)dy$$
$$=\int^{R}_{-R}\left(- g \sigma \sqrt{R^2-y^2}\right)d(R^2-y^2) +\int^{R}_{-R}\left( 2g \sigma \frac{R^2}{4}\sqrt{1-\frac{y^2}{R^2}}\right)dy$$
$$=\left[-\frac{ 2g \sigma}{3} {\sqrt{R^2-y^2}}^3\right]^{R}_{-R}+\int^{R}_{-R}\left( 2g \sigma \frac{R^2}{4}\sin \theta \right)dy$$

$$\frac{y}{R}=\cos\theta \implies \frac{dy}{d\theta}=-R\sin \theta$$
$$=0+\int^{R}_{-R}\left( 2g \sigma \frac{R^3}{4}(-\sin^2 \theta) \right)d\theta$$

$$=\int^{R}_{-R}\left( 2g \sigma \frac{R^3}{4}×\frac{\cos(2 \theta)-1}{2} \right)d\theta$$

$$=\frac{g \sigma R^3}{4}\left[ \sin \theta \cos \theta-\theta\right]^{R}_{-R}$$

$$=\frac{g \sigma R^3}{4}\left[ \frac{y}{R}\sqrt{1-\frac{y^2}{R^2}}-\cos^{-1}(y/R)\right]^{R}_{-R}$$

$$=\frac{g \sigma R^3}{4}\left[ \frac{y}{R}\sqrt{1-\frac{y^2}{R^2}}-\cos^{-1}(y/R)\right]^{R}_{-R}$$
$$=\frac{\sigma g R^3 \pi}{4}=\frac{mgR}{4}$$
 
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