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

  • #1
Aurelius120
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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?
 
Last edited:
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  • #2
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 😊
 
  • #3
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.
 
  • #4
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
 
  • #5
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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  • #6
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.
 
  • #7
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$$

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

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

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

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

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

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

What is linear speed in the context of rotational motion?

Linear speed in rotational motion refers to the tangential speed of a point or particle situated at a certain distance from the axis of rotation. It is the rate at which the particle covers distance along the circular path.

How do you calculate the linear speed of a particle in a rotating system?

The linear speed (v) of a particle in a rotating system can be calculated using the formula: v = r * ω, where r is the radius (distance from the axis of rotation) and ω (omega) is the angular speed in radians per second.

What units are used for linear speed in rotational systems?

Linear speed in rotational systems is typically measured in meters per second (m/s).

How does the radius affect the linear speed of a particle in rotational motion?

The radius directly affects the linear speed; as the radius increases, the linear speed increases proportionally if the angular speed remains constant. This is because the particle has to cover a larger circumference in the same amount of time.

What is the relationship between angular speed and linear speed?

The relationship between angular speed (ω) and linear speed (v) is given by the equation v = r * ω. This means that linear speed is the product of the radius and the angular speed. If either the radius or the angular speed increases, the linear speed will also increase.

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