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Falling rod, finding velocity of CM

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  • #1
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Homework Statement


A rod of mass m and length l is held vertically on a smooth horizontal floor. If the rod begins to fall from this position, find the speed of its CM when it makes an angle ##\theta## with vertical.


Homework Equations





The Attempt at a Solution


This is a solved problem in my book and it is solved using the energy method. I was wondering if this can be solved through Newton's laws.

The forces acting on the rod are its own weight and normal reaction from the floor.
From Newton second law:
$$mg-N=ma \,\,\,\, (*)$$

Taking torque about the CM:
$$N\frac{l}{2}\sin\theta=\frac{ml^2}{12}\alpha \,\,\,\, (**)$$
where ##\alpha## is the angular acceleration about CM. Since ##\alpha=a/(l/2)##, I substitute this in (**) and solve for N. From here I get a relation between N and a. Substituting this relation in (*), I get:
$$a=\frac{3g\sin\theta}{1+3\sin\theta}$$
Since, ##a=v(dv/dy)## and ##y=l/2(1-\cos\theta)##,
$$a=2v\frac{dv}{l\sin\theta d\theta}$$

Therefore,
$$2vdv=\frac{3gl\sin^2\theta}{1+3\sin\theta}d\theta$$
The next step is to integrate both the sides. I could not integrate the right hand side so I used Wolfram Alpha.
http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=integrate sin^2(x)/(1+3sin(x)) dx

Wolfram Alpha gives a very strange answer and it does not match with the book's final answer.

Are my equations wrong? :confused:

Any help is appreciated. Thanks!
 

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Answers and Replies

  • #2
tiny-tim
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Hi Pranav-Arora! :smile:
Since ##\alpha=a/(l/2)## …
nooo :redface:

(btw, your use of τ = Iα is valid, since this is a 1D case, and d/dt rc.o.m x vc.o.m = 0)
 
  • #3
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Hi tiny-tim! :)

nooo :redface:
Why? I don't see any error there. :confused:
 
  • #4
tiny-tim
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a is the (vertical) acceleration of the centre of mass

a = rα is correct when the point at distance r (in this case, the end of the rod) is fixed (and then the centre of mass is moving in a circle)

here, the end of the rod is moving, the centre of mass is not moving in a circle, y = l/2 cosθ, and a = y'' :smile:
 
  • #5
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a is the (vertical) acceleration of the centre of mass

a = rα is correct when the point at distance r (in this case, the end of the rod) is fixed (and then the centre of mass is moving in a circle)

here, the end of the rod is moving, the centre of mass is not moving in a circle, y = l/2 cosθ, and a = y'' :smile:
I am not sure if I get it. How do I find N then? Since ##\alpha=a/(l/2)## is invalid, do I have to substitute ##\alpha=d^2\theta/dt^2##?
 
  • #6
tiny-tim
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yes :smile:
 
  • #7
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yes :smile:
Solving for N, I get:
$$N=\frac{ml}{6\sin\theta}\frac{d^2\theta}{dt^2}$$
Substituting in (*)
$$mg-\frac{ml}{6\sin\theta}\frac{d^2\theta}{dt^2}=m\frac{d^2y}{dt^2} \,\,\,\, (***)$$

Since ##y=(l/2)(1-\cos\theta)##, ##dy/dt=(l/2)\sin\theta d\theta/dt##, therefore,
$$\frac{d^2y}{dt^2}=\frac{l}{2}\sin\theta \frac{d^2\theta}{dt^2}+\frac{l}{2}\cos\theta\frac{d\theta}{dt}$$
Substituting the above in (***) gives a second order differential equation but solving this gives ##\theta## as a function of time and I need the velocity as a function of ##\theta##. :confused:
 
  • #8
tiny-tim
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Substituting the above in (***) gives a second order differential equation but solving this gives ##\theta## as a function of time and I need the velocity as a function of ##\theta##. :confused:
(your dθ/dt should be squared)

you can eliminate t by putting ω = dθ/dt,

and d2θ/dt2 = dω/dt = dω/dθ dθ/dt = ω dω/dθ :smile:
 
  • #9
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(your dθ/dt should be squared)

you can eliminate t by putting ω = dθ/dt,

and d2θ/dt2 = dω/dt = dω/dθ dθ/dt = ω dω/dθ :smile:
But I need velocity as a function of angle, not the angular velocity. :confused:
 
  • #10
tiny-tim
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i didn't say it was easy! :smile:

this is why it's so much easier to use energy :wink:
 
  • #11
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i didn't say it was easy! :smile:

this is why it's so much easier to use energy :wink:
Yes, you are right. I recognised the complexity when it lead me to a second order differential equation.

But still, is there no way to find velocity from angular velocity?
 
  • #12
tiny-tim
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you can solve that differential equation for ω as a function of θ,

and then use v = l/2 ω sinθ :wink:
 
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  • #13
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you can solve that differential equation for ω as a function of θ,

and then use v = l/2 ω sinθ :wink:
Thanks a lot tiny-tim! :smile:
 

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