Where to drill a 1.0 m stick to get the shortest pendulum period?

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A student wants to use a L = 1.0 m stick as a pendulum. She plans to drill a small hole through the meter stick and suspend it from a smooth pin attached to the wall (Fig. 14-35). Where in the meter stick should she drill the hole to obtain the shortest possible period? Answer in number of meters from the upper end

How short an oscillation period can she obtain with a meter stick in this way?

http://www.webassign.net/gianpse3/14-35alt.gif

My guess is 0 meter from the upper end will give the shortest period. Agree?
 
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physicsss said:
My guess is 0 meter from the upper end will give the shortest period. Agree?
No. But why guess? Figure it out. What's the period of a physical pendulum?
 
T=2*pi*sqrt(I/mg(0.5-x))
T=2*pi*sqrt( (1/12mL^2+m(0.5-x)^2) / (12*m*g*(0.5-x) )

Canceling the m
T=2*pi*sqrt( (1/12*L^2+(0.5-x)^2) / (12*g*(0.5-x) )

So if x=0, the period is the smallest...why is this wrong?
 
Last edited:
physicsss said:
T=2*pi*sqrt(I/mg(0.5-x))
T=2*pi*sqrt( (1/12mL^2+m(0.5-x)^2) / (12*m*g*(0.5-x) )
OK, but better use (L/2 - x) instead of (0.5 - x). Also, it seems that there's an extraneous 12 in your denominator.
Canceling the m
T=2*pi*sqrt( (1/12*L^2+(0.5-x)^2) / (12*g*(0.5-x) )


So if x=0, the period is the smallest...why is this wrong?
What makes you think that the period is smallest when x=0?
 
Because it gives me the smallest T possible when I plug it in...0.50 would give me 0/0
 
physicsss said:
Because it gives me the smallest T possible when I plug it in...
How do you know that? Did you check T for every possible value of x?
 
The problem is...i don't...and I don't know how to either. :confused:
 
If you know some calculus, you can find the value of x that minimizes the period by taking a derivative and setting it equal to zero.
 
OK, I did that and got x=1.5 or -0.5. So -0.5 is the answer? But that doesn't make much sense since they want distance from the upperend of the stick...
 
No, those answers are not correct. Two suggestions:
(1) Correct your expression as I advised in post #4. Your answer should be in terms of L.
(2) Redo the derivative.
 
I'm still getting the same answers I got before. Is there any other way to do this since the derivative of 2*pi*sqrt( (L^2+(L/2-x)^2) / (12*g*(L/2-x) ) with respect to x is really difficult for me.
 
physicsss said:
I'm still getting the same answers I got before. Is there any other way to do this since the derivative of 2*pi*sqrt( (L^2+(L/2-x)^2) / (12*g*(L/2-x) ) with respect to x is really difficult for me.
That expression should be:
[tex]T = 2 \pi \sqrt{\frac{L^2/12 + (L/2 - x)^2}{g(L/2 - x)}}[/tex]
This will be a minimum when the expression within the square root is a minimum. So the only thing you need to take the derivative of is this:
[tex]\frac{L^2/12 + (L/2 - x)^2}{(L/2 - x)}[/tex]
Use the quotient (or product) rule. It's not as bad as it looks. :smile: