Pendulum Timer Question: Comparing the Speed of Weighted Pendulums

  • Thread starter Thread starter Engelwood
  • Start date Start date
  • Tags Tags
    Pendulum Timer
AI Thread Summary
The discussion centers on the effect of weight distribution on the speed of pendulum timers. A pendulum with weight closer to the base is expected to swing faster due to a lower moment of inertia, making it easier to move. The concept of moment of inertia is crucial, as it indicates that mass farther from the pivot point increases resistance to motion. An experiment with a baseball bat illustrates this principle, showing that gripping the bat near the handle allows for easier movement compared to gripping it near the barrel. Ultimately, the pendulum with weight towards the base will indeed swing faster than the one with weight at the end.
Engelwood
Messages
3
Reaction score
0

Homework Statement


If I have two exact pendulum timers (like the one in the picture below) and I place a weighted object at the end of the one stick, and towards the bottom (near the base) on the other, which one will swing faster?


http://wwwdelivery.superstock.com/WI/223/1613/PreviewComp/SuperStock_1613R-8223.jpg

 
Last edited by a moderator:
Physics news on Phys.org
Any intuitive ideas as to what will happen? How do you think the concepts of physics apply to this problem? What changes when you add the weight?
 
Nabeshin said:
Any intuitive ideas as to what will happen? How do you think the concepts of physics apply to this problem? What changes when you add the weight?

Nabe, the weight distribution is the obvious difference. I think the timer with the weight towards the base will swing faster. The reason being is because of the experiment that I conducted with my 31oz Demarini baseball bat: with one hand, I grabbed the bat around the barrel (fat part) and continued to supinate and pronate my wrist and few times. I did the same thing again, the only difference with the subsuquent test was that I grabbed the bat towards the handle (bottom) and then preformed the wrist supination and pronation. Moviing the bat was 'easier' when I was holding it by the handle as opposed to around the barrel.

The 'end' of the bat, the barrel, has a greater amount of weight distributed at its end compared to the handle. Therefore, I conclude that the pendulum timer with the weight towards the base will move faster than the timer with the weight loaded towards the end of the stick.

Am I right?
 
Your conclusion is correct, and I like the fact that you tried an experiment to figure it out!

However, saying "I did this with a baseball bat..." is not sufficient explanation to draw conclusions about a different system. To add a little rigor to the conclusion, think specifically about moment of inertia.
 
Thanks. I have not had any Physics classes in quite some time. When in doubt, try to see if you can exprapolate abstract ideas from specific examples. :)

Nabe, could you please elaborate a bit on how inertia relates to this specific scenario.

Thanks in advance,

E
 
Well, like I said, the key concept here is moment of inertia. When the mass is distributed farther away from the pivot point, the moment of inertia increases (it is proportional to r^2). This is what you are seeing in your baseball bat example. The higher the moment of inertia, the harder it is to move, so, given an equal force, the slower it will move.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top