How Does Increasing the Frequency Affect Accuracy in a Circular Motion Lab?

AI Thread Summary
Increasing the frequency of the rubber stopper's revolution in a circular motion lab leads to decreased accuracy in measuring the frequency. As the stopper spins faster, it becomes challenging to count revolutions and time them precisely due to the rapid motion. The operator's hand movement, which has the same period as the stopper, complicates observation, making it harder to focus on the stopper itself. Additionally, while the tension force must balance the weight of the stopper, the angle of the string remains constant despite increased speed. Overall, higher frequencies result in greater difficulty in achieving accurate measurements.
msimard8
Messages
58
Reaction score
0
This is a question pertaining to a lab we did. Ill try to describe it the best i can. What we pratically did was attach a rubber stopper to an end of a string, put the string to a hollow tube (like a torn apart pen), and tied a mass on the bottom. We had to hold on the tube and start swinging the rubber stopper horizontally by rotating the tube. The mass tied to the string caused a tension force. It is impossible to have the stopper on the string spinning perfectly horizontal because perpindicular forces are independent of each other. We then had to calculate the frequency of the stopper while changing some variables.

Here is my question,

For the greatest accuracy in this investigation, the tension force acting on the tsopper should be horizontal. In this context, what happens to the accuracy as the frequency of revolution of the stopper increases (with the other variables held constant)?


i said as the frequency increases, the accuracy decreases, but i don't know how to explain why.
 
Physics news on Phys.org
Oh, i did a very similar lab to this one. Anyways, your answers is correct and i think it is so because as the stopper spins faster, it gets harder to count the revolutions and stop the timer at the exact time of the revolution occurring.
 
To make the stopper rotate you need to wave your hand to and fro. This motion will have the same period as that of the stopper. It is much easier to watch the motion of the operator's hand than to try and see what the stopper is doing (which is moving too quickly to see - the eye has great difficulty in focussing on it). It becomes much more difficult to count the revolutions as the speed increases.

Yes, you are correct in thinking that the tension will point a bit upwards. The vertical component need to balance the weight of the stopper while the horizontal component supplies the centripetal force.

The angle of the string below the horizontal should not increase with speed since the needed vertical component of the tension need not increase at a greater radius (weight of the stopper stays the same). The stopper will swing at a lower height though at a greater radius.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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...
Back
Top