Centripetal Force Lab Questions?

[master x964]

So we did a lab in physics today and it was a contraption where we had a stopper on one get of the string, and weights on the other, and we spun the stopper around, and there's 4 questions that i don't under stand, so if you could help me with this, and please explain as i want to know how to do this, that'd be awesome!

Questions:

1. In an experiment using this apparatus the mass and centripetal force are kept constant, by how much must the velocity change to increase the radius a factor of four(4X)?




2. You calculated the acceleration of the stopper for both lengths. What factors determine the acceleration and rank the factors from greatest to least.


3. Suppose you find that a centripetal force of 12 Newtons is required to keep a given object in a particular circular path when it is moving with uniform speed. Assume you are able to double the mass, velocity, or radius individually at will.
a) What will the magnitude of the centripetal force be for each of these individual changes?



b) Which change modifies the centripetal force the most and why?


4) Write the equation for centripetal force in terms of: mass, revolutions, radius, and period (using any constants needed).
a) What effect would an error in time measurement have on the F in time was too large?
b) What effect on the calculated F would a radius have if the value used for radius was too small?

 

[sophiecentaur]

To my mind this is a really poor experiment but it's the one everyone gets told to do. It must be that the equipment is so easy to make. The problem is that the weight of the stopper is constant so you have to vary string length AND rotation rate to fit the same centripetal force. It would be much better if you could have the string of a fixed length and measure the force for a given rotation rate. The way you did it is (imo) more demanding and it gives 'them' a chance to make you think more (God forbid!)
Look at this link. It tells you the relationship between V and r and string tension (centripetal force = stopper mass times acceleration).
That equation is what you need in order to answer the questions. Re-arranging it to put what you want on one side will tell you how to answer.
Come back with some thoughts on this as I don't want just to give you the answer.

 

[master x964]

Here's what i got for my answers
1. 8 times

2.

3a.

3b. The mass, and the velocity, the higher the mass the more higher of velocity you need to maintain the speed

4. F(of centripetal force) = m4(pi^2)rf^2

and I'm working on 2 and 3a ... still

 

[master x964]

moved to: https://www.physicsforums.com/showthread.php?p=4603818#post4603818

 

[HalfLostAndSearching]

1. To increase the radius by a factor of four, the velocity must decrease by a factor of four. This is because the centripetal force is directly proportional to the mass, velocity, and square of the radius. So if the radius is increased by four, the velocity must decrease by four in order for the centripetal force to remain constant.

2. The acceleration of the stopper is determined by the mass, velocity, and radius. The greatest factor is the velocity, followed by the radius, and then the mass. This is because the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius. The mass also plays a role, but it is not as significant as the other two factors.

3. a) If the mass is doubled, the centripetal force will also double. If the velocity is doubled, the centripetal force will quadruple. If the radius is doubled, the centripetal force will decrease by a factor of four.

b) Doubling the velocity will have the greatest effect on the centripetal force because it is squared in the formula. This means that any change in the velocity will have a greater impact on the centripetal force compared to changes in the mass or radius.

4) a) The equation for centripetal force is F = (m*v^2)/r. If there is an error in the time measurement, the value for velocity will be incorrect and therefore the calculated centripetal force will also be incorrect.

b) If the value used for radius is too small, the calculated centripetal force will be too large. This is because the centripetal force is inversely proportional to the radius, so a smaller radius will result in a larger force.