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## Homework Statement

The pre-lab is based on the following lab: http://iws.collin.edu/mbrooks/physics/experiments/PHYS 2425 Exp 7 Manual.pdf

Moderator edit: Extract of an image from the PDF showing the mechanism under discussion:

1. Draw two free-body diagrams for the rotating mass in teh experiment, showing all the forces acting on it when (a) it is revolving at radius r and is attached to the Center Post spring, and (b) it is revolving while hanging but not attached to the spring.

2. Why is it necessary to level the rotating platform? How would the motion of the revolving mass be affected if the base were not level? Explain.

3. Why is it important that the string attached to the revolving mass be parallel to the platform? Explain.

4. A mass m=100g is attached to a spring connected to a rotating platform. The platform is then rotated such that the centripetal force acting on the mass is supplied entirely be the spring. The period of rotation T is measured for each of a series of rotating raii r, while keeping the spring force constant. The data is displayed in the graph below ( y=.8626x, the y is T^2 in s^2 and the x is r in meters).

a) From this graph, what is the force of the spring?

b) What would be the slope of the graph if the rotating mass was reduced to 50 g?

## Homework Equations

F

_{c}= ma

_{c}

If T is the period of rotation, then v = 2πr/T

## The Attempt at a Solution

1. For the string, would there only be a centripetal force that would be the tension of the string, and for the spring would there only be the centripetal force caused by the spring?

2. I understood that the rotating platform needed to be leveled since the lab tables are not leveled. Would the motion be affected because it would be at an angle because of gravity?

4.

Solution attempt: To find the force, I thought that the problem was looking for the centripetal force. So F

_{c}= ma

_{c}= mv

^{2}/r = m(2πr/T)

^{2}/r = 4π

^{2}rm/T

^{2}

I then used the function, and set r = 1, so T

^{2}= .8626.

I solved for the force to be 183.02 N

Solution attempt: I used the same equation, but substituted T2 for sr, where s is the slope of the function.

So,

F

_{c}= 4π

^{2}rm/T2 = 4π

^{2}rm/(sr) = 4π

^{2}m/s ⇔ s= 4π

^{2}m/F

_{c}

I used the same force I found in a and used a mass of .05 kg to get .011 s2/m

Thank you in advance

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