Questions on science?

Question

1 What is the force on a .35kg CD .2m from the center if it takes .15s to complete a rotation
2 With 3N force you can swingb your .35m key chain around in a circle in .85s.What is the mass of your keys

Answer 1

1)
This is a poorly-worded question. You are probably asking for the acceleration, not the force, at that distance from the center. The mass of the CD is not important in this context.
The centripetal acceleration of points on the disc a distance r from the center is:
ac = v²/r
Where v is the linear velocity at that point. You can find it like this:
v = 2πr/T = 2π(0.2)/0.15 = 8.38 m/s
So
ac = (8.38)²/(0.2) = 350 m/s²
*Note: you can't just multiply ac by the mass of the CD to get the 'force'. It wouldn't mean anything. But if you placed an object on the CD at this point, and multilpied ac by the OBJECT's mass, that would be the centripetal forced on the object.

2)
F = m*v²/r ──► m = r*F / v²
Again, find the linear velocity, v, like this:
v = 2πr/T = 2π(0.35)/(0.85) = 2.59 m/s
So
m = (0.35)(3)/(2.59)² = 0.157 kg

Answer 2

Good luck on your test/homework.