Using centripetal force, determine the acceleration of gravity

[spike spiegel]

Homework Statement

This question relates centripetal acceleration and gravity. The general question is to determine the acceleration of gravity using centripetal force. A coin is tied to a string and swung in a circle over a person’s head. There is also a plastic tube attached to the string. The number of full rotations in a certain amount of time is counted, and the length of the string is measured as well. My question is, how would I go about finding the acceleration of gravity. I know that the centripetal force, (mv^2)/r, is equal to the force of gravity, F(g). I know the radius and the velocity, but what do I do for the mass, and how to I solve the overall problem for gravity? Is the mass just the mass of the coin, or the mass of the tube plus that of the coin?

Homework Equations

Centripetal force = (mv^2)/r
(mv^2)/r = F(c) = F(t) = F(g)
Mass of coin - 58.1 g
Mass of tube - 20.1 g

The Attempt at a Solution

One line of data
Using mass=coin + tube
20 revolutions in 18.91 s at a radius of 1.03 m
m = 58.1 + 20.1 = 78.2 g
r = 1.03 m
v = d/t
2pir = 2pi(1.03)/ 18.91 = 0.342 m/s
(78.2*.342^2)/1.03= 8.88 m/s^2
Pretty close to 9.8, but is this the right way to solve it?

 

[Doc Al]

I know that the centripetal force, (mv^2)/r, is equal to the force of gravity, F(g).

That won't work. For one thing, assuming you are swinging the coin in a horizontal circle, the centripetal acceleration (and force) act horizontally. But gravity acts vertically.

Instead, examine the forces acting on the object. Draw a free body diagram and apply Newton's 2nd law to both vertical and horizontal directions. Hint: The string isn't horizontal. What angle does it make?

I know the radius and the velocity, but what do I do for the mass, and how to I solve the overall problem for gravity? Is the mass just the mass of the coin, or the mass of the tube plus that of the coin?

If you are studying forces on the coin, then it's the mass of the coin that matters. But, as is often the case, you might not need the actual mass. Just call it 'm' and see if it drops out.