Centripetal motion problem

1. Nov 2, 2005

BrimmZERO

"A mass of 3.900 kg is suspended from a 1.450 m long string. It revolves in a horizontal circle.
The tangential speed of the mass is 3.247 m/s. Calculate the angle between the string and the vertical (in degrees)."

Here is a diagram, labelling angle theta I'm supposed to solve for: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype11/prob03_pendulum.gif

I've deduced so far that: sin(theta) = opp/hyp

In this case, opp = radius and hyp = length of string (L) or (1.450 m)

So sin(theta) = r/L

In turn, r can be solved by means of Fc = mv^2/r, as r = mv^2/Fc

so sin(theta) = mv^2/FcL

M is given, V is given, and L is given. Fc I'm sort of puzzled on, since I can't use the MV^2/R equation again. Any clues?

Last edited by a moderator: Apr 21, 2017
2. Nov 2, 2005

Staff: Mentor

Hint: What provides the centripetal force?
Hint: Consider both vertical and horizontal forces acting on the mass.

3. Nov 2, 2005

BrimmZERO

Okay, I think I'm getting somewhere. Would it be right to say the Fc is equal to the x-factor of the weight of the mass on the string? I could say this therefore:

Fc = mgsin(theta)

and then

sin(theta) = mv^2/mgsin(theta)L

which would cancel mass, giving me:

sin(theta) = v^2/gsin(theta)L

would that be right? or is mass even significant in this problem?

4. Nov 3, 2005

Staff: Mentor

Not right. For one thing, the weight of the mass acts down--it has no horizontal component.

Try this: Identify all the forces acting on the mass. (There are two forces.) Then consider horizontal and vertical components.