# Homework Help: Circular Motion, an object attached by a string to a motor

1. Oct 24, 2016

### SDTK

1. The problem statement, all variables and given/known data

A pendulum traveling at constant speed along a circular path
Determine the angle that the string makes with the vertical.

T
he length of the string was measured to be 60 cm.
The period was measured to be 1.41 s
The frequency was found to be 0.709

2. Relevant equations

Is it possible to find the radius with only the Period T and frequency f?

3. The attempt at a solution

I believe that it should be possible to use the period and frequency to find the velocity, and from there the radius. Once I have the radius, I can find the angle.

Using:
v = 2(pi)r / T, v = 2 (pi)fr,
v = (2 x pi x radius)/period, v = 2 x pi x frequency x radius

a = v^2/r = (2(pi)f)^2r =(2(pi)/T)^2 r
a = velocity squared / radius = (2 x pi x frequency) squared x radius
= ((2 x pi) / T))squared x radius

I have tried solving for "r" and substituting the solution in order to find velocity, .... but I end up circling back the an equation with both the two unknown variables that I am trying to find, v and r.

D.T.

2. Oct 24, 2016

### Staff: Mentor

How about drawing a free body diagram of the pendulum? Identify all forces acting and apply Newton's 2nd law.

3. Oct 24, 2016

### Staff: Mentor

Referring to the thread title, I'm not seeing where the motor comes in....

4. Oct 24, 2016

### Staff: Mentor

I'm guessing the string of the pendulum is tied to some spinning axis--driven by the motor.

5. Oct 24, 2016

### Staff: Mentor

Ah. Alrighty then.

6. Oct 24, 2016

### SDTK

Air resistance is not considered

I understand that that force (net) = ma = (mv^2)/r
and that mass "cancels out" so Force(net) = a = v^2 / r.

Velocity is not known. Neither is the radius, or angle

(the length of the string providing the tension, the period and the frequency are known.)

I thought that I could solve for velocity, and from velocity, solve for radius, ... but am now wondering if there is enough info to so do

Last edited: Oct 24, 2016
7. Oct 24, 2016

### Staff: Mentor

Apply Newton's 2nd law to vertical and horizontal force components separately.

8. Oct 24, 2016

thank you