# Centripetal Force - trying to find time

1. Dec 16, 2012

### cdornz

1. The problem statement, all variables and given/known data
A physicist is standing 10 feet from the center of a platform that can rotate about a vertical axis - the platform is horizontal. She is holding a weight that is hanging freely from a string. The platform starts rotating and eventually she notices that the weight is no longer hanging straight down but is making an angle of 10° with the vertical. How fast is she moving at that time?

2. Relevant equations
Normally, for centripetal equations I would use a=v2/R
I've also used the equation Fcentripetal=4∏2Mf2R

3. The attempt at a solution
I've never had to solve a centripetal force equation with knowing the mass of either the person or the weight or without a velocity. I'm honestly not quite sure what route to go to solve this problem.

2. Dec 16, 2012

### ap123

Draw a free-body diagram for the weight.
Once you take x- and y-components, you should be able to cancel the mass out.

3. Dec 16, 2012

### cdornz

So I'm not sure if I drew this correctly or not, but drawing out the free body diagram for the weight, would those components cancel out with the tension of the string?

4. Dec 16, 2012

### ap123

It should be clear from your diagram that the y-component of the tension cancels out the weight.
The x-component of the tension will be the only horizontal force on the object - this provides the centripetal force.

Have you written out the components? They should simplify down to a simple expression.

5. Dec 17, 2012

### cdornz

For components, I would have:

Ftension; the y-component pointing upwards canceled out with the weight of the mass pointing downwards. There is no Ffriction since the object is in the air. Otherwise, I'm not sure what other forces.

$\Sigma$Fx = Fnet
Ftension = ma

Problem I run into with this, is that I don't have a way of calculating the mass or the acceleration. All I know is that the angle of the mass is 280° from the +x axis.

6. Dec 17, 2012

### ap123

So, for the y-components, you have
Tcos(10°) - mg = 0
→ Tcos(10°) = mg

and for the x-components,
Tsin(10°) = ma
where a is the centripetal acceleration

If you insert the expression for centripetal acceleration and combine these 2 equations, you can get rid of T and m.

7. Dec 17, 2012

### Staff: Mentor

Sneaky approach:

If you consider the string and weight to be stationary in a rotating frame of reference (think of taking a snapshot of the rotating platform at some instant of time when you have a profile view of the happenings) then it becomes fair to speak about centrifugal forces and accelerations in that frame of reference. It turns out that these centrifugal effects have the same magnitude but opposite direction to our usual centripetal ones. Then, using Newton's second law f = ma, we can drop the constant m's and label a diagram with accelerations rather than forces:

Does this give you any ideas about solving for ac?

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8. Dec 17, 2012

### cdornz

So I went with this and worked with the axes a bit to make it make more sense to me. I ended up with this as a result:

$\Sigma$Fx=max
Tx+(mg)x = max
T(cos100°) +mg(cos270°) = ma(cos180°)
T(cos100°) = -ma
T(cos100°) = -mv2/R

T(sin100°) - mg = 0
T(sin100°)/g = m

T(cos100°) = -T(sin100°)/g * v2/R

My final equation was v2 = (T(cos100°))gR/-T(sin100°)
v2= -55.56t/-0.98t = 56.47 which I then took the square root of and so the velocity ended up being 7.51ft/s.