# Centripetal Acceleration question

1. Nov 1, 2009

### DavidDishere

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

Consider a conical pendulum that consists of a bob
on one end of a string of negligible mass with the other end of the
string attached to a point on the ceiling, as shown. Given the proper
push, this pendulum can swing in a circle at a given angle, maintaining
the same distance from the ceiling throughout its swing. If the mass of
the bob is , the length of the string is d, and the angle at which it
swings is θ, what is the speed (v) of the mass as it swings? [Hint: Find
the vertical and inward radial components of the string’s tension.]

2. Relevant equations

FR = (mv2)/r

3. The attempt at a solution

For $$\Sigma$$Fy I got FT = mg/cos θ

I said r = dsinθ

And for $$\Sigma$$FR I got FTdsin2 θ = v2

Plug in mg/cos θ for FT and sqrt(gd tanθ sin θ)= v

Is that right?

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2. Nov 1, 2009

### Delphi51

I don't follow your equation for the radial forces. You have the Ft*sinθ component equaling the centripetal force, don't you? Surely there has to be an m in it. Why is the sin squared?

3. Nov 2, 2009

### DavidDishere

At the start I have
FTsin θ = mv2/r

Put in dsin θ for r and move to the other side

FTsin2 θ = mv2

Put in mg/cos θ from the y forces for FT

mgdsin2 θ/cos θ = mv2

The masses cancel and 1 of the sin2/cos becomes tangent

gd tan θ sin θ = v2

sqrt(gdtan θ sin θ) = v

Did I do something wrong? r is equal to dsin θ right?

4. Nov 2, 2009

### ApexOfDE

:( from your image, I find that $$sin \Theta = \frac{d}{R}$$

5. Nov 2, 2009

### DavidDishere

sin is opp/hyp right? r would be opp and d would be hyp. sin θ = r/d So dsin θ = r

6. Nov 2, 2009

### Staff: Mentor

No, your solution is perfectly fine. And yes, r = d sinθ. (In your first post you did leave off an m, as Delphi51 points out.)