Centripetal Acceleration question

[DavidDishere]

Homework Statement

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.]

Homework Equations

F_R = (mv²)/r

The Attempt at a Solution

For \SigmaF_y I got F_T = mg/cos θ

I said r = dsinθ

And for \SigmaF_R I got F_Tdsin² θ = v²

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

Is that right?

 

[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?

 

[DavidDishere]

At the start I have
F_Tsin θ = mv²/r

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

F_Tsin² θ = mv²

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

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?

 

[ApexOfDE]

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

 

[DavidDishere]

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

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

 

[Doc Al]

sqrt(gdtan θ sin θ) = v

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

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.)