# Spinning Mass

## Homework Statement

A mass of 3.50 kg is suspended from a 1.59 m long string. It revolves in a horizontal circle as shown in the figure.
The tangential speed of the mass is 2.99 m/s. Calculate the angle between the string and the vertical.

a_c = v^2/r
F = m*a_c
tanθ=sinθ/cosθ
sin^2θ=1-cos^2θ

## The Attempt at a Solution

Vertical: T*cos(θ)-m*g=0 → T=(m*g)/cos(θ)
Horizontal: T*sin(θ) = m*a_c
Combined the two: ((m*g)/cos(θ))sin(θ)=m*a_c → m*g*tan(θ)=m*a_c → g*tan(θ)=a_c
we know that a_c=v^2/r, plug that in to get g*tan(θ)=v^2/r
r=L*sin(θ) → g*tan(θ)=v^2/(L*sin(θ))

I cannot get past this point and the notes on the problem say that a quadratic equation needs to be solved at some point.

#### Attachments

• prob03_pendulum.gif
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Nathanael
Homework Helper

## Homework Statement

A mass of 3.50 kg is suspended from a 1.59 m long string. It revolves in a horizontal circle as shown in the figure.
The tangential speed of the mass is 2.99 m/s. Calculate the angle between the string and the vertical.

a_c = v^2/r
F = m*a_c
tanθ=sinθ/cosθ
sin^2θ=1-cos^2θ

## The Attempt at a Solution

Vertical: T*cos(θ)-m*g=0 → T=(m*g)/cos(θ)
Horizontal: T*sin(θ) = m*a_c
Combined the two: ((m*g)/cos(θ))sin(θ)=m*a_c → m*g*tan(θ)=m*a_c → g*tan(θ)=a_c
we know that a_c=v^2/r, plug that in to get g*tan(θ)=v^2/r
r=L*sin(θ) → g*tan(θ)=v^2/(L*sin(θ))

I cannot get past this point and the notes on the problem say that a quadratic equation needs to be solved at some point.

You wrote r in terms of θ and proceeded to try to solve for θ. It left to a bit of a dead end, so now try to solve for r

See if you can write tan(θ) in terms of known quantites and r (it can be done) then you'll be able to solve it

• 1 person
TSny
Homework Helper
Gold Member
Good work so far. Continue from your last equation. Try isolating the trig functions on one side of the equation and all the constants on the other side. Let x = cosθ and see if you can get a quadratic equation in x.

• 1 person
Nathanael
Homework Helper
Good work so far. Continue from your last equation. Try isolating the trig functions on one side of the equation and all the constants on the other side. Let x = cosθ and see if you can get a quadratic equation in x.

I don't think that will work. The equation becomes:

$\frac{v^{2}}{gL}=\frac{sin^{2}(θ)}{cos(θ)}$

There is another way to solve for θ (involving the quadratic equation) if you solve for r and then use r to find θ (once you know r, θ is trivial)

EDIT: actually you're right you could solve it that way... (solution removed) sorry about that you are correct

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I don't think that will work. The equation becomes:

$\frac{v^{2}}{gL}=\frac{sin^{2}(θ)}{cos(θ)}$

I understand where this equation comes from, but I am not sure how you would go about solving that for θ.

There is another way to solve for θ (involving the quadratic equation) if you solve for r and then use r to find θ (once you know r, θ is trivial)

I don't understand at all what I would use to solve for r or how this would result in a quadratic equation.
I am struggling with every aspect of this problem.

Nevermind!!!! I figured it out, thank you all so much for you guidance in this problem.

Nathanael
Homework Helper
I know you said you figured it out now, but I'll still show you the other methods.

I understand where this equation comes from, but I am not sure how you would go about solving that for θ.
You could rewrite $sin^2(θ)$ as $1-cos^2(θ)$ (from pythagorean theorem) and then the equation becomes:
$\frac{v^2}{gL}=\frac{1-cos^2(θ)}{cos(θ)}=\frac{1}{cos(θ)}-cos(θ)$
Multiply both sides by cos(θ) and rearrange and you'll get a quadratic equation (with x=cos(θ)) which can be solved for cos(θ):
$cos^2(θ)+\frac{v^2}{gL}cos(θ)-1=0$
There will only be one root with an absolute value of less than one, so simply take the arccosine of it and you'll have θ
I don't understand at all what I would use to solve for r or how this would result in a quadratic equation.
You had written the equation $g*tan(θ)=\frac{v^2}{r}$
Since $L^2=r^2+y^2$ (where y is the unknown side of the triangle) you could use the equation $L^2=r^2+(r*cot(θ))^2$
(... r*cot(θ) is the unknown side ...) to rewrite $tan(θ)$ as $\sqrt{\frac{r^2}{L^2-r^2}}$
Plug that expression into the "tan(θ)" part of your expression (and then square both sides and rearrange) and you'll get a quadratic equation (with $x=r^2$) which you can solve for $r^2$
The quadratic equation would be:
$g^2r^4+v^4r^2-v^4L^2=0$
There will be only 1 positive root, so take the square root of that to find r, then take the arcsin of (r/L) to find θ

Using either method, you'll find θ=41.1° (I double checked)

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