K&K Question 3.5 - Mass and Axle

[Radarithm]

Homework Statement

A mass m is connected to a vertical revolving axle by two strings of length l, each making an angle of 45 degrees with the axle, as shown. Both the axle and mass are revolving with constant angular velocity ω. Gravity is directed downward.

(a) Draw a clear force diagram for m
(b) Find the tension in the upper string, $T_{up}$ and the tension in the lower string, $T_{low}$.

Homework Equations

$ma=mr\omega^2=ml\omega^2\sin{\varphi}$
Where phi is 45 degrees.

The Attempt at a Solution

Image: http://gyazo.com/8626cda317d6dca7dc1bbe751b643247
FBD: http://postimg.org/image/tez986p19/

The equations for the x and y directions, respectively:
$T_u\sin{\varphi}-T_d\sin{\varphi}=ml\omega^2\sin{\varphi}$
$T_u\cos{\varphi}-mg-T_d\cos{\varphi}=ml\omega^2\sin{\varphi}$

Solving for the tension in the upper cord, I get this:
$T_u=-mg$

Assuming that the tangent of 45 is 1. Am I on the right track or am I making a mistake right now? If it has to do with reference frames, I wouldn't know what to do; I know that:
$F_{apparent}=F_{true}+F_{fictitious}$
I don't know how to apply it though. Should I continue solving for the tension in the lower rope, or do I need to correct something?

 

[haruspex]

$T_u\sin{\varphi}-T_d\sin{\varphi}=ml\omega^2\sin{\varphi}$

Check the signs

$T_u\cos{\varphi}-mg-T_d\cos{\varphi}=ml\omega^2\sin{\varphi}$

Check the term on the right hand side. Is this a cut-and-paste error?

 

[Radarithm]

Check the signs
Check the term on the right hand side. Is this a cut-and-paste error?

You're right. Fixed it and got the equation:
$$T_u=\frac{m\sin{\varphi}(\ell\omega^2+g)}{1+\sin{\varphi}}$$

When I apply this to the hint: If $\ell\omega^2=\sqrt{2}g$ then $T_u=\sqrt{2}mg$.

All that's left is solving for the other tension. Thanks for the help.
It's funny how easy it is to make dumb mistakes :tongue:

 

[Tanya Sharma]

You're right. Fixed it and got the equation:
$$T_u=\frac{m\sin{\varphi}(\ell\omega^2+g)}{1+\sin{\varphi}}$$

This is not the correct value of tension in the upper string.

When I apply this to the hint: If $\ell\omega^2=\sqrt{2}g$ then $T_u=\sqrt{2}mg$.

If you apply the hint i.e ω²l =√2g in the expression of T_u you have obtained,you get mg not √2mg

 

[Radarithm]

This is not the correct value of tension in the upper string.



If you apply the hint i.e ω²l =√2g in the expression of T_u you have obtained,you get mg not √2mg

In fact I get 2mg when using a calculator. I'm currently working on it. Am I off with the forces or the algebra?

 

[Tanya Sharma]

Did you fix the errors pointed out by haruspex in post#2 ?

If yes ,what are the two equations you get ?

 

[Radarithm]

I did, and here's what I got:
$$T_u\sin{\varphi}+T_d\sin{\varphi}=m\ell\omega^2\sin{\varphi}$$
$$T_u\cos{\varphi}-mg-T_u\cos{\varphi}=0$$

 

[Tanya Sharma]

I did, and here's what I got:
$$T_u\sin{\varphi}+T_d\sin{\varphi}=m\ell\omega^2\sin{\varphi}$$

Correct

$$T_u\cos{\varphi}-mg-T_u\cos{\varphi}=0$$

You have denoted both the tensions by T_u .

 

[Radarithm]

Correct
You have denoted both the tensions by T_u .

Sorry, typing error.
So what am I doing wrong here? There is no centripetal acceleration in the y direction, is there? The mass is rotating in a horizontal plane.

 

[Tanya Sharma]

Sorry, typing error.

Okay.But what is the correct equation ?

 

[Radarithm]

No problem.But what is the correct equation ?

$$T_u\cos{\varphi}-mg-T_d\cos{\varphi}=0$$

 

[Tanya Sharma]

Good .

So now you have got two equations .You just need to be more careful while doing algebra . Go slowly .

Replace sinø and cosø with 1/√2 . What is the value of T_u you get?

 

[Radarithm]

Here it goes:
$$\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0$$
$$T_d=T_u-mg$$

Substituting $T_d$ into the next equation, we get:
$$\frac{T_u}{\sqrt{2}}+\frac{T_u-mg}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}$$
Solving for $T_u$, I finally get the following:
$$T_u=\frac{m(\ell\omega^2+g)}{2}$$

I still get the wrong answer which is off by 4.
I've tried another way but I get $\sqrt{2}g$ instead of $\sqrt{2}mg$
and another substitution also didn't do the trick.

 

[Tanya Sharma]

Here it goes:
$$\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0$$
$$T_d=T_u-mg$$

Please recheck .

 

[Radarithm]

Please recheck .

I got it:
$$T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}$$

This is what I get for not practicing. Thanks for the help!

 

[Tanya Sharma]

I got it:
$$T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}$$

This is incorrect.

 

[Radarithm]

This is incorrect.

How? I applied the hint using a calculator with the values m = 2 and g = 9.8
I got something around 27.718 which is equal to the root of two times 2 times 9.8.

 

[jbunniii]

I just did this problem last week, and looking at my answer vs. yours, I'm having trouble understanding how you got $1 + \sqrt{2}$ in the denominator. I don't think it's right. Can you show your calculation leading up to the final answer?

 

[Tanya Sharma]

$$T_u-T_d=\sqrt{2}mg$$

$$T_u+T_d=mω^2l$$

Just add the two equations . What do you get ?

 

[Radarithm]

I just did this problem last week, and looking at my answer vs. yours, I'm having trouble understanding how you got $1 + \sqrt{2}$ in the denominator. I don't think it's right. Can you show your calculation leading up to the final answer?

$$\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0$$
$$T_d=T_u-\sqrt{2}mg$$
$$\frac{T_d}{\sqrt{2}}+\frac{T_d}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}$$
$$T_u+\sqrt{2}T_u-2mg=m\ell\omega^2$$
$$T_u(1+\sqrt{2})=m(\ell\omega^2+2g)$$
$$T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}$$

 

[Radarithm]

$$T_u-T_d=\sqrt{2}mg$$

$$T_u+T_d=mω^2l$$

Just add the two equations . What do you get ?

$$m(\sqrt{2}g+\omega^2\ell)=0$$

 

[jbunniii]

$$\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0$$
$$T_d=T_u-\sqrt{2}mg$$
$$\frac{T_d}{\sqrt{2}}+\frac{T_d}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}$$

Agree so far, except on the last line, one of those $T_d$'s should be $T_u$.

$$T_u+\sqrt{2}T_u-2mg=m\ell\omega^2$$

How did you get this from the above?

 

[Tanya Sharma]

$$\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0$$
$$T_d=T_u-\sqrt{2}mg$$
$$\frac{T_d}{\sqrt{2}}+\frac{T_d}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}$$
$$T_u+\sqrt{2}T_u-2mg=m\ell\omega^2$$
$$T_u(1+\sqrt{2})=m(\ell\omega^2+2g)$$
$$T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}$$

It should be $T_u+T_u-\sqrt{2}mg=m\ell\omega^2$

 

[Tanya Sharma]

$$m(\sqrt{2}g+\omega^2\ell)=0$$

No...that should be $$2T_u=m(\sqrt{2}g+\omega^2\ell)$$.

Now divide both sides of the equation by 2 .You get the value of T_u.

 

[Radarithm]

Agree so far, except on the last line, one of those $T_d$'s should be $T_u$.

How did you get this from the above?

It was a typo.
And I recognized that dumb mistake (I multiplied the entire thing by the root of 2). It should have went like this:
$$2T_u-\sqrt{2}mg=m\ell\omega^2$$
$$T_u=\frac{m(\ell\omega^2+g\sqrt{2})}{2}$$
This should be correct.

 

[jbunniii]

$$T_u=\frac{m(\ell\omega^2+g\sqrt{2})}{2}$$
This should be correct.

I got the same answer, so either we're both right or both wrong.

 

[Tanya Sharma]

$$2T_u-\sqrt{2}mg=m\ell\omega^2$$
$$T_u=\frac{m(\ell\omega^2+g\sqrt{2})}{2}$$
This should be correct.

:thumbs: