Work Done in slowly pulling a thread

[Bling Fizikst]

Homework Statement

ok

Relevant Equations

ok

Say some external agent does the above work $W_{\text{ext}}$ , say the inital depth was $h$ , then final depth $=\frac{h}{2}$ . Now , we don't know the radius of the new circular path , as well as the new velocity of the particle . Employing the work energy theorem : $$W_{\text{ext}}-mg(h/2)=m\frac{v^2}{2}-m\frac{v_{\circ}^2}{2}$$. I think i am unable to exploit the fact that the process was done slowly and the change in kinetic energy during that process $\approx 0$ . Also , since there doesn't seem to be any non zero external moment so we can write $$\triangle H_{O/I}=0$$ where $O$ is the center of the circular path at all times . THis means $$mv_{\circ} R=mvr$$ . Stuck here .

 

[kuruman]

Can you find an expression that relates the radius $r$ of the horizontal circle and the vertical "depth" $h$?
Hint: Draw a free body diagram of the bead.

 

[Bling Fizikst]

Can you find an expression that relates the radius $r$ of the horizontal circle and the vertical "depth" $h$?
Hint: Draw a free body diagram of the bead.

From the FBD , i assumed the angle between the thread and the vertical is $\theta$ . Now , writing the force equations : $$\frac{mv_{\circ}^2}{R}=T\sin\theta $$ $$mg=T\cos\theta$$ From here we can find : $$T=\frac{m\sqrt{v_{\circ}^4+R^2g^2}}{R}$$ and $$\tan\theta=\frac{v_{\circ}^2}{Rg}$$ . But writing it after the thread is pulled everything changes ,i.e, $\theta$ , $r$ . How do i deal with it?

 

[erobz]

You can eliminate $\tan \theta$ for an expression of $h,R$.

Use the inextensible sting relationship of some length $L$ to figure out an expression for the displacement of the string "at the hand" $y$

Then the work should be $\int T dy$ as $y$ goes from 0 to the value it has at $h/2$.

 

[kuruman]

But writing it after the thread is pulled everything changes ,i.e, $\theta$ , $r$ . How do i deal with it?

Like I said

Can you find an expression that relates the radius $r$ of the horizontal circle and the vertical "depth" $h$?

I see that @erobz agrees with my recommendation.

 

[erobz]

Like I said

I see that @erobz agrees with my recommendation.

Yeah, they didn't quite have the follow through you were looking for, so I thought I'd double down.

 

[haruspex]

From the FBD , i assumed the angle between the thread and the vertical is $\theta$ . Now , writing the force equations : $$\frac{mv_{\circ}^2}{R}=T\sin\theta $$ $$mg=T\cos\theta$$ From here we can find : $$T=\frac{m\sqrt{v_{\circ}^4+R^2g^2}}{R}$$ and $$\tan\theta=\frac{v_{\circ}^2}{Rg}$$ . But writing it after the thread is pulled everything changes ,i.e, $\theta$ , $r$ . How do i deal with it?

Intriguing question.
Can’t think of a sound argument, but I suspect the system would move so as to minimise the work done. However, that gives me a cubic (in $r^2$) to solve.

the process was done slowly and the change in kinetic energy during that process $\approx 0$

Why would that be? If the mass were on a frictionless table and h=0 the KE would increase.

 

[haruspex]

You can eliminate $\tan \theta$ for an expression of $h,R$.

Use the inextensible sting relationship of some length $L$ to figure out an expression for the displacement of the string "at the hand" $y$

Then the work should be $\int T dy$ as $y$ goes from 0 to the value it has at $h/2$.

How will you find T as a function of y?

 

[erobz]

How will you find T as a function of y?

I would get everything $(T,dy)$ in terms of $h$ via the inextensible rope constraint $y + f( h,R) = L$ ( $R$ and $h$ are related through the result for $\tan \theta$).

 

[haruspex]

I would get everything in terms of $h$ via the inextensible rope constraint $y + f( h,R) = L$.

Still don’t see how you will do that. T is a function of v.
For a given y, we can choose different values of r and solve to find $\theta$, v, T, whatever, without contradiction.
Maybe a calculus approach resolves it. Consider dy over which T can be taken as constant.

 

[erobz]

Still don’t see how you will do that. T is a function of v.
For a given y, we can choose different values of r and solve to find $\theta$, v, T, whatever, without contradiction.
Maybe a calculus approach resolves it. Consider dy over which T can be taken as constant.

I was taking the OP's result for $T$ and $\tan \theta$ to be correct? $T(R)$, $R(h)$, etc...

Solving that integral... I don't make any claims on having completed that. but it looks like it can be worked over with some constant manipulation ( and multiplying through by 1 in terms of factors of a function of $h$) to make the numerator in the expresson for $T(h)$ cancel with the denominator in the expression for $dy = g(h) dh$.

 

[Lnewqban]

Now , we don't know the radius of the new circular path , as well as the new velocity of the particle .

What could change the value of the tangential velocity of the bead?

 

[haruspex]

What could change the value of the tangential velocity of the bead?

As the radius diminishes the tangential speed must increase to conserve angular momentum. Think, spinning skater.

 

[Lnewqban]

As the radius diminishes the tangential speed must increase to conserve angular momentum. Think, spinning skater.

Isn't the angular velocity what must increase to compensate for the reduction of the moment of inertia?

 

[haruspex]

Isn't the angular velocity what must increase to compensate for the reduction of the moment of inertia?

Both increase, one linearly, one quadratically: $L=mrv=mr\omega^2$.

 

[Manish_529]

Homework Statement: ok
Relevant Equations: ok

View attachment 360891
Say some external agent does the above work $W_{\text{ext}}$ , say the inital depth was $h$ , then final depth $=\frac{h}{2}$ . Now , we don't know the radius of the new circular path , as well as the new velocity of the particle . Employing the work energy theorem : $$W_{\text{ext}}-mg(h/2)=m\frac{v^2}{2}-m\frac{v_{\circ}^2}{2}$$. I think i am unable to exploit the fact that the process was done slowly and the change in kinetic energy during that process $\approx 0$ . Also , since there doesn't seem to be any non zero external moment so we can write $$\triangle H_{O/I}=0$$ where $O$ is the center of the circular path at all times . THis means $$mv_{\circ} R=mvr$$ . Stuck here .

Here, we can assume that the radius and velocity of the circular path remain same because, in the question it has mentioned that only the depth of the circular path has decreased and nothing such about the circular path changing has been mentioned.

 

[Bling Fizikst]

Based on the above discussions , i wrote a general relation for $v,r,\alpha$ : $$T_x=T\sin\alpha=\frac{mv^2}{r}$$ $$T_y=T\cos\alpha =mg$$ From here , $$\tan\alpha=\frac{r}{y} =\frac{v^2}{rg}$$ $$\implies y\equiv y(r,v)=\frac{r^2g}{v^2}$$ Now , if i were to compute $$\int \vec{T}\cdot \vec{dy} =\int T_y dy=mg\frac{h}{2}$$ . Not sure what to make of this , all these are functions of $(r,v)$. And yeah , idt you use the $\tan\alpha$ relation as it changes corresponding to the change in $r$ and $y$ , it ain't fixed imo . @erobz

I actually then used conservation of angular momentum about $O$ : $$mv_{\circ}R=mvr\implies r=\frac{v_{\circ}R}{v}$$ Using this in $$\frac{h}{2}=\frac{r^2g}{v^2}$$ We can find $$v^2=\sqrt{2}v_{\circ}^2$$ From the work energy equation , i wrote first : $$W_{\text{ext}}= m\frac{v^2}{2}-m\frac{v_{\circ}^2}{2} +mg\frac{h}{2}$$ Using the fact that $h,v$ are known we substitute it's values in terms of the known parameters to find : $$W_{\text{ext}}=\frac{mv_{\circ}^2}{2}\left[\frac{g^2R^2}{v_{\circ}^4}+\sqrt{2}-1\right]$$

 

[Manish_529]

The limit shall be from h to h/2 because while calculating the work done by variable forces, we perform a line integral as the limit goes from the bodies initial position to the final position and not it's displacement

 

[Bling Fizikst]

The limit shall be from h to h/2 because while calculating the work done by variable forces, we perform a line integral as the limit goes from the bodies initial position to the final position and not it's displacement

You are right , but here $mg$ is a constant force .

 

[Manish_529]

You are right , but here $mg$ is a constant force .

yes but the exernal forces acting is variable and depends on y

 

[Manish_529]

Based on the above discussions , i wrote a general relation for $v,r,\alpha$ : $$T_x=T\sin\alpha=\frac{mv^2}{r}$$ $$T_y=T\cos\alpha =mg$$ From here , $$\tan\alpha=\frac{r}{y} =\frac{v^2}{rg}$$ $$\implies y\equiv y(r,v)=\frac{r^2g}{v^2}$$ Now , if i were to compute $$\int \vec{T}\cdot \vec{dy} =\int T_y dy=mg\frac{h}{2}$$ . Not sure what to make of this , all these are functions of $(r,v)$. And yeah , idt you use the $\tan\alpha$ relation as it changes corresponding to the change in $r$ and $y$ , it ain't fixed imo . @erobz

I actually then used conservation of angular momentum about $O$ : $$mv_{\circ}R=mvr\implies r=\frac{v_{\circ}R}{v}$$ Using this in $$\frac{h}{2}=\frac{r^2g}{v^2}$$ We can find $$v^2=\sqrt{2}v_{\circ}^2$$ From the work energy equation , i wrote first : $$W_{\text{ext}}= m\frac{v^2}{2}-m\frac{v_{\circ}^2}{2} +mg\frac{h}{2}$$ Using the fact that $h,v$ are known we substitute it's values in terms of the known parameters to find : $$W_{\text{ext}}=\frac{mv_{\circ}^2}{2}\left[\frac{g^2R^2}{v_{\circ}^4}+\sqrt{2}-1\right]$$

Here you have considered T_y to be equal to mg, but it's not, because if that had been the case, then the vertical ascent of the bead wouldn't have happened

 

[Bling Fizikst]

Here you have considered T_y to be equal to mg, but it's not, because if that had been the case, then the vertical ascent of the bead wouldn't have happened

I just used work energy theorem which includes $W_{\text{ext}}+W_{mg}=\triangle KE$ , all i need is $W_{\text{ext}}$ . Morever , i don't even think you can find tension in that way explicitly . So , yeah just ignore that work done by tension thing , it's useless imo .

 

[Manish_529]

I just used work energy theorem which includes $W_{\text{ext}}+W_{mg}=\triangle KE$ , all i need is $W_{\text{ext}}$ . Morever , i don't even think you can find tension in that way explicitly . So , yeah just ignore that work done by tension thing , it's useless imo .

Actually you also need to consider the tension force since, it's vertical component is performing work. I have posted the solution in one of my replies you can check it`

 

[haruspex]

Here, we can assume that the radius and velocity of the circular path remain same because, in the question it has mentioned that only the depth of the circular path has decreased and nothing such about the circular path changing has been mentioned.

That is unphysical.
In order to raise the mass the tension must be increased a little. Since the tension has a horizontal component, this will lead to a reduction in the radius and thus an increase in speed.
This becomes ever clearer as the mass approaches the roof.

 

[haruspex]

I would get everything $(T,dy)$ in terms of $h$ via the inextensible rope constraint $y + f( h,R) = L$ ( $R$ and $h$ are related through the result for $\tan \theta$).

I think you might be right after all.
For a given y, we have four unknowns: T, r, h and theta.
We have force equations in two directions, conservation of angular momentum and the geometric relationship between r, h, y and theta. So that should be enough to get T as a function of y.
Messy though.

 

[erobz]

$$ \frac{R}{h} = \frac{v_o^2}{R g} $$

$R$ is a function of $h$

Eliminate $R$ from $T$ in the following relationship you found.

$$T=\frac{m\sqrt{v_{\circ}^4+R^2g^2}}{R}$$

Inextensible Rope Constraint:

$$ y = L - \sqrt{R^2 + h^2} $$

Again, eliminate $R$. Take the derivative$\frac{dy}{dh}$:

When you put $T$ and $dy$ in the integrand you can massage the integrand to cancel the factors involving the root in the numerator of $T$, and the root in the denominator of $dy$.

The result looks pretty clean to me.

 

[erobz]

I now see that the expression for $T$ and $\tan \theta$ aren't correct in post 3. those should be instantaneous velocity of the bead $v$, not $v_o$.

$$ T = m\frac{ \sqrt{v^4 + R^2 g^2}}{R} $$

$$ \tan \theta = \frac{R}{h} = \frac{v^2}{Rg} $$

They did correct the expressions in post no. 17.

But I still think it's ok because of the conservation of angular momentum relationship $m v_o R_o = mv R$, I'm sure it is more manipulation. I haven't re-done the integral though, it might be a mess as @haruspex says.

 

[erobz]

By combining:

$$ T \sin \theta = m \frac{v^2}{r} $$

$$ T \cos \theta \approx mg $$

$$ mv_o r_o = m v r $$

$$y = L - \sqrt{h^2 + r^2}$$

I get the following relationships in terms of the instantaneous height $h$:

$$ r = \frac{\sqrt{g} v_o }{g} \left( h_o h \right)^{1/4}$$

$$ v^4 = v_o^4 h_o h^{-1} $$

Differentiate $y$ w.r.t. $h$ ( after substitution for $r$):

$$ dy = -\frac{1}{2} \left( h^2 +\frac{v_o^2}{g}( h_o h )^{1/2} \right)^{-1/2} \left( 2h + \frac{v_o^2 h_o^{1/2} }{2g} h^{-1/2} \right) dh $$

$$ T(h) = \frac{m \sqrt{ v_o^4 h_o h^{-1} + v_o^2g ( h_o h)^{1/2} )} }{ \frac{g^{1/2}}{g} v_o ( h_o h )^{1/4} } $$

The work would in going from $h_o$ to $h_o/2$:

$$ W = \int T dy = -\frac{1}{2} \int_{h_o}^{h_o/2} \frac{m \sqrt{ v_o^4 h_o h^{-1} + v_o^2g ( h_o h)^{1/2} } }{ \frac{g^{1/2}}{g} v_o ( h_o h )^{1/4} } \left( h^2 +\frac{v_o^2}{g}( h_o h )^{1/2} \right)^{-1/2} \left( 2h + \frac{v_o^2 h_o^{1/2} }{2g} h^{-1/2} \right) dh $$

So its a wildfire of an integral...

The question is, does this really coincide with the expected result for this exercise? Is the intended result?

From the work energy equation , i wrote first : $$W_{\text{ext}}= m\frac{v^2}{2}-m\frac{v_{\circ}^2}{2} +mg\frac{h}{2}$$ Using the fact that $h,v$ are known we substitute it's values in terms of the known parameters to find : $$W_{\text{ext}}=\frac{mv_{\circ}^2}{2}\left[\frac{g^2R^2}{v_{\circ}^4}+\sqrt{2}-1\right]$$

 

[Steve4Physics]

The question is, does this really coincide with the expected result for this exercise? Is the intended result?

FWIW I get the same answer as the OP gives in Post #17:
$W = \frac {mv_0^2}{2}\left[\frac{g^2R^2}{v_0^4}+\sqrt{2}-1\right]$

No calculus was needed. Just some algebra based on three equations:

Initial: $\frac {v_0^2}{gR} = \frac Rh$ (because both sides equal $\tan \theta_0$)

Final: $\frac {v^2}{gr} = \frac r{(h/2)}$ (because both sides equal $\tan \theta$)

$Rv_0 = rv$ (from conservation of angular momentum)

A calculus based approach should, of course, give the same result but looks much harder.

Various edits to correct Latex - it's not rendering correctly on my PC.