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# Physics Question!!!

A block of mass $$m$$ is attached to a spring and kept in horizontal position in a vertical plane. When the block is allowed to then, find the maximum extension in the spring.

Details--

1) Spring's constant=$$k$$

2) Natural length of spring= $$l$$

img

I was trying to make a problem using some concepts of Work Energy and Circular motion and I came up with this. Here is my approach, please tell me if I am right or not.

First of all, the extension in the spring will be maximum when radial velocity ($$v_{r}$$) will be zero. Now the work is only done by conservative forces in this situation so we can conserve energy.

$$mg(l+x)cos \theta=\frac{mV_{0}^{2}}{2}+\frac{kx^{2}}{2}$$

Now $$V_{0}^{2}=V_{t}^{2}+V_{r}^{2}$$

The extension will be maximum when $$V_{r}=0$$. So this does not implies that the velocity of the block is zero at the lower most position. Also we can see that only a component of $$mg$$ is acting in the tangential direction which is responsible for tangential velocity and tangential acceleration. So $$V_{t}$$ is not going to be zero at the lower most position.

If we consider torque about point $$O$$ then

$$mgsin\theta (l+x)=\frac { dL }{ dt }$$

$$mgsin\theta (l+x)=\frac { dI }{ dt } \omega +I\frac { d\omega }{ dt }$$

$$I=m{ (l+x) }^{ 2 }$$

$$\frac { dI }{ dt } =m(2x{ v }_{ r }+2l{ v }_{ r })$$

$$mgsin\theta (l+x)=2m(x+l){ v }_{ r }\omega +I\frac { d\omega }{ dt }$$

I got this equation. There is a term of $$v_{r}$$ in the equation that has to related with $$\omega$$ or $$\theta$$ to solve the problem.

Although the motion is not circular but if consider the motion of the block for a very small interval of time then we can assume that it's a circular motion so $$v_{t}=(l+x) \omega$$. Can we do so?

Using this we can express $$V_{r}$$ in term of angular velocity. But still there is term containing $$x$$ there. How can this be solved?

Am I overlooking some thing very obvious? Because using calculus in spring problem often gets very complicated.

Note by Satvik Pandey
2 years, 2 months ago

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I think you cannot assume that the motion is circular, however small the interval you take. I for one think that maximum extension will occur at bottom-most position. · 2 years, 2 months ago

@satvik pandey Check out the 2-D spring simulation on this page. · 2 years, 2 months ago

Thanks for the link! This proves that the extension is maximum at the bottom most position. :D

But what about the calculation and explanation?? · 2 years, 2 months ago

Is that simulation removed. · 2 years, 2 months ago

You have to scroll down page to see it. · 2 years, 2 months ago

Even if the motion is not completely circular then I think we can use $$V=R \omega$$. Like we did here. Can we ??

I too thought that the extension will be maximum at the bottom most position. But when I thought about that after watching Kushal comment, deeply then I don't see a way to justify that. :( · 2 years, 2 months ago

Satvik Please answer my to my comment in this · 2 years, 2 months ago

The total velocity is sum of the squares of radial and tangential velocities, just like x and y components in Cartesian co-ordinates.

Then use $$V_{t}=R \omega$$. :) · 2 years, 2 months ago

What is $$V_{\phi}$$ ? What is it responsible for ? · 2 years, 2 months ago

It is radial velocity. It is responsible for the radial movement of the particle. · 2 years, 2 months ago

Thank you. · 2 years, 2 months ago

You are welcome!!! :) · 2 years, 2 months ago

In the sketch both $$V_r$$ look the same. · 2 years, 2 months ago

My handwriting is not very good. Velocity of the block along the spring is $$v_{r}$$ and velocity of the block tangential to the spring is $$v_{t}$$. · 2 years, 2 months ago

There are two $$V_r$$s ? Why should radial velocity be zero in all positions ? · 2 years, 2 months ago

$$V_{r}$$ and $$v_{r}$$ both represents radial velocity. In order to achieve maximum extension in the spring the radial velocity of the block should be zero at a particular point. It is not zero through out the motion. · 2 years, 2 months ago

We can use $$g sin \theta =da_{t}/dt$$ and substitute $$a_{t}=(l+x) \alpha$$ by approximation but still we get the same situation. · 2 years, 2 months ago

I think that should be $$\frac{dv_r}{dt} = gcos\theta-kx/m$$

And $$\frac{dv_t}{dt} = gsin \theta$$

You should go with energy conservation , thats easier · 2 years, 2 months ago

I wrote $$\frac { { da }_{ t } }{ dt }$$ by mistake.

I tried to use energy conservation. But here entire gravitational energy in not getting converted to spring potential energy only. A part of it is also converted to kinetic energy. I think the extension will be maximum when the radial velocity would be zero.right? But this does not implies that tangential velocity would also be zero at that moment.

Let $$x$$ be tha maximum extension in the spring then-

$$mg(l+x)=\frac{mv_{t}^{2}+kx^{2}}{2}$$. There are two variables $$v_{t} and x$$. And we have only one equation.

What do you think? @Kushal Patankar · 2 years, 2 months ago

Cetripetal accleration equation will give the relation between $$v_t$$ and x

%6

I made Vr zero. · 2 years, 2 months ago

How could I miss that!!!!!!!!!!!

Thank you Kushal for helping. · 2 years, 2 months ago

I have one more confusion.

The motion of the block is not a pure circular motion which increases the intricacy of the problem.

I think we can not just equate $$kx$$ and $$mgcos \theta$$ to centripetal force. There is acceleration other that centripetal acceleration in radial direction because the block is performing radial motion also.

I think the equation should be

$$mgcos\theta -kx=m\left\{ \frac { d\vec { { V }_{ r } } }{ dt } -\frac { { V }_{ t }^{ 2 } }{ (l+x) } \right\}$$

Taking radially outward as +ve direction.

What do you think, Kushal?? · 2 years, 2 months ago

What is the need for that equation , I mean that mechanical energy conservation does all our work. And I think that the equation u made is correct for intermediate radial acceleration. · 2 years, 2 months ago

How you got the second equation??

The radial velocity is zero at the lower most position but this doesn't implies that radial acceleration $$\frac { \vec { { dV }_{ r } } }{ dt }$$ is also zero at that point. Am I missing some thing very obvious? :( · 2 years, 2 months ago

@satvik pandey i suppose whatever you have said till now is correct..!!
It is a case of Coriolis force.. When the particle has radial speed as well then the acceleration is written as

Therefore we cannot write the equation of centripetal force as discussed before..!! · 2 years, 1 month ago

Could you post whole analysis(solution) of the problem (along with calculation). · 2 years, 1 month ago

Thank you ! Sir. Sorry for delay in replying. You got equation using polar co-ordinates. right?