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= \(l\)

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.

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## Comments

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TopNewest@Mvs Saketh @Ronak Agarwal @Shashwat Shukla @karan shekhawat @Kushal Patankar @Sudeep Salgia @Tanishq Varshney @Raghav Vaidyanathan @Gautam Sharma @Azhaghu Roopesh M @Everyone . Please reply.

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@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..!!

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Yes Carioles component comes in to when the sliding part is curved. Can this problem be transformed in a such a mechanism ?

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

Sir, could you please help me here.

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Could you post whole analysis(solution) of the problem (along with calculation).

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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.

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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

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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

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%6

I made Vr zero.

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Thank you Kushal for helping.

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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??

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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? :(

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There are two \(V_r\)s ? Why should radial velocity be zero in all positions ?

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\(V_{r}\) 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.

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In the sketch both \(V_r \) look the same.

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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 \(v_{t}\).

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Thank you.

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You are welcome!!! :)

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Satvik Please answer my to my comment in this

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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 \(V_{t}=R \omega\). :)

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What is \(V_{\phi}\) ? What is it responsible for ?

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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.

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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. :(

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@satvik pandey Check out the 2-D spring simulation on this page.

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

But what about the calculation and explanation??

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Try to use lagrangian mechanics. Find the total kinetic energy and take the total potential energy and apply the E-L equations. There are too many forces to equate. A spring pendulum is a hard mechanical system to crack.

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There is a good video on youtube comparing Newtonian and Lagrangian mechanics on the example of the elastic pendulum, which the example here is.

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