# Need Help in Mechanics A horizontal disc of mass $M$ and radius $R$ is pivoted along one of its diameter and is free to rotate about it.A mass $m$ falls through height $h$ and sticks perfectly inelastically to the disc at point $A$. A spring of force constant $k$ is attached to point B of the disc as shown in figure. Assume $A,B$ are diametrically opposite points.Assume no heat dissipation ans the spring is initially in relaxed state.

I created this situation and I wish to examine it as practice. Can anyone please verify these equations?

Energy conservation: $mgh=\dfrac{1}{2}kx^2-mgx+\dfrac{1}{2}Iw^2$ (where $x$ is the distance through which the spring compresses and $w$ is angular velocity of disc)

Angular momentum conservation: $m\sqrt{2gh}R=Iw$ (since net torque zero initially)

$I=mR^2+\dfrac{MR^2}{4}$

Can somebody please improvise this and find more results? I think it may exhibit simple harmonic motion too (not able to prove)... Note by Ram Avasthi
2 years, 1 month ago

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Your main problem is that the spring does not remain vertical as the disc rotates. Suppose that the mass $m$ makes an angle $\theta$ with the horizontal. Then the length of the spring (assuming that its natural length is $L$) is $\sqrt{R^2(1-\cos\theta)^2 + (L-R\sin\theta)^2}$ and hence conservation of energy tells us that $\tfrac12k\big( \sqrt{L^2 - 2RL\sin\theta + 2R^2(1-\cos\theta)} - L\big)^2 - mgR\sin\theta + \tfrac12I\dot{\theta}^2$ is constant. For small oscillations, this means that $\tfrac12kR^2\theta^2 - mgR\theta + \tfrac12I\dot{\theta}^2$ is constant. Differentiating this gives you SHM in $\theta$.

You will get SHM, but only as an approximation for small values of $h$ (so that the oscillations are small).

- 2 years ago

Oh right, I indeed forgot that spring won't remain vertical. Thanks for correcting. Also, what do you reckon with the angular momentum? Is it conserved?

- 2 years ago

Conservation of angular momentum is fine, as you set it out, to describe the initial motion. As I said, though, either $h$ will be very small so that you only get small oscillations, or else you will need to solve the general differential equation in $\theta$ numerically - it is not likely that it will have a nice exact solution.

- 2 years ago

- 2 years, 1 month ago