# 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
9 months, 2 weeks 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).

- 9 months, 2 weeks 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?

- 9 months, 2 weeks 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.

- 9 months, 2 weeks ago

- 9 months, 2 weeks ago