Need Help in Mechanics

A horizontal disc of mass MM and radius RR is pivoted along one of its diameter and is free to rotate about it.A mass mm falls through height hh and sticks perfectly inelastically to the disc at point AA. A spring of force constant kk is attached to point B of the disc as shown in figure. Assume A,BA,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=12kx2mgx+12Iw2 mgh=\dfrac{1}{2}kx^2-mgx+\dfrac{1}{2}Iw^2 (where xx is the distance through which the spring compresses and ww is angular velocity of disc)

Angular momentum conservation: m2ghR=Iwm\sqrt{2gh}R=Iw (since net torque zero initially)

I=mR2+MR24I=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
1 year, 5 months ago

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Your main problem is that the spring does not remain vertical as the disc rotates. Suppose that the mass mm makes an angle θ\theta with the horizontal. Then the length of the spring (assuming that its natural length is LL) is R2(1cosθ)2+(LRsinθ)2 \sqrt{R^2(1-\cos\theta)^2 + (L-R\sin\theta)^2} and hence conservation of energy tells us that 12k(L22RLsinθ+2R2(1cosθ)L)2mgRsinθ+12Iθ˙2 \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 12kR2θ2mgRθ+12Iθ˙2 \tfrac12kR^2\theta^2 - mgR\theta + \tfrac12I\dot{\theta}^2 is constant. Differentiating this gives you SHM in θ\theta.

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

Mark Hennings - 1 year, 5 months ago

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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?

Ram Avasthi - 1 year, 5 months ago

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

Mark Hennings - 1 year, 5 months ago

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