A Hint Question to Robo-space volley

Classical Mechanics Level 5

I had posted a question Robo-space Volley probably a month ago but found that not many could solve it. This is a question but simpler one on the same principle. A volleyball can be molded as a non-stretchable but flexible spherical envelope of mass m=260g and radius R=20cm filled with air at excess pressure △ P = 13 a t m \triangle P=13atm This is a hypothetical volleyball. The excess pressure remains unchanged with small deformations in the volleyball when it is hit or when it strikes some rigid surface. Mass of the air inside the volleyball can be neglected. Such a volleyball strikes the wall and bounces back without loosing speed let the speed of the volleyball be v=3m/s. How long will the volleyball remain in contact with the wall. If the time taken is x then find units digit of [1000x] where [.] represents GIF


The answer is 2.

Milun Moghe by Milun Moghe , 25, India

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1 solution

Milun Moghe
Mar 15, 2014

When the volleyball gets deformed it takes the shape of a dome with circular base. the restoring force is given by

F = m a = A △ P = π ( 2 R x − x ) 2 △ P = − ω 2 x F=ma=A\triangle P=\pi(\sqrt{2Rx-x})^{2}\triangle P=-\omega^{2}x

t = π ω = π m 2 R △ P t=\frac{\pi}{\omega}=\sqrt{\frac{\pi m}{2R\triangle P}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

But, you apply F = m a F = ma on center of mass. You will first have to find the position of center of mass. I actually got △ x c m = x 2 \displaystyle \triangle x_{cm} = \frac{x}{2} , using standard integration technique and hence the answer as 1 2 π × m R △ P \displaystyle \frac{1}{2} \sqrt{\frac{\pi \times m}{R \triangle P}} .

jatin yadav - 7 years, 2 months ago

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I didn't understand how you got x/2 will you please elaborate give some explanation. Not offencive

Milun Moghe - 7 years, 2 months ago

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Consider a ring at an angle θ \theta from the center(not the center of mass, but the center if the dome is completed in a sphere.) Find the distance of center of mass from the center using x c m = ∫ ϕ π R cos ⁡ θ × σ × 2 π R 2 sin ⁡ θ d θ σ × 4 π R 2 x_{cm} = \frac{\int_{\phi}^{\pi} R \cos \theta \times \sigma \times 2 \pi R^2 \sin \theta d \theta}{\sigma \times 4 \pi R^2}

= − R 4 sin ⁡ 2 ϕ - \frac{R}{4} \sin^2 \phi .

Here, sin ⁡ ϕ = 2 R x − x 2 R \sin \phi = \frac{\sqrt{2Rx - x^2}}{R}

Hence, x c m ≈ − x 2 x_{cm} \approx -\frac{x}{2} .

But the center is itself R − x R-x from the wall.

Hence, distance of cm from wall = R − x + x 2 = R − x 2 R-x+\frac{x}{2} = R - \frac{x}{2}

Hence, △ x c m = R − ( R − x 2 ) = x 2 \triangle x_{cm} = R - (R - \frac{x}{2}) = \frac{x}{2} .

jatin yadav - 7 years, 2 months ago

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@Jatin Yadav – I guess here you forgot the base of the dome which is a disc

Milun Moghe - 7 years, 2 months ago

I hope now reading this solution you will be able to solve robo space volley

Milun Moghe - 7 years, 2 months ago

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