Use conservation of energy to justify where the maximum speed of the sphere occurs

Question

Suppose a solid sphere of mass $M$ rolls along the surface given by $z=x^2/2-R$ without slipping. Also suppose that it is released from height $h$ and that the centre of the sphere remains in the $x-z$ plane.

I have found that the kinetic energy is given by $$T=\frac{7}{10}M\dot{x}^2(1+x^2)$$ and the potential is given by $$V=Mg\left(\frac{x^2}{2}-R\right)$$

I'm first asked to find the maximum speed, which I know will occur when $x=0$. So this gives me $$Mgh=\frac{7}{10}M\dot{x}^2$$ and so the maximum speed is $$\sqrt{\frac{10gh}{7}}$$

Next I'm asked to justify that the maximum speed occurs at $x=0$ using conservation of energy. Physically it's clear to me that it should occur at $x=0$, how do I show it though?

David Quinn · Accepted Answer · 2017-03-25T18:13:43.167

1

Write down the conservation of energy equation $T+V=C$ and differentiate this with respect to time $t$

So you have $$\frac 95m\dot{x}\ddot{x}(1+x^2)+\frac{9}{10}m\dot{x}^2(2x\dot{x})+mgx\dot{x}=0$$

For maximum $v$ set acceleration $\ddot{x}$ equal to zero and it follows immediately that since $\dot{x}\neq0$ then $x=0$

edited Mar 25 '17 at 18:13

answered Mar 25 '17 at 16:22

David Quinn

34,121

Would you mind demonstrating this? I'm having some trouble concluding that $x=0$ after taking the derivative. – MHW Mar 25 '17 at 16:29
Please see my edit – David Quinn Mar 25 '17 at 17:03
Should it not be $+2mgx\dot{x}$ since we're looking at the quantity $T+V$? – MHW Mar 25 '17 at 17:13
Yes sorry a typo... – David Quinn Mar 25 '17 at 18:13
...amd you have changed the expression for the potential energy? – David Quinn Mar 25 '17 at 18:15

Use conservation of energy to justify where the maximum speed of the sphere occurs

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