# How much energy can be storage in solid sphere with a condition “centripetal acceleration <= $a_{max}$”

How much Rotational energy can be storage in solid sphere(radius = r, mass = m and uniform density distribution) with a condition that "centripetal acceleration of any point in the sphere and at the surface of the sphere mustn't more than $a_{max}$" ?

If I rotate sphere around only one axis, from a condition "centripetal acceleration <= $a_{max}$" I will get

$a_{max} = \omega^2 r$

$\omega = \sqrt{\frac{a_{max}}{r}}$

$I_{solid sphere} = \frac{2}{5}mr^2$

$E = \frac{1}{2} * \frac{2}{5}mr^2 * \frac{a_{max}}{r} = \frac{1}{5}mra_{max}$

But if I rotate sphere around more than one axis, it can storage more energy because when I rotate sphere around only one axis, centripetal acceleration at two poles is 0.

If I rotate sphere around more than one axis, for example

I rotate sphere around x-axis with $\omega_2 = \frac{1}{2}\sqrt{\frac{a_{max}}{r}}$ and I rotate that rotation axis around y-axis with the same $\omega_2$. (I am not sure that it doesn't violate the condition.)

There is no point that centripetal acceleration is 0.

If I rotate sphere around x-axis (called x1-rotation-axis) with $\omega_n = \frac{\omega_1}{n}$, and then I rotate x1-rotation-axis around y-axis (called y1-rotation-axis) with the same $\omega_n$, and then I rotate y1-rotation-axis around z-axis (called z1-rotation-axis) with the same $\omega_n$, and then I rotate z1-rotation-axis around x-axis (called x2-rotation-axis) with the same $\omega_n$, and then I rotate x2-rotation-axis around y-axis (called y2-rotation-axis) with the same $\omega_n$, and ...(limit n to infinity)

Is this is a solution of this problem ? If it is, "How much Rotational energy can be storage by using this way" ? (I don't know how to calculate. It so complex. Please help me.)

• I think your answer is correct. – Anurag B. Apr 24 '18 at 12:44
• But I don't know how much energy it can storage. – Ro Theory Apr 24 '18 at 13:50

According to Leonhard Euler's rotation theorem a solid body has at any time only one rotation axis. Therefore you need only to consider rotation about one axis. And any rotation axis of your sphere is equivalent. Thus it suffices to consider one rotation axis and to consider the angular velocity $\omega_{max}$ that makes the centripetal acceleration at the radius $r$ maximal, i.e. $r \omega_{max}^2=a_{max}$. With the moment of inertia of the sphere, $I_{sphere}$, you obtain the maximum energy stored in the sphere $$E_{max}= \frac {I_{sphere} \omega_{max}^2}{2}$$.