Question

A small body of mass $m=0.30kg$ starts sliding down from the top of a smooth sphere of radius $R=1.00m$. The sphere rotates with a constant angular velocity $\omega =6.0rad/s$ about a vertical axis passing through its centre. The Coriolis force (in N) at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere is (10+x)N. The value of x is:

Answer 1

The equation of motion in the rotating coordinate system is,
$m\overrightarrow{w}=\overrightarrow{F}+m{\omega }^2\overrightarrow{R}+2m(\overrightarrow{v}\times \overrightarrow{w})$
Now, $\overrightarrow{v}=R\theta \overrightarrow{e_{\theta }}+R\sin \theta \dot{\phi }\overrightarrow{e_{\phi }}$
and $\overrightarrow{w}=w^{\prime }\cos \theta \overrightarrow{e_r}-w^{\prime }\sin \theta \overrightarrow{e_{\theta }}$
$\frac{1}{2m}\overrightarrow{F_{cor}}=\left|\begin{array}{ccccc}\overrightarrow{e_r}& & \overrightarrow{e_{\theta }}& & \overrightarrow{e_{\phi }}\\ 0& & R\theta & & R\sin \theta \dot{\phi }\\ \omega \cos \theta & & -\omega \sin \theta & & 0\end{array}\right|$
$=\overrightarrow{e_r}\left(\omega R{\sin }^2\theta \dot{\phi }\right)+\omega R\sin \theta \cos \theta \dot{\phi }\overrightarrow{e_{\theta }}-\omega R\theta \cos \theta \overrightarrow{e_{\theta }}$
Now on the sphere,
$\overrightarrow{v}=(-R{\dot{\theta }}^2-R{\sin }^2\theta {\dot{\phi }}^2)\overrightarrow{e_r}$
$+(R\dot{\theta }-R\sin \theta \cos \theta {\dot{\phi }}^2)\overrightarrow{e_{\theta }}$
$+(R\sin \theta \ddot{\phi }-2R\cos \theta \dot{\theta }\dot{\phi })\overrightarrow{e_{\phi }}$
Thus the equation of motion are,
$m(-R{\dot{\theta }}^2-R{\sin }^2\theta {\dot{\phi }}^2)=N-mg\cos \theta +m{\omega }^{2}R{\sin }^2\theta +2m\omega R{\sin }^2\theta \dot{\phi }$
$m(R\dot{\theta }-R\sin \theta \cos \theta {\dot{\phi }}^2)=mg\sin \theta +m{\omega }^{2}R\sin \theta \cos \theta +2m\omega R\sin \theta \cos \theta \dot{\phi }$
$m(R\sin \theta \ddot{\phi }-2R\cos \theta \dot{\theta }\dot{\phi })=-2m\omega R\dot{\theta }\cos \theta$
From the third equation, we get, $\dot{\phi }=-\omega$
A result that is easy to understand by considering the motion in non-rotating frame. The eliminating $\dot{\phi }$ we get,
$mR{\dot{\theta }}^2=mg\cos \theta -N$
$mR\dot{\theta }=mg\sin \theta$
Integrating the last equation
$\frac{1}{2}mR{\dot{\theta }}^2=mg(1-\cos \theta )$
Hence $N=(3-2\cos \theta )mg$
So the body must fly off for $\theta ={\theta }_0={\cos }^{-1}\frac{2}{3}$, exactly as if the sphere were non-rotating.
Now, at this point $F_{cf}=$ centrifugal force $=m{\omega }^{2}R\sin {\theta }_0=\sqrt{\frac{5}{9}}m{\omega }^{2}R$
$F_{cor}=\sqrt{{\omega }^2{R}^2{\theta }^2{\cos }^2\theta +{\left({\omega }^2{R}^2\right)}^2{\sin }^2\theta }\times 2m$
$=\sqrt{\frac{5}{9}{\left({\omega }^{2}R\right)}^2+{\omega }^2{R}^2\times \frac{4}{9}\times \frac{2g}{3R}}\times 2m=\frac{2}{3}m{\omega }^{2}R\sqrt{5+\frac{8g}{3{\omega }^{2}R}}$ = $17.25$