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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. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere. |
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Answer» Solution :The EQUATION of motion in the rotating COORDINATE system is, `mvecw=VECF+momega^2vecR+2m(vecvxxvecomega)` Now, `vecv=Rthetavece_0+Rsin thetaoverset(.)varphivece_(varphi)` and `vecw=w^'cos thetavece_r-w^'sin thetavece_0` `(1)/(2m)vecF_(cor)=|{:(vece_r,+vece_0,vece_varphi,,),(0,Rtheta,Rsinthetaoverset(.)varphi,,),(omegacostheta,-omegasintheta,0,,):}|` `=vece_r(omegaRsin^2thetaoverset(.)varphi)+omegaRsinthetacosthetaoverset(.)varphivece_(theta)-omegaRcosthetavece_(theta)` Now on the sphere, `vecv=(-Roverset(.)theta^2-Rsin^2thetavarphi^2)vece_r` `+(Roverset(.)theta-Rsin thetacos thetaoverset(.)varphi^2)vece_(theta)` `+(Rsin thetaoverset(..)varphi+2Rcos theta OVERSET(.)thetaoverset(.)varphi)vece_vaprhi` Thus the equation of motion are, `m(-Roverset(.)theta^2-Rsin^2thetaoverset(.)varphi^2)=N-mgcostheta+momega^2Rsin^2theta+2momegaRsin^2thetaoverset(.)varphi` `m(Roverset(.)theta-Rsinthetacosthetaoverset(.)varphi^2)=mg sin theta+momega^2Rsin thetacostheta+2momegaRsinthetacosthetaoverset(.)varphi` `m(Rsin thetaoverset(..)varphi+2Rcos thetaoverset(.)thetaoverset(.)varphi)=-2momegarRoverset(.)thetacostheta` From the third equation, we get, `underset(.)varphi=-omega` A result that is easy to understand by considering the motion in non-rotating frame. The ELIMINATING `overset(.)varphi` we get, `mRunderset(.)theta^2=mgcostheta-N` `mRunderset(.)theta=mgsintheta` Integrating the last equation, `1/2mRunderset(.)theta^2=mg(1-costheta)` Hence `N=(3-2costheta)mg` So the body must fly off for `theta=theta_0=cos^-12/3`, exactly as if the sphere were nonrotating. Now, at this point `F_(ef)=` centrifugal force `=momega^2Rsintheta_0=sqrt(5/9)momega^2R` `F_(cor)=sqrt(omega^2R^2theta^2cos^2theta+(omega^2R^2)^2sin^2theta)xx2m` `=sqrt(5/9(omega^2R)^2+omega^2R^2xx4/9xx(2g)/(3R))xx2m=2/3momega^2Rsqrt(5+(8g)/(3omega^2R))` 
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