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 ω=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:
The equation of motion in the rotating coordinate system is,
m→w=→F+mω2→R+2m(→v×→w)
Now, →v=Rθ→eθ+Rsinθ˙φ→eφ
and →w=w′cosθ→er−w′sinθ→eθ
12m→Fcor=∣∣
∣
∣∣→er→eθ→eφ0RθRsinθ˙φωcosθ−ωsinθ0∣∣
∣
∣∣
=→er(ωRsin2θ˙φ)+ωRsinθcosθ˙φ→eθ−ωRθcosθ→eθ
Now on the sphere,
→v=(−R˙θ2−Rsin2θ˙φ2)→er
+(R˙θ−Rsinθcosθ˙φ2)→eθ
+(Rsinθ¨φ−2Rcosθ˙θ˙φ)→eφ
Thus the equation of motion are,
m(−R˙θ2−Rsin2θ˙φ2)=N−mgcosθ+mω2Rsin2θ+2mωRsin2θ˙φ
m(R˙θ−Rsinθcosθ˙φ2)=mgsinθ+mω2Rsinθcosθ+2mωRsinθcosθ˙φ
m(Rsinθ¨φ−2Rcosθ˙θ˙φ)=−2mωR˙θcosθ
From the third equation, we get, ˙φ=−ω
A result that is easy to understand by considering the motion in non-rotating frame. The eliminating ˙φ we get,
mR˙θ2=mgcosθ−N
mR˙θ=mgsinθ
Integrating the last equation
12mR˙θ2=mg(1−cosθ)
Hence N=(3−2cosθ)mg
So the body must fly off for θ=θ0=cos−123, exactly as if the sphere were non-rotating.
Now, at this point Fcf= centrifugal force =mω2Rsinθ0=√59mω2R
Fcor=√ω2R2θ2cos2θ+(ω2R2)2sin2θ×2m
=√59(ω2R)2+ω2R2×49×2g3R×2m=23mω2R√5+8g3ω2R = 17.25