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A small body A starts sliding off the top of a smooth sphere of radius R. Find the angle theta (figure) corresponding to the point at which the body breaks off the sphere, as well as the break-off velocity of the body. |
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Answer» Solution :Let us depict the forces acting on the body A (which are the force of GRAVITY `mvecg` and the normal reaction `VECN`) and write equation `vecF=mvecw` via projection on the unit vectors `hatu_t` and `hatu_n` (figure). From `F_t=mw_t` `mg sin THETA=m(dv)/(dt)` `=m(vdv)/(DS)=m(vdv)/(Rd theta)` or, `gRsin theta d theta=vdv` Integrating both sides for obtaining `v(theta)` `UNDERSET(0)overset(theta)int gR sin theta d theta=underset(0)overset(v)int v dv` or, `v^2=2gR(1-cos theta)` (1) From `F_n=mw_n` `mg cos theta-N=mv^2/R` (2) At the moment the body loses contact with the surface, `N=0` and therefore the Eq. (2) becomes `v^2=gRcos theta` (3) where `v` and `theta` corresponds to the moment when the body loses contact with the surface. Solving Eqs. (1) and (3) we obtain `cos theta=2/3`or, `theta=cos^-1(2//3)` and `v=sqrt(2gR//3)`. 
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