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A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released. Find: (a) the total acceleration of the sphere and the thread tension as a function of theta, the angle of deflection of the thread from the vertical, (b) the thread tension at the moment when the vertical component of the sphere's velocity is maximum, (c) the angle theta between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally. |
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Answer» Solution :Let us depict the forces acting on the small sphere m, (at an arbitrary position when the thread makes an angle `theta` from the vertical) and write equation `vecF=mvecw` VIA projection on the unit vectors `hatu_t` and `hatu_n`. From `F_t=mw_t`, we have `mg sin theta=m(DV)/(dt)` `=m (vdv)/(ds)=m(vdv)/(l(-d theta))` (as vertical is reference line of ANGULAR position) or `vdv=gl sin theta d theta` Integrating both the sides: `UNDERSET(0)overset(v)INT v dv=-gl underset(x//2)overset(0)int sin theta d theta` or, `v^2/2=gl cos theta` Hence `v^2/l=2 g cos theta =w_n` .(1) (Eq. (1) can be easily obtained by the conservation of mechanical energy). From `F_n=mw_n` `T-mg cos theta=(mv^2)/(l)` Using (1) we have `T=3 mg cos theta` (2) Again from the Eq. `F_t=mw_t`: `mg sin theta=mw_1` or `w_t=g sin theta` (3) Hence `w=sqrt(w_t^2+w_n^2)=sqrt((gsin theta)^2+(2 g cos theta)^2)` (using 1 and 3) `=gsqrt(1+3cos^2theta)` (b) Vertical component of velocity `v_y=v sin theta` So, `v_y^2=v^2 sin^2 theta=2gl cos theta sin^2 theta` (using 1) For maximum `v_y` or `v_y^2`, `(d(cos theta sin^2 theta))/(d theta)=0` which yields `cos theta=1/sqrt3` Therefore from (2) `T=3mg1/sqrt3=sqrt3mg` (c) We have `vecw=w_thatu_t+w_nhatu_n` thus `w_y=w_(l(y))+w_(n(y))` But in accordance with the problem `w_y=0` So, `w_(t(y))+w_(n(y))=0` or, `g sin theta sin theta +2 g cos^2 theta(-cos theta)=0` or , `cos theta=1/sqrt3` or, `theta=54*7^@` 
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