question_answer

A glass marble slides from rest from the topmost point of a vertical circle of radius r along a smooth chord. Does the time of descent depend upon the chord chosen?

Answer:

                As shown in Fig. let C be the centre of the circle. Suppose chord AB makes angle \[\theta \] with the vertical. If the marble takes time t to slide along chord AB, then                 \[AB=0\times t+\frac{1}{2}g\cos \theta \times {{t}^{2}}\] [\[\because \] Acceleration along \[AB=g\cos \theta \]] or  \[{{t}^{2}}=\frac{2AB}{g\cos \theta }=\frac{2\times 2r\cos \theta }{g\cos \theta }=\frac{4r}{g}\] or  \[t=2\sqrt{\frac{r}{g}}\] As t is independent of \[\theta \], so the time of descent does not depend on the chord chosen.