InterviewSolution
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InterviewSolution
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A light rigid wire of length 1 m is attached to a ball A of mass 500 g to one end. The other end of the wire is fixed, so that the wire can rotate freely in the vertical plane about its fixed and At the lowest point of the circular motion, the ball is given a horizontal velocity 6 m/s. Determined the radial component of the acceleration of the ball, when this rigid wire makes an angle 60^@ with the upward vertical (Takeg = 10 m//s^2) |
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Answer» `10m//s^2` length of a rigid wire, l= 1 m mass of the ball, m = 500 g=0.5kg VELOCITY of ball at a point, C =v m/s  Motion of body in vertical plane is shown in the figure. In` DELTA BOC` ` cos 60^2 = (OB)/(OC)` `1/2 = h/1` ` RARR h = 1/2 m` ` therefore AB = OA + OB = l + h = 1+ 1/2 ` `AB= 3/2 m`....(i) According to the laws of CONSERVATION of energy, (KE+PE) at a point A= (KE + PE) at point C `1/2 "mu"^2 + 0 + 1/2 mv^2 + mg (AB)` `u^2 = v^2 + 2g . 3/2 ` ` 6^2 = v^2 + 2 xx 10 xx 3/2 ` ` v^2 = 6` ` therefore ` Radial component of acceleration of the ball ` a_C = (v^2)/(l) = 6/1 =6m//s^2` |
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