A stone at the end of a sling is whirled in a vertical circle ofradius 1.20 m at

Question

A stone at the end of a sling is whirled in a vertical circle ofradius 1.20 m at a constant speedv0 = 2.00 m/s asin Figure P4.57. The center of the string is 1.50 m above theground. Figure P4.57 (a) What is the range of the stone if it is released when the slingis inclined at 30.0 degree with the horizontal at A? 1 (b) What is the range of the stone if it is released when the slingis inclined at 30.0 degree with the horizontal at B? 2 (c) What is the acceleration of the stone just before it isreleased at A? Magnitude (absolute value) 3 Direction in the direction ofv0 upward downward toward the center of thecircle (d) What is the acceleration of the stone just after it is releasedat A? Magnitude (absolute value) 5 Direction toward the center of thecircle upward in the direction ofv0 downward

Explanation / Answer

   change the values according to the given    for the simplification let us choose that thepositive y direction as upwards and positive    x direction
   as the leftwards so the vertical height of the stonewhen released from A or B will be
   yi = (2.40 + 1.00 sin 30.0o)m
      = ........ m
(a)
   the equations of motion after release at A are
   vy = vi sin60.0o - gt
        = (2.40 m / s)(sin60.0o) - (9.80 t)
        = (2.07 - 9.80 t) m /s
   vx = vicos60.0o
   = (2.40 m / s) (cos60.0o)
   = 1.2 m / s   
   so we get
   y = (yi + 2.07 t - 4.90 t2)m
   xA = (1.2 t) m
   using the above quadratic form we get
   when y = 0
   t = - (2.07) ± [(2.07)2 + 4(4.90) (yi)] / - 9.80
     = ........ s
   xA = (1.2) (t)
          = ........ m
(b)
   now the equation of motion after release at point Bwill be
   vy = vi (- sin60.0o)- g t
       = (- 2.07 - 9.80 t) m /s
   vx = vi cos60.0o
       = 1.2 m / s
   yi = (yi - 2.07 t - 4.90t2) m
   when y = 0
   t = + (2.07) ± [(- 2.07)2 + 4(4.90) (yi)] / - 9.80
     = ........ s
   xB = (1.2) (t)
          = ........ m  
(c)
   towards the center
(d)
   downward

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