A slingshot consisting of a 0.250 Kg stone attached to the end of a 0.750 meter

Question

A slingshot consisting of a 0.250 Kg stone attached to the end of a 0.750 meter long string was whirled in a vertical circle with a radius of 0.750 meters. The stone had a frequency of 1.25 revolutions/second. Calculate the stone's centripetal acceleration. Calculate the centripetal force acting on the stone. Calculate the tension in the string when the stone is at the top of the loop. Calculate the tension in the string when the stone is at the bottom of the loop. A 700. newton stunt pilot flying a 980. kg plane in a vertical loop with a radius of 600. meters is "weightless" at the top of the loop. Calculate the plane's speed. Week 7. assignment 2 A series of satellites are to be orbited around the earth's moon to establish a GPS system. When answering the questions below, assume it takes the moon exactly 29 earth days to rotate once on its axis. Additional information required to answer this question can be found in the text. Calculate the altitude of these satellites. Calculate the velocity of these satellites.

Explanation / Answer

Given Mass of the stone ,m = 0.25 kg Length of the string is , l = 0.75 m Radius of the circle , r = 0.75 m Angular velocity of the stone , = 1.25 rev/s = 7.85 rad/s a) The centripetal acceleration is , a = ^2 r = (7.85 rad/s)^2 *(0.75m)     a = 46.2 m/s^2 b) Centripetal force acting on the stone is     Fc = ma = m( ^2 r ) = 0.25kg ( 46.2 m/s^2)     Fc = 11.55 N c) When the stone is at the top, the tension is             FT = m( ^2 r ) - mg = 11.55N - 0.25kg*9.8 m/s^2             FT = 9.1 N d) When the stone is at thebottom , the tension is           FT' = m( ^2 r ) + mg = 11.55N + 2.45 N             FT' = 14 N

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