Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

52739 In her hand a softball pitcher swings a ball of mass 0.251 kg around a vertical circular path of radius 59.5 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.2 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 14.4 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release? Answer: The problem can be solved easily by using the work energy rule i.e. the work done on an object = gain in its kinetic energy. Here the force component in tangential direction is parallel to direction of motion and hence angle q between the small displacement ds and the force component is zero. Hence the work done is given by dW F ds Fds cos Fds But the ball is moving on a semicircular distance from the top to the bottom and hence the distance s is changes from 0 to R. So the total work done is given by R R 0 0 W Fds F ds FR 30.2 3.14 0.595 56.42 J (This result can be obtained directly using Fs because the force is always in the direction of motion.) If the final velocity at the end of path be v then increase in the kinetic energy is(1/2)m(v2 – u2) Hence using the work energy rule we have or or (1/2)0.251(v2 – 14.42) = 56.42 v2 = 656.92 v = 25.63 m/s