A 4.5 kg ball is whirled it constant speed 2 3 m/s in a horizontal circle as sho

Question

A 4.5 kg ball is whirled it constant speed 2 3 m/s in a horizontal circle as shown at right The speed is such that the ball maintains the same distance above the ground. On the ball shown below, draw the forces, including components where appropriate. Write a 2nd law equation for the x-direction. Write a 2nd law equation for the y-direction Compute the tension in the string. Compute the angle theta shown. The person lets go of the rope. Compute the horizontal distance traveled by the ball before it hits the ground.

Explanation / Answer

The mass of the ball m = 4.5kg

the speed of the ball v = 2.3m/s

the radius of the circle r = 0.7m

(b) From Newton's laws

      Tcos = mv^2/r

(c) along the y-axis

       Tsin = mg

From above equations

         tan = mg/mv^2/r

                = gr/v^2

                = (9.8)(0.7)/(2.3)^2

             = 52.4 deg

(d) the tension in the string

         T = mg/sin

            = (4.5)(9.8) /sin52.4

            = 55.66N

(e) The angle = 52.4 deg

(f) The time taken to reaah the ground from height h = 1.5m

           t = 2h/g

             = 2(1.5)/9.8

              = 0.55sec

therefore the horizontal distance

         x = (2.3)(0.55) = 1.265 m

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