A ball of mass m = 1.5 kg at the end of a thin cord of length r = 0.89 m revolve

Question

A ball of mass m = 1.5 kg at the end of a thin cord of length r = 0.89 m revolves in a vertical circle about point O, as shown in the figure. During the time we observe it, the only forces acting on the ball are gravity and the tension in the cord. The motion is circular but not uniform because of the force of gravity. The ball increases in speed as it descends and decelerates as it rises on the other side of the circle. At the moment the cord makes an angle ? = 39? below the horizontal, the ball's speed is 6.0 m/s. Take ? increasing downward as shown.

At this point, determine the tangential acceleration.
m/s2 (tangent to the path)

At this point, determine the radial acceleration.
_____________ m/s2 (toward the center)

At this point, determine the tension in the cord, FT.
_____________ N

A ball of mass m = 1.5 kg at the end of a thin cord of length r = 0.89 m revolves in a vertical circle about point O, as shown in the figure. During the time we observe it, the only forces acting on the ball are gravity and the tension in the cord. The motion is circular but not uniform because of the force of gravity. The ball increases in speed as it descends and decelerates as it rises on the other side of the circle. At the moment the cord makes an angle ? = 39 below the horizontal, the ball's speed is 6.0 m/s. Take ? increasing downward as shown. At this point, determine the tangential acceleration. m/s2 (tangent to the path) At this point, determine the radial acceleration. _____________ m/s2 (toward the center) At this point, determine the tension in the cord, FT. _____________ N

Explanation / Answer

tangential acc =   g cos? =

              =   9.8 cos39 =    7.616 m/s2

radial acc = v2/ r = 62 / 0.89 =     40.449 m/s2

tension incord...

    T - mgsin?   = m ar

    T =   mg sin? + mar   = 1.5 * 9.8 * sin39 + 1.5 * 40.449 =

             =    69.92 N

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