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A 0.160 kg ball attached to a light cord is swung in a vertical circle of radius 70.0 cm. At the top of the swing, the speed of the ball is 3.26 m/s. The centre of the circle is 1.50 m above the floor.

Calculate the speed of the ball when the cord is 30.0̊ below the horizontal.

Verified Answer :

Everest2017

When the cord carrying the ball is at 30 deg below horizontal, the ball is at
       1.5 m - 0.70 sin 30 = 1.15 meters above ground

Energy of the ball at the topmost point of swing :
       KE + PE  = 1/2 m v² + m g h
      = 1/2 0.160 3.26²  +  0.160 9.81 (1.50 + 0.70)  = 8.767 Joules
Energy of the ball when the cord at 30 deg from horizontal
       = 1/2 m V² + m g h  =
       = 1/2 0.160 V² + 0.160 9.81 * 1.15 = 0.08V² + 1.805 Joules

Conservation of energy :  0.08 V² + 1.805 = 8.767
         V² = 87.025

 V = 9.329 m/sec

Answer:

The speed of the ball is 5.59 m/s.

Explanation:

Given that,

Mass = 0.160 kg

Radius = 70 cm =0.70 m

Distance = 1.50

Speed at top = 3.26 m/s

Angle = 30.0°

We need to calculate the speed of the ball

The total energy at the top

\(K.E_{i}+P.E=\dfrac{1}{2}mv_{i}^2+mgh\)

The final kinetic energy of ball at that point when the cord is 30° below the horizontal

\(K.E_{f}=\dfrac{1}{2}mv_{f}^{2}\)

Using conservation of energy

\(K.E_{f}=K.E_{i}+P.E\)

\(\dfrac{1}{2}mv_{f}^2=\dfrac{1}{2}mv_{i}^2+mgh\)

\(\dfrac{1}{2}\times0.160\times v_{f}^2=\dfrac{1}{2}\times0.160\times(3.26)^2+0.160\times9.8\times(0.70+0.70\times\sin30^{\circ})\)

\(v_{f}=\sqrt{(3.26)^2+2\times9.8\times1.05}\)

\(v_{f}=5.59\ m/s\)

Hence, The speed of the ball is 5.59 m/s.

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