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 center of the circle is 1.50 m above the floor.
Calculate the speed of the ball when the cord is 30.0° below the horizontal.
Answer (3)
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
The speed of the ball is 5.59 m/s. ;
The speed of the ball when the cord is 30° below the horizontal is approximately 5.99 m/s. This is calculated using the conservation of mechanical energy by comparing the kinetic and potential energies at two points in the swing. By determining the height and applying the relevant equations, we find the new speed.
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