Question

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.

Answer 1

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 2

The speed of the ball is 5.59 m/s.

Step-by-step 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=(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)=(1)/(2)mv_(f)^(2)`

Using conservation of energy

`K.E_(f)=K.E_(i)+P.E`

`(1)/(2)mv_(f)^2=(1)/(2)mv_(i)^2+mgh`

`(1)/(2)*0.160* v_(f)^2=(1)/(2)*0.160*(3.26)^2+0.160*9.8*(0.70+0.70*\sin30^(\circ))`

`v_(f)=√((3.26)^2+2*9.8*1.05)`

`v_(f)=5.59\ m/s`

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