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Motion in a vertical circle and conservation of energy

A stone  tied to a string of length l is whirled around a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at the lowest position and has a speed u. What is the magnitude of change in its velocity as it reaches a position where the string is horizontal?

Answer:

Let’s assume that the potential energy at the lowest position be zero. So, when the string is horizontal, the stone has risen by a vertical height l, the length of the string which is also the radius of the vertical circle.

If v is the magnitude of velocity at the horizontal position, then according to the law of conservation of energy,

KE+PE at the lowest position = KE+PE at the horizontal position

\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgl

From the equation above, v-u can be calculated.

v=\sqrt{u^{2}-2gl}

 

The following links will help you for deeper understanding and you can browse through some solved problems from the topic too.

Current electricity – Numerical problems

  1. In a potentiometer arrangement a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell, the balance point shifts to 63 cm. What is the emf of the second cell?

E1= 1.25 V

l1 = 35 cm

l2 = 63 cm

E1/E2 = l1/l2

E2 = l2E1/l1

E2 = 63 x 1.25/35

E2 = 2.25 V (Ans)

A Problem for Practice

A storage battery of emf 8 V and internal resistance 0.5 ohm is charged by 120V dc supply using a resistor 15.5 ohm in series. What is the terminal voltage of the battery during charging?

(Try to solve this problem and post your responses as comment to this post)

Hint: During charging; V = E +Ir

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