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Standard XII

Physics

Finding Average Velocity

A particle of...

Question

A particle of mass 10 gms performs a vertical circular motion of radius = 10 cm. The velocity at the lowest point of the circle is 5 m/s. What will be its velocity when it reaches at the middle point of the circle (i.e makes an angle zero degrees with the radius vector)

v_{2} = \sqrt{2}

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v_{2} = \sqrt{5}

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v_{2} = \sqrt{23}

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v_{2} = \sqrt{32}

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Solution

The correct option is C $v_{2} = \sqrt{23}$
At the lowest point, total energy = $m v^{2} / 2 + 0$
At the horizontal position, total energy = $m v_{2}^{2} / 2 + m g (0.1)$
Using Conservation of energy, we have , $m v_{2}^{2} / 2 = m v^{2} / 2 - m g (0.1)$
solving, we get, $v_{2} = \sqrt{23}$

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A particle of mass 10 gms performs a vertical circular motion of radius = 10 cm. The velocity at the lowest point of the circle is 5 m/s. What will be its velocity when it reaches at the middle point of the circle (i.e makes an angle zero degrees with the radius vector)

The correct option is C v2=√23At the lowest point, total energy = mv2/2+0At the horizontal position, total energy = mv22/2+mg(0.1)Using Conservation of energy, we have , mv22/2=mv2/2−mg(0.1)solving, we get, v2=√23

The correct option is C $v_{2} = \sqrt{23}$
At the lowest point, total energy = $m v^{2} / 2 + 0$
At the horizontal position, total energy = $m v_{2}^{2} / 2 + m g (0.1)$
Using Conservation of energy, we have , $m v_{2}^{2} / 2 = m v^{2} / 2 - m g (0.1)$
solving, we get, $v_{2} = \sqrt{23}$