A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to by the end of the second revolution after the beginni

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Solution

Q.
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{- 4}$ $J$ by the end of the second revolution after the beginning of the motion [2016]

1 $0.18 {m/s}^{2}$

2 $0.2 {m/s}^{2}$

3 $0.1 {m/s}^{2}$

4 $0.15 {m/s}^{2}$

Ans.
(3)

Here, $m = 10$ $g = 10^{- 2} kg, R = 6.4 cm = 6.4 \times 10^{- 2} m,$

$K_{f} = 8 \times 10^{- 4} J, K_{i} = 0, a_{t} = ?$

Using work energy theorem,

Work done by all the forces = Change in KE

$W_{tangential force} + W_{centripetal force} = K_{f} - K_{i}$

$\Rightarrow a_{t} = \frac{K_{f}}{4 π R m} = \frac{8 \times 10^{- 4}}{4 \times \frac{22}{7} \times 6.4 \times 10^{- 2} \times 10^{- 2}} = 0.099 \approx 0.1 {ms}^{- 2}$

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