A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant
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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 x 10^-4 J by the end of the second revolution after the beginning of the motion?
kinetic energy = 8 × 10^-4 J
or, 1/2 mv² = 8 × 10^-4
or, 1/2 × 10 × 10^-3 v² = 8 × 10^-4
or, v² = 16 × 10^-2 => v = 0.4 m/s
initial velocity of particle, u = 0 m/s
we have to find Tangential acceleration at the end of 2nd revolution.
total distance covered, s = 2(2πr) = 4πr
so, v² = 2as
a = v²/2s = (0.4)²/2(4πr)
= 16 × 10^-2/(8 × 3.14 × 6.4 × 10^-2)
= 0.0995 m/s² ≈ 0.1 m/s²
kinetic energy = 8 × 10^-4 J
or, 1/2 mv² = 8 × 10^-4
or, 1/2 × 10 × 10^-3 v² = 8 × 10^-4
or, v² = 16 × 10^-2 => v = 0.4 m/s
initial velocity of particle, u = 0 m/s
we have to find Tangential acceleration at the end of 2nd revolution.
total distance covered, s = 2(2πr) = 4πr
so, v² = 2as
a = v²/2s = (0.4)²/2(4πr)
= 16 × 10^-2/(8 × 3.14 × 6.4 × 10^-2)
= 0.0995 m/s² ≈ 0.1 m/s²
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Answer:
so,
the answer is a=0.099 = 0.1 m/s^2
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