A particle is performing uniform circular motion withspeed 7 m/s. If r

1.	A particle is performing uniform circular motion withspeed 7 m/s. If radius of circular path is 2 m thenmagnitude of average velocity of particle in timeinterval, At = 1.5 s is
Answer» We have,Angular momentum = L = IωWhere, I is the moment of inertia and ω is the angular velocity.So frequency of revolution is, f = ω/2π=> ω = 2πfKinetic energy, K = ½ Iω2 = ½ Lω = ½ (2πfL) = πfL …………..(1)Now, frequency is doubled, so, let f/ = 2f = 2(ω/2π) = ω/πThe kinetic energy is halved, so, K/ = K/2 = ½ (πfL)If L/ is the new angular momentum, then using (1) we can write,K/ = πf/L/=> ½ (πfL) = π(2f)L/=> L/ = L/4So, angular momentum becomes one fourth of its original value. l/4 is the andwer of the question L/4is the correct answer 1/4 is the correct answer L/4 is the correct answer 1l/4 is the correct answer

A particle is performing uniform circular motion withspeed 7 m/s. If radius of circular path is 2 m thenmagnitude of average velocity of particle in timeinterval, At = 1.5 s is

Answer»

We have,Angular momentum = L = IωWhere, I is the moment of inertia and ω is the angular velocity.So frequency of revolution is, f = ω/2π=> ω = 2πfKinetic energy, K = ½ Iω2 = ½ Lω = ½ (2πfL) = πfL …………..(1)Now, frequency is doubled, so, let f/ = 2f = 2(ω/2π) = ω/πThe kinetic energy is halved, so, K/ = K/2 = ½ (πfL)If L/ is the new angular momentum, then using (1) we can write,K/ = πf/L/=> ½ (πfL) = π(2f)L/=> L/ = L/4So, angular momentum becomes one fourth of its original value.

l/4 is the andwer of the question

L/4is the correct answer

1/4 is the correct answer

L/4 is the correct answer

1l/4 is the correct answer

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