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 × 10 - 4   J by the end of the second revolution after the beginning of the motion?

Step-by-step solution

Tangential acceleration a t = r α = constant = K α = K r At the end of second revoluation angular velocity is ω then ω 2 - ω 0 2 = 2 α θ ω 2 = 2 ( K r ) ( 4 π ) ω 2 = 8 π K r K.E. of the particle is = K . E . = 1 2 m v 2 K . E . = 1 2 m r 2 ω 2 K . E . = 1 2 m ( r 2 ) ( 8 πK r ) = 1 2 m r ( 8 π K ) 8 × 10 − 4 = 1 2 × 10 × 10 − 3 × 6.4 × 10 − 2 × 8 × 3.14 × K K = 2 6.4 × 3.14 = 0.1 &#1