A particle of mass 10 ~g moves along a circle of radius 6.4 ~cm with a constant tangential acceleration. If the kinetic energy of the particle becomes 8 10⁻⁴ ~J by the end of the second revolution after the beginning of the motion, the magnitude of the tangential acceleration is

Step-by-step solution

Consider the following diagram: The tangential acceleration is given by, aligned & a_t= d v ~d t = Constant & d v ~d t =a_t & d s ~d t d v ~d s =a t & v= d v ~d t =a_t & ₀^v v^2 2 ~d v= ₀^ (4 r) a_t ~d s aligned We know, Now, kinetic energy is given by, K E= m v^2 2 Now, using eq ^ n (1) &(2) aligned & ( 2 K E m )=(8 r) a_t & a_t= ( K E (4 r) m ) aligned Given, K E=8 10⁻⁴ ~J , r=6.4 10⁻² ~m and m=1 10⁻² ~kg . a_t= (8 10⁻⁴ ) J 4 (6.4 10⁻² ~m ) (10⁻² ~kg ) =0.1 ~m / s ^2