A particle of mass 10 g moves along a circle of radius 6.4 cm with constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal 8xx10^(-4)J by the end of the second revolution after the beginning of the motion?

A particle of mass 10 g moves along a circle of radius 6.4 cm with con

1.	A particle of mass 10 g moves along a circle of radius 6.4 cm with constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal 8xx10^(-4)J by the end of the second revolution after the beginning of the motion?
Answer» `0.15m//s^(2)` `0.18m//s^(2)` `0.2m//s^(2)` `0.1m//s^(2)` Solution :`m=0.01kg` `r=6.4cm` `(1)/(2)MV^(2)=8xx10^(-4)J` `THEREFORE v^(2)=(16xx10^(-4))/(m)` `=(16xx10^(-4))/(0.01)` `therefore v^(2)=16xx10^(-2)(m//s)^(2)` `2a_(T)S=16xx10^(-2)` Distance covered in second REVOLUTION `S=2pir+2pir` `a_(T)=(16xx10^(-2))/(2S)` `=(16xx10^(-2))/(2xx4pir)` `=(2xx10^(-2))/(3.14xx6.4xx10^(-2))` `=(2)/(3.14xx6.4)` `=0.099~~0.1m//s^(2)`