A particle moves along a circle of radius #20/π# m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has began the tangential acceleration is ??

Question

45392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-13T13:58:26

The tangential acceleration is #=40ms^-2#

Assuming that the particle starts from rest

The initial angular velocity is #omega_0=0rads^-1#

The radius of the circle is #r=20/pim#

The final velocity of the particle is #v=80ms^-1#

The final angular velocity is

#omega=v/r=80/(20/pi)=4pirads^-1#

The angle is #theta=2*2pi=4pi#

Applying the equation

#omega^2=omega_0^2+2alphatheta#

Where #alpha # is the angular acceleration

#alpha=(omega^2-omega_0^2)/(2theta)#

#=((4pi)^2-0)/(2*4pi)=(16pi^2)/(8pi)=(2pi)rads^-2#

The tangential acceleration is

#a_T=alpha*r=2pi*20/pi=40ms^-2#