A particle begins to move with a tangential acceleration of constant magnitude #0.6m/s^2# in a circular path. If it slips when it's total acceleration becomes #1m/s^2#, then the angle through which it has turned before slipping is?

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Answer 1 · 2018-04-25T03:17:05

#2/3# radians

The angular acceleration #alpha# is related to the tangential acceleration #a_t# by

#alpha = a_t/R#

where #R# is the radius of the circular orbit.

The angular velocity at a time #t# is given by #omega = alpha\ t = a_t/R t#.

At this instant, the two components of the acceleration are

The magnitude of the acceleration is

# sqrt(a_t^2+a_t^4/R^2t^4)#

The time at which the particle slips is given by

#a_t^2+a_t^4/R^2t^4 = a_0^2#

where #a_0=1\ "ms"^-2#

Thus

#t^4 = (a_0^2-a_t^2)/a_t^4 R^2implies t^2 = sqrt(a_0^2/a_t^2-1) R/a_t#

and the angle through which the particle has turned at this time is

#theta = 1/2 alpha t^2 = 1/2 a_t/R t^2 = 1/2sqrt((a_0/a_t)^2-1)#

Substituting the values of #a_0# and #a_t#, we get

#theta = 1/2sqrt((5/3)^2-1)= 1/2times 4/3 = 2/3#