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?

1 Answer
Apr 25, 2018

2/3 radians

Explanation:

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 tangential acceleration a_t
  • the centripetal acceleration omega^2 R = a_t^2/R t^2

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