A particle starts moving at t = 0 in a circle of radius R = 2 m with constant angular acceleration of 3 rad/sec^2. Initial angular speed of the particle is 1 rad/sec . At the instant when the angle between the acceleration vector and the velocity vector of the particle is 37º, calculate the value of ‘t’ at this moment

1
by juttuc4

2014-09-26T03:52:56+05:30

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
I assume that the particle is always moving in the circle of radius 2m.

Radius of circle = r = 2m  ;    Angular acceleration = α = 3 rad/sec²

Angular velocity at time t = ω = ω₀ + α t = ( 1 + 3 t ) rad/sec

Linear velocity = v = r ω = 2(1+3t) m/sec.
Linear velocity is always perpendicular to radius and is tangential.

Linear acceleration (tangential and in the direction of velocity) = a = r α = 6 m/sec²

Centripetal acceleration (radially inwards) = a' = r ω² = 2 * (1+3 t)²

Net acceleration of the particle is inclined to tangent to circle by angle 37⁰.

tan 37⁰  =  a' / a    = 2 * (1+3t)² / 6

(1+3t)² = 2.26

1+ 3 t = 1.5035

t = 0.5035 / 3 = 0.1678 sec

thanx n u r welcom