Question

A particle starts moving at t = 0 in a circle of radius R = 2m with constant angular acceleration of a - 3 "rad/sec"^(2). Initial angular speed of the particle is 1 rad/sec. At time t_(0) the angle between the acceleration vector and the velocity vector of the particle is 37^(@). What is the value of t_(0) ?

Answer 1

Answer:

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

Explanation:

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