Question

The angular position of a point on a rotating wheel is given by θ = 2.78 + 1.02t² + 4.96t³, where θ is in radians and t is in seconds. At t = 0, what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at t = 7.77 s? (d) Calculate its angular acceleration at t = 1.36 s. (e) Is its angular acceleration constant?

Answer 1

Angular velocity, w = dθ / dt

And, angular accleration, a = dw / dt = d^2 θ / dt^2

(a) put t = 0 in the given expression -

θ = 2.78 radians.

(b) w = dθ / dt = 0 + (2*1.02)t + (3*4.96)t^2 = 2.04t + 14.88t^2

put, t = 0

we get, w = 0

(c) put, t = 7.77 s

w = 2.04*7.77 + 14.88*7.77^2 = 15.85 + 898.35 = 914.2 rad/s

(d) a = dw / dt = 2.04 + (2*14.88)t = 2.04 + 29.76t

put, t = 1.36 s

a = 2.04 + 29.76*1.36 = 42.51 rad/s^2

(e) Look the expression for angular accleration -

angular acceleration constant = 2.04