Question

A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by θ=cos2t, where θ is in radians and t is in seconds.

Determine the magnitude of the acceleration of the particle when θ= 35 ∘

Answer 1

The angular acceleration is -2.44 rad/s², while the linear acceleration is -14.66 in/s².

Step-by-step explanation:

First we need to find the time, at the given position. W e are given the position of particle to be:

θ = 35°

Converting it to radians because, the given equation is in radians:

θ = (35°) (π radians/180°)

θ = 0.611 radians

Now, we have the equation:

θ = Cos(2t)

2t = Arc Cos (θ)

2t = Arc Cos (0.611 radians)

t = 0.91/2

t = 0.457 sec

Now, to determine angular acceleration of the particle, we must derivate the equation twice with respect to 't'

Angular Velocity = ω = dθ/dt = -2Sin(2t)

Angular Acceleration = α = -4Cos(2t)

Now, we use the value of t:

α = -4Cos(2 x 0.457)

α = -2.44 rad/s² (negative sign shows decceleration)

Now for linear acceleration, we know that:

a = rα

a = (6 in)(-2.44 rad/s²)

a = -14.66 in/s² (negative sign shows decceleration)