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During a certain period of time, the angular displacement of a swinging door is described by 𝜃 = 5.00 + 10.0𝑡 + 2.00𝑡2 where θ is radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration, (a) at time t = 0 s, (b) at time t = 3.00 s.
alphonsio

Graphical representation to the angular displacement of 𝜃 = 5.00 + 10.0𝑡 + 2.00𝑡2

  • At t=0t = 0 s:

    • Angular Displacement: θ=5\theta = 5 radians
    • Angular Speed: ω=10\omega = 10 rad/s
    • Angular Acceleration: α=4\alpha = 4 rad/s²

  • At t=3t = 3 s:

    • Angular Displacement: θ=53\theta = 53 radians
    • Angular Speed: ω=22\omega = 22 rad/s
    • Angular Acceleration: α=4\alpha = 4 rad/s²

Given the equation for the angular displacement θ(t)\theta(t) as a function of time tt:

θ(t)=5+10t+2t2\theta(t) = 5 + 10t + 2t^2

where θ\theta is in radians and tt is in seconds. We need to determine the angular displacement, angular speed, and angular acceleration at times t=0t = 0 s and t=3.00t = 3.00 s.

1. Angular Displacement, θ(t)\theta(t)

This is given directly by the equation :

θ(t)=5+10t+2t2\theta(t) = 5 + 10t + 2t^2

2. Angular Speed, ω(t)\omega(t)

The angular speed is the first derivative of the angular displacement with respect to time:

ω(t)=dθ(t)dt\omega(t) = \frac{d\theta(t)}{dt}

Let's differentiate the given equation:

ω(t)=ddt(5+10t+2t2)=0+10+4t\omega(t) = \frac{d}{dt} \left(5 + 10t + 2t^2\right) = 0 + 10 + 4t

So,

ω(t)=10+4t(in radians per second)\omega(t) = 10 + 4t \quad \text{(in radians per second)}

3. Angular Acceleration, α(t)\alpha(t)

The angular acceleration is the second derivative of the angular displacement with respect to time, or the first derivative of the angular speed with respect to time:

α(t)=dω(t)dt\alpha(t) = \frac{d\omega(t)}{dt}

Let's differentiate ω(t)\omega(t) :

α(t)=ddt(10+4t)=0+4=4(in radians per second squared)\alpha(t) = \frac{d}{dt} (10 + 4t) = 0 + 4 = 4 \quad \text{(in radians per second squared)}

So the angular acceleration α(t)\alpha(t) is constant and equal to 4 radians per second squared.

Now, let's calculate the values at t=0t = 0 s and t=3t = 3 s.

(a) At t=0t = 0 s:

  1. Angular Displacement θ(0)\theta(0):

θ(0)=5+10(0)+2(0)2=5 radians\theta(0) = 5 + 10(0) + 2(0)^2 = 5 \text{ radians}

  1. Angular Speed ω(0)\omega(0):

ω(0)=10+4(0)=10 radians per second\omega(0) = 10 + 4(0) = 10 \text{ radians per second}

  1. Angular Acceleration α(0)\alpha(0):

α(0)=4 radians per second squared\alpha(0) = 4 \text{ radians per second squared}

(b) At t=3t = 3 s:

  1. Angular Displacement θ(3)\theta(3) :

θ(3)=5+10(3)+2(3)2=5+30+18=53 radians\theta(3) = 5 + 10(3) + 2(3)^2 = 5 + 30 + 18 = 53 \text{ radians}

  1. Angular Speed ω(3.00)\omega(3.00) :

ω(3)=10+4(3)=10+12=22 radians per second\omega(3) = 10 + 4(3) = 10 + 12 = 22 \text{ radians per second}

  1. Angular Acceleration α(3)\alpha(3):

α(3)=4 radians per second squared\alpha(3) = 4 \text{ radians per second squared}

Summary:

  • At t=0t = 0 s:

    • Angular Displacement: θ=5\theta = 5 radians
    • Angular Speed: ω=10\omega = 10 rad/s
    • Angular Acceleration: α=4\alpha = 4 rad/s²

  • At t=3t = 3 s:

    • Angular Displacement: θ=53\theta = 53 radians
    • Angular Speed: ω=22\omega = 22 rad/s
    • Angular Acceleration: α=4\alpha = 4 rad/s²