Point A of the circular disk is at the angular position θ at time t = 0. The disk has angular velocity ω = 4 rad/s at t = 0 and experiences angular acceleration α = 2 rad/s². Deter... Point A of the circular disk is at the angular position θ at time t = 0. The disk has angular velocity ω = 4 rad/s at t = 0 and experiences angular acceleration α = 2 rad/s². Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1 s.

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Understand the Problem

The question addresses the dynamics of a circular disk in terms of angular motion. It specifically asks for the velocity and acceleration of a given point on the disk at a certain time, requiring the use of angular velocity and acceleration equations. The overall aim is to analyze the motion of point A given the initial conditions and the changes due to angular acceleration.

Answer

Velocity: $$ \vec{v_A} = 1.2 \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix} $$ Acceleration: $$ \vec{a_A} = 0.4 \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix} + 7.2 \begin{pmatrix} -\sin(\theta) \\ \cos(\theta) \end{pmatrix} $$
Answer for screen readers

The velocity of point A at $t=1$ s is given by:

$$ \vec{v_A} = 1.2 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} $$

The acceleration of point A at $t=1$ s is:

$$ \vec{a_A} = 0.4 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} + 7.2 \begin{pmatrix} -\sin(\theta) \ \cos(\theta) \end{pmatrix} $$

Steps to Solve

  1. Determine the angular velocity at time t = 1 s

To find the angular velocity at time $t = 1 \text{ s}$, we use the equation:

$$ \omega(t) = \omega_0 + \alpha t $$

where
$\omega_0 = 4 , \text{rad/s}$ (initial angular velocity)
$\alpha = 2 , \text{rad/s}^2$ (angular acceleration)
Substituting the values:

$$ \omega(1) = 4 + 2(1) = 6 , \text{rad/s} $$

  1. Calculate the tangential velocity of point A

The tangential velocity ($v_t$) at a radius $r$ is given by:

$$ v_t = r \omega $$

Given that the radius ( r = 0.2 , \text{m} ) (200 mm), we find:

$$ v_t = 0.2 \times 6 = 1.2 , \text{m/s} $$

  1. Express the velocity vector in terms of unit vectors

Since point A starts at angle $\theta$ and moves with angular velocity, we can express the final position in unit vectors:

$$ \vec{v_A} = v_t \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} $$

At $t=1s$, with $\theta$ determined, the velocity is:

$$ \vec{v_A} = 1.2 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} $$

  1. Calculate the centripetal acceleration

The centripetal acceleration ($a_c$) is given by:

$$ a_c = \frac{v_t^2}{r} $$

Substituting values:

$$ a_c = \frac{(1.2)^2}{0.2} = 7.2 , \text{m/s}^2 $$

  1. Calculate the total acceleration of point A

The total acceleration ($\vec{a_A}$) consists of tangential acceleration ($a_t$) and centripetal acceleration ($a_c$):

$$ a_t = r \alpha = 0.2 \times 2 = 0.4 , \text{m/s}^2 $$

The total acceleration vector is:

$$ \vec{a_A} = \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} a_t + \begin{pmatrix} -\sin(\theta) \ \cos(\theta) \end{pmatrix} a_c $$

Substituting values:

$$ \vec{a_A} = 0.4 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} + 7.2 \begin{pmatrix} -\sin(\theta) \ \cos(\theta) \end{pmatrix} $$

The velocity of point A at $t=1$ s is given by:

$$ \vec{v_A} = 1.2 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} $$

The acceleration of point A at $t=1$ s is:

$$ \vec{a_A} = 0.4 \begin{pmatrix} \cos(\theta) \ \sin(\theta) \end{pmatrix} + 7.2 \begin{pmatrix} -\sin(\theta) \ \cos(\theta) \end{pmatrix} $$

More Information

This problem illustrates how to analyze the motion of a point on a rotating disk by applying the concepts of angular velocity, tangential velocity, and centripetal acceleration. The importance of distinguishing between linear and angular quantities is emphasized.

Tips

  • Forgetting to convert units (e.g., mm to meters).
  • Mixing up tangential and centripetal acceleration.
  • Not properly applying trigonometric functions based on the angle.

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