Chapter 15, Problem 15.248RP

A wheel moves in the xy plane in such a way that the location of its center is given by the equations x_O = 12t³ and y_O = R = 2, where x_O and y_O are measured in feet and t is measured in seconds. The angular displacement of a radial line measured from a vertical reference line is θ = 8t⁴, where θ is measured in radians. Determine the velocity of point P located on the horizontal diameter of the wheel at t = 1 s.

Fig. P15.248

To determine

Find the velocity of the point P located on the horizontal diameter at time $t=1\text{s}$.

Answer to Problem 15.248RP

The velocity of the point P located on the horizontal diameter at time $t=1\text{s}$ is $\underset{\_}{\left(\text{36}\text{i}-64\text{j}\right)\text{ft/s}}$.

Explanation of Solution

Given information:

Show the location of the center of the wheel as follows:

$\begin{array}{l}{x}_0=12t^3\\ y_0=R=2\end{array}$

Here, $x_0$ and $y_0$ are in feet and t are in seconds.

The angular displacement of the radial line measured from vertical reference line is denoted by $\theta$.

Show the angular displacement $\left(\theta \right)$ as follows:

$\theta =-8t^4$

Here, $\theta$ is in radians, and t is in seconds.

Calculation:

Consider the position of the point P with respect to point O is denoted by ${\text{r}}_{P/O}=2\text{i}\text{ft}$.

Calculate the angular velocity of the wheel at time $t=1\text{s}$ using the relation:

${\omega }_{OP}=\frac{d\theta }{dt}\text{k}$

Substitute $-8t^4$ for $\theta$.

$\begin{array}{c}{\omega }_{OP}=\frac{d\left(-8t^4\right)}{dt}\text{k}\\ \text{=}\left(-8\times 4t^3\right)\text{k}\\ \text{=}-32t^3\text{k}\end{array}$

Substitute $1\text{s}$ for t.

$\begin{array}{c}{\omega }_{OP}\text{=}-32{\left(1\right)}^3\text{k}\\ \text{=}-32\text{k}\text{rad/s}\end{array}$

Calculate the velocity at point O using the relation:

${\text{v}}_O=\left(\frac{d{x}_0}{dt}\right)\text{i}$

Substitute $12t^3$ for $x_0$.

$\begin{array}{c}{\text{v}}_O=\left[\frac{d\left(12t^3\right)}{dt}\right]\text{i}\\ \text{=}\left(\text{12}\times {\text{3t}}^2\right)\text{i}\\ ={\text{36t}}^2\text{i}\end{array}$

Substitute $1\text{s}$ for t.

$\begin{array}{c}{\text{v}}_O=\text{36}\times {\left(1\right)}^2\text{i}\\ =\text{36}\text{i}\text{ft/s}\end{array}$

Show the relation between the velocity at point P and O as follows:

${\text{v}}_P={\text{v}}_O+{\omega }_{OP}\times {\text{r}}_{P/O}$

Substitute $\text{36}\text{i}\text{ft/s}$ for ${\text{v}}_O$, $-32\text{k}\text{rad/s}$ for ${\omega }_{OP}$, and $2\text{i}\text{ft}$ for ${\text{r}}_{P/O}$.

$\begin{array}{c}{\text{v}}_P=\text{36}\text{i}-32\text{k}\times 2\text{i}\\ =\left(\text{36}\text{i}-64\text{j}\right)\text{ft/s}\end{array}$

Thus, the velocity of the point P located on the horizontal diameter at time $t=1\text{s}$ is $\underset{\_}{\left(\text{36}\text{i}-64\text{j}\right)\text{ft/s}}$.