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 $x_0=12t^3$ and $y_0=R=2$ , where $x_0$and $y_0$ are measured in feet and t is measured in seconds. The angular displacement of a radial line measured from a vertical reference line is $\theta =8t^4$, where $\theta$ is measured in radians. Determine the velocity of point P located on the horizontal diameter of the wheel at $t=1s$ .

Fig. P15.248

To determine

Velocity of point P at $t=1s.$

Answer to Problem 15.248RP

$t=1s$

The velocity at point P is equal to $\left(36ft/s\right)i-\left(64ft/s\right)j.$

Explanation of Solution

Given information:

Point P is located on horizontal diameter.

The angular displacement is given as:

$\theta =8t^4$

The linear displacement is given as:

$x_0=12t^3$

The angular velocity $\omega$ is defined as:

$\omega =\frac{d\theta }{dt}$

The linear velocity $v_0$ is defined as:

$v_0=\frac{d{x}_0}{dt}$

Calculation:

According to given information:

$\theta =8t^4$

Differentiate

$\omega =\frac{d\theta }{dt}=32t^3$

At $t=1s$

$\omega =32{\left(1s\right)}^3=32rad/s$

Therefore

$v^1=R\omega =\left(2ft\right)\left(32rad/s\right)=64ft/s$

Then

$x_0=12t^3$

Differentiate

$v_0=\frac{d{x}_0}{dt}=36t^2$

At $t=1s$

$v_0=36{\left(1s\right)}^2=36ft/s$

Therefore, the velocity of point P is:

$v=\left(36ft/s\right)i-\left(64ft/s\right)j$

Conclusion:

At $t=1s$

The velocity at point P is equal to $\left(36ft/s\right)i-\left(64ft/s\right)j$.