Question

A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2, where xo and yo 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 the point P located on the horizontal diameter of the wheel at t = 1.4 s.

Answer 1

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

• the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

• the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

$-V_y = R* \omega$

where $\omega$(angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   $P = 104.04 \hat{i} -314.432 \hat{j}$