Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
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 [tex](-V_y)[/tex] along the y-axis
- the linear velocity [tex](V_x)[/tex] that is directed along the x-axis
Now;
[tex]V_x = \frac{d}{dt}(12t^3+2) = 36 t^2[/tex]
[tex]V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s[/tex]
Also,
[tex]-V_y = R* \omega[/tex]
where [tex]\omega[/tex](angular velocity) = [tex]\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)[/tex]
[tex]-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s[/tex]
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is [tex]P = 104.04 \hat{i} -314.432 \hat{j}[/tex]
