Question:

If the rope is drawn toward the motor M at a speed of ${v}_{M}$ = (5${t}^{3/2}$) m/s , where t is in seconds, determine the speed of cylinder A when t = 1 s .

Step-by-step

position Coordinates: By referring to Fig. a, the length of the rope written in terms of the position coordinates $s_{A}$ and $s_{M}$ is

3 $s_{A}$+$s_{M}$=l

Time Derivative: Taking the time derivative of the above equation,
(+$\downarrow$)         3 $v_{A}$+$v_{M}$=0
Here, $v_{M}$=$\left(5 t^{3 / 2}\right)$ m/s. Thus,

3 $v_{A}$+5 $t^{3 / 2}$=0
$v_{A}$=$\left(-\frac{5}{3} t^{3 / 2}\right)$ m/s=$\left(\frac{5}{3} t^{3 / 2}\right)$ m /$\left .s\right|_{t=1 s}$=1.67 m/s