66Consider the following diagram:
The x and y directions are shown with the origin of these axes being shown by short lines perpendicular to these axes. Note that the origin of the y coordinate moves with the bigger block.
Since the length of the thread remains constant. Therefore,
4x_M+y_m=\ \mathrm{constant}
In writing this equation, I have used xM to denote the position of mass M w.r.t the origin (neglecting the sizes of the pulleys) and ym as the position of m w.r.t the origin of the y coordinate.
Differentiating once w.r.t time t, we get
4v_{Mx}+v_{my}^\prime=0
Here, v_{Mx} is the velocity of M along the x axis w.r.t the ground and v_{my}^\prime is the velocity of m along the y direction w.r.t M. It is easy to see that if v_{Mx} is positive then v_{my}^\prime is negative. To obtain the velocity v_m of m w.r.t the ground we use the vector relation
\vec{v}_{mG}=\vec{v}_{mM}+\vec{v}_{MG}
We have the following figure
Taking into account that |\vec{v}_{MG}|=|v_{Mx}|=v_M (say)
|\vec{v}_{mG}|=v_m
and |\vec{v}_{mM}|=|v_{my}^\prime|=|-4v_{Mx}|=4v_M
we get
\cos(\pi-\theta)=\dfrac{v_M^2+|\vec{v}_{mM}|^2-v_m^2}{2|\vec{v}_{mM}|v_M}
i.e.
-\cos\theta=\dfrac{v_M^2+16v_M^2-v_m^2}{2\cdot 4v_M^2}=\dfrac{17}{8}-\dfrac{1}{8}\left(\dfrac{v_m}{v_M}\right)^2
Solving, we get
\boxed{\dfrac{v_m}{v_M}=\dfrac{9}{2}}
