Question

A wheel $A$ is connected to a second wheel $B$ by means of inextensible string, passing over a pulley $C$, which rotates about a fixed point horizontal axle $O$, as shown in the figure. The system is released from rest. The wheel $A$ rolls down the inclined plane $OK$ thus pulling up wheel $B$ which rolls along the inclined plane $ON$.
Determine the velocity (in $m/s$) of the axle of wheel $A$, when it has travelled a distance $s=3.5m$ down the slope. Both wheels and the pulley are assumed to be homogeneous disks of identical weight and radius. Neglect the weight of the string. The string does not slip over $C$. (Take $\alpha ={53}^o$ and $\beta ={37}^o$)

Answer 1

First, draw the fbd on A, B and C.
And $\omega$ be the angular acceleration of A and B.
On A,
$mgsin\left(\alpha \right)-T_1-f_1=ma$---------->1 {where , $T_1$ is tension on A due to string, and $f_1$ is frictional force on A .}
On B,
$T_2-mgsin\left(\beta \right)-f_2=ma$.------>2
On C,
$\left[\right(T_1-T_2\left)\omega \right]R=M\frac{R^2}{2}\alpha$----------->3
Apply Torque equations;
ON A,
at COM;
$f_{1}R=M\frac{R^2}{2}\omega$--------->4
at point of contact to inclined,
$\left[mgsin\right(\alpha )-T_1]R=3M\frac{R^2}{2}\alpha$----------->5
ON B;
At COM;
$f_{2}R=\frac{M{R}^2}{2}\omega$------------>6
At the point of contact to inclined,
$[T_2-mgsin(\beta \left)\right]R=3I\omega$---------->7
AND we know that,
$a=\omega$R----------->8
By solving 4 and 5 ,
we get $f_1=\frac{mgsin\left(\alpha \right)-T_1}{3}$
by solving 6 and 7,
we get,
$f_2=\frac{T_2-mgsin\left(\beta \right)}{3}$
by substituting above got equations in 1 and 2 .and solving them we get,
$mg\left(sin\right(\alpha )-sin(\beta \left)\right)-[T_2-T_1]=\frac{3}{2}ma$
substitute 3 in above equation,
we get,
$a=\frac{2g}{35}$
By substituting it in $v=\sqrt{2as}=2\frac{m}{s}$