Question

A thread is wound around two discs on either sides. The pulley and the two discs have the same mass and radius. There is no slipping at the pulley and no friction at the hinge. The linear acceleration of disc $1$, disc $2$ and the angular acceleration of pulley be $a_1$, $a_2$ and $\alpha$ respectively. Then

• A. $a_1=a_2=\frac{g}{3}$
• B. $a_1=a_2=\frac{2g}{3}$
• C. $\alpha =0$
• D. $a_1=\frac{g}{3},a_2=\frac{2g}{3}$

Answer 1

Answer: (B), (C)

$\alpha =0$

Let $\text{R}$ be the radius of the discs and $T_1$ and $T_2$ be the tensions in the left and right segments of the rope.


From free body diagram, acceleration of disc $1$

$a_1=\frac{mg-T_1}{m}...\left(1\right)$

Acceleration of disc $2$

$a_2=\frac{mg-T_2}{m}...\left(2\right)$

Angular acceleration of disc $1$,
${\alpha }_1=\frac{\tau }{I}$
where,
$\tau =$torque about the centre$=T_{1}R$
$I=$Moment of inertia$=\frac{1}{2}m{R}^2$

Substituting the values, we get
${\alpha }_1=\frac{T_{1}R}{\frac{1}{2}m{R}^2}=\frac{2T_1}{mR}...\left(3\right)$

Similarly, angular acceleration of disc $2$,
${\alpha }_2=\frac{2T_2}{mR}...\left(4\right)$
Both ${\alpha }_1$ and ${\alpha }_2$ are clockwise

Angular acceleration of pulley:


From the above figure,
$\alpha =\frac{(T_2-T_1)R}{\frac{1}{2}m{R}^2}=\frac{2(T_2-T_1)}{mR}...\left(5\right)$

For no slipping,
$a_1-R{\alpha }_1=R{\alpha }_2-a_2=R\alpha ...\left(6\right)$

From equation $\left(3\right)$, $\left(4\right)$ and $\left(5\right)$, we have,

$\alpha =\frac{2(\frac{mR{\alpha }_2}{2}-\frac{mR{\alpha }_1}{2})}{mR}$

$\Rightarrow \alpha ={\alpha }_2-{\alpha }_1...\left(7\right)$

From equation $\left(1\right)$ and $\left(6\right)$, we have
$\frac{mg-T_1}{m}-R{\alpha }_1=R\alpha$

Substituing the value of $T_1$ from equation $\left(3\right)$, we get

$\frac{mg-\frac{mR{\alpha }_1}{2}}{m}-R{\alpha }_1=R\alpha$

$\Rightarrow {\alpha }_1=\frac{g-R\alpha }{\frac{3R}{2}}...\left(8\right)$

Similarly, from equation $\left(2\right)$, $\left(6\right)$ and $\left(4\right)$,

$\Rightarrow {\alpha }_2=\frac{g+R\alpha }{\frac{3R}{2}}...\left(9\right)$

Now, from equation $\left(7\right)$, $\left(8\right)$ and $\left(9\right)$, we get
$\alpha =\frac{g+R\alpha }{\frac{3R}{2}}-\frac{g-R\alpha }{\frac{3R}{2}}$

$\Rightarrow \alpha =\frac{4\alpha }{3}$

$\Rightarrow -\frac{\alpha }{3}=0$

$\therefore \alpha =0$

Now sustituting $\alpha =0$ in equation $\left(8\right)$, we get

${\alpha }_1=\frac{2g}{3R}$

Now, from equation $\left(6\right)$ we get,

$a_1=a_2=\frac{2g}{3R}\left(R\right)=\frac{2g}{3}$

Hence, option (b) and (c) are correct.

Why this question: This question will help the students to learn how to properly apply Newton's second law in translational and rotational form along with no slipping condition.