Linear acceleration of cylinder of mass $m_{2}$ is $a_{2}$. Then angular acceleration $\alpha_{2}$ is (given…
- $\frac{a_{2}}{R}$
- $\frac{\left(a_{2}+\mathrm{g}\right)}{R}$
- $\frac{2\left(a_{2}+g\right)}{R}$
- None of these
Solution
Let the tension in the string connected with mass \(m_1\) is \(T_1\) and the tension in the string connecting both the masses is \(T_2-\alpha_1\) is the acceleration of the mass \(m_1, a_2\) is the linear acceleration of the mass \(m_2\) and \(a_2\) is the angular acceleration of the mass \(\mathrm{m}_2\).
For mass \(\mathrm{m}_2\) writing the Newton's 2nd law of motion,
$\begin{aligned}
F_{\text{nat}} &= m_a \\
\Rightarrow m_{2g}-T_2 &= m_{2} a_{2} \ldots \ldots (i)
\end{aligned}$
The moment of inertia of the circular ring about its centre of mass is given as $I=M R^{2}$. Here, $M$ is the mass and $R$ is the radius of the ring. So,
$\begin{aligned}
I_{1} &= m_{1} R^{2} \\
I_{2} &= m_{2} R^{2}
\end{aligned}$
Net torque acting about the centre of mass \(\mathrm{m}_2\) is \(T_2 R\) and the direction is clockwise. For no slipping condition,
\(a_2-a_2 R\)
So, the angular acceleration will be,
\(\alpha_2=\frac{a_2}{R}\)
For mass \({m}_1\),
Using no slipping condition,
$\begin{aligned}
& c \alpha_{1}-\alpha_{2} \\
& \therefore a_{1}=\frac{a_{2}}{R}
\end{aligned}$
Asked in: JEE Mains - Rotational Motion - Test 2