Rotation Ques 10
10 A rigid massless rod of length $3L$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be
(2019 Main, 10 Jan II)
(a) $\frac{g}{13,l}$
(b) $\frac{g}{2 l}$
(c) $\frac{7 g}{3 l}$
(d) $\frac{g}{3 l}$
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Answer:
Correct Answer: 10.( a )
Solution:
Torque $(\tau)$ about $\quad P=\mathbf{r}_1 \times \mathbf{F}_1+\mathbf{r}_2 \times \mathbf{F}_2$
$ \begin{array}{lc} \Rightarrow & \tau=l \times 5 M_{0} g \text { (outwards) }-2 l \times 2 M_{0} g \\ \Rightarrow & \tau=5 M_0 g l-4 M_0 g l \\ \Rightarrow & \tau=M_0 g l \text { (outwards) } \\ \text { or } & \tau=M_0 g l \end{array} $
Now we know that torque is also given by
$ \tau=I \alpha $
Here, $I=$ moment of inertia (w.r.t. point $P$ ) of rod and $\alpha=$ angular acceleration.
For point $P, I=\left(5 M_0\right) \times l^2+\left(2 M_0\right)(2 l)^2 \quad\left[\because I=M R^2\right]$
$ \Rightarrow \quad I=13 M_0 l^2 $
Putting value of $I$ from Eq. (iv) in Eq. (iii), we get
$ \tau=\left(13 M_0 l^2\right) \alpha $
From Eqs. (ii) and (v), we get
$ M_0 g l=13 M_0 l^2 \alpha \Rightarrow \alpha=\frac{g}{13 l} $
BETA
