Question

A uniform rigid rod of mass $m$and length $l$ is released from vertical position on rough surface with sufficient friction for lower end not to slip as shown in figure. When rod makes angle $60°$ with vertical then find correct alternatives.

• A. $\alpha =\frac{2g}{l}$
• B. $\alpha =\sqrt{\frac{3g}{2l}}$
• C. $N=mg/16$
• D. $a_{radical}=3g/4$

Answer 1

Answer: (B), (C), (D)

$a_{radical}=3g/4$

Step 1 : Given data

Mass of the rigid rod $=m$

Length of the rod $=l$

Rod making angle with vertical $={60}^0$

Let when the rod making an angle with vertical then changed Kinetic Energy be $∆K$ and Potential Energy be $∆U$.

Let the Radial acceleration be $\alpha$ and vertical acceleration be ${\alpha }_v$.

Step 2 : When the rod making angle with vertical then find it's alternatives

As there is no dissipation of energy then the total energy will be conserved,

$∆K+∆U=0$

$∆K=-∆U$

We know that,

The Kinetic Energy of a rotational rod is, $K=\frac{1}{2}{I}_0{\omega }^2$ (where, $I_0$ is Moment of Inertia, $\omega$ is angular velocity.)

When the rod rotates the vertical position of the center of gravity $△l$ changes as shown in the figure,

$\Rightarrow △l=\frac{l}{4}$

Therefore, the Potential Energy of the rotational rod is, $U=mg△l$ (where $△l$ is length of the vertical position of the center of gravity and $g$ is gravity.)

Therefore,

$$\begin{gathered} ∆K=-∆U \\ \Rightarrow \frac{1}{2}{I}_0{\omega }^2=-mg∆l \\ \Rightarrow \frac{1}{2}{I}_0{\omega }^2=-mg\frac{l}{4} \end{gathered}$$

(Negative potential energy means that the work done is smaller as we approach the gravitational field. Hence we can ignore the sign).

Step 3. Find Angular velocity $\omega$,

We know that Moment of inertia for a rod $I_0=\frac{1}{3}m{l}^2$,

$$\begin{gathered} \Rightarrow \frac{1}{2}\left(\frac{1}{3}m{l}^2\right){\omega }^2=mg\frac{l}{4} \\ \Rightarrow {\omega }^2=\frac{3\times 2mgl}{4m{l}^2} \\ \Rightarrow {\omega }^2=\frac{3g}{2l} \\ \Rightarrow \omega =\sqrt{\frac{3g}{2l}} \end{gathered}$$

Hence, the Angular Velocity $\omega =\sqrt{\frac{3g}{2l}}$.

Step 4. Find Radial acceleration $\alpha$ and Vertical acceleration ${\alpha }_v$,

Now we know that the Torque $\tau =r·F\sin \theta$ (where, $F$ is force, $r$ is length of rotation and $\theta$ is angle),

Here, $r=\frac{l}{2}$from figure, and we know that $F=mg$,

Also Torque in terms of Moment of Inertia $I_0$ and radial acceleration $\alpha$ is given by,

$\tau =\alpha I_0$

Then we can write,

$$\begin{gathered} \tau =I_0\alpha \\ \Rightarrow r·F\sin \theta =I_0\alpha \\ \Rightarrow \left(\frac{l}{2}\right)mg\sin 60=I_0\alpha \\ \Rightarrow mg\frac{l}{2}\times \frac{\sqrt{3}}{2}=\frac{1}{3}m{l}^2\times \alpha \\ \Rightarrow \frac{3\sqrt{3}mgl}{4m{l}^2}=\alpha \\ \Rightarrow \alpha =\frac{3\sqrt{3}g}{4l} \end{gathered}$$

Radial acceleration $\alpha =\frac{3\sqrt{3}g}{4l}$.

Now,

Vertically downward acceleration ${\alpha }_v,$

$$\begin{gathered} {\alpha }_v=\alpha r\sin \theta -{\omega }^2\cos \theta \\ \Rightarrow {\alpha }_v=\frac{3\sqrt{3}g}{4l}\left(\frac{l}{2}\right)\sin 60°-{\left(\sqrt{\frac{3g}{2l}}\right)}^2\cos 60° \end{gathered}$$

$$\begin{gathered} \Rightarrow a_v=\frac{3\sqrt{3}g}{8}\left(\frac{\sqrt{3}}{2}\right)+\frac{3g}{4}\left(\frac{1}{2}\right) \\ \Rightarrow a_v=\frac{9g}{16}+\frac{3g}{8} \\ \Rightarrow a_v=\frac{9g+6g}{16} \\ \Rightarrow a_v=\frac{15g}{16} \end{gathered}$$

Hence, Vertically downward acceleration ${\alpha }_v=\frac{15g}{16},$

Step 5. Find Normal Force $N,$

Apply law of motion in vertical direction then Normal Force $N,$

$$\begin{gathered} mg-N=m{a}_v \\ \Rightarrow mg-N=\frac{15}{16}mg \\ \Rightarrow N=mg-\frac{15}{16}mg \\ \Rightarrow N=\frac{mg}{16} \end{gathered}$$

Hence the correct answers are Option B ,C, D.