Question

In Figure, a small $\mathbf{50}\mathbf{g}$ block slides down a frictionless surface through height $\mathbf{h}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{}\mathbf{cm}$and then sticks to a uniform rod of mass $\mathbf{100}\mathbf{}\mathbf{g}$and length$\mathbf{40}\mathbf{}\mathbf{cm}$ . The rod pivots about point Othrough angle $\mathbf{\theta }$before momentarily stopping. Find$\mathbf{\theta }$

Short answer

The angle through which the rod pivots about point Obefore momentarily stopping is $32°$.

Step-by-step solution

Step 1: Given

1. $m=5\text{0g}$
2. $M=1\text{00g}$
3. $\mathrm{h}=2\text{0cm}$

Determining the concept

Use the law of conservation of angular momentum and energy to find the angle.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

$\begin{array}{c}{\mathbf{L}}_{\mathbf{i}}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}\\ \mathbf{PE}\mathbf{=}\mathbf{KE}\end{array}$

Where, L is angular momentum, PE is potential energy and KE is kinetic energy.

 Step 3: Determining the angle through which the rod pivots about point O before momentarily stopping

First use conservation of energy to find velocity as follow:

$\mathrm{mgh}=0.5{\mathrm{mv}}^2$

So,

$\mathrm{v}=\sqrt{2\mathrm{gh}}$

Now, write the angular momentum before the collision is,

${\mathrm{L}}_{\mathrm{i}}=\mathrm{mvd}$

Here, $d$ is the length of the rod from the point $O$, about which the rod would be rotating.

Now, conservation of angular momentum,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ \mathrm{mvd}=({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{md}}^2)\mathrm{\omega }\end{array}$

Rotational inertia of the rod is about the axis passing through its end can be written as,

$\begin{array}{c}\mathrm{I}=\frac{1}{3}{\mathrm{Md}}^2\\ \mathrm{mvd}=\left(\frac{1}{3}{\mathrm{Md}}^2+{\mathrm{md}}^2\right)\mathrm{\omega }\end{array}$

So,

$\begin{array}{l}\mathrm{\omega }=\frac{\left(\mathrm{md}\sqrt{2\mathrm{gh}}\right)}{\frac{{\mathrm{Md}}^2}{3}+{\mathrm{md}}^2}\\ \mathrm{\omega }=\frac{\left(\mathrm{m}\sqrt{2\mathrm{gh}}\right)}{\mathrm{d}\left(\frac{\mathrm{M}}{3}+\mathrm{m}\right)}\end{array}$

From trigonometry,

Now, write,

$\begin{array}{c}\mathrm{H}=\mathrm{d}(1-\mathrm{cos\theta })\\ \frac{1}{2}({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{md}}^2){\mathrm{\omega }}^2=\mathrm{mgH}+\frac{\mathrm{MgH}}{2}\end{array}$

Plugging value of $\omega$found above,

$\frac{1}{2}({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{md}}^2){\mathrm{\omega }}^2=\mathrm{mg}\text{d}(1-\text{cosθ})+\frac{\mathrm{Mg}}{2}\text{d}(1-\text{cosθ})$

For rod rotating about its end,

$\begin{array}{c}{\mathrm{I}}_{\mathrm{rod}}=\frac{1}{3}{\mathrm{Md}}^2\\ \frac{1}{2}\left(\frac{{\mathrm{Md}}^2}{3}+{\mathrm{md}}^2\right)\frac{{\left(\mathrm{m}\sqrt{2\mathrm{gh}}\right)}^2}{{\mathrm{d}}^2{\left(\frac{\mathrm{M}}{3}+\mathrm{m}\right)}^2}=\mathrm{mg}\text{d}(1-\text{cosθ})+\frac{\mathrm{Mg}}{2}\text{d}(1-\text{cosθ})\\ \frac{1}{2}\frac{{\left(\mathrm{m}\sqrt{2\mathrm{gh}}\right)}^2}{\left(\frac{\mathrm{M}}{3}+\mathrm{m}\right)}=\left[\mathrm{mg}\text{d}+\frac{\mathrm{Mg}}{2}\text{d}\right](1-\text{cosθ})\end{array}$

Since, all the values are known except $\theta$, solve this for$\theta$.

Solving this for angle,

$\mathrm{\theta }={32}^0$

Hence, the angle through which the rod pivots about point Obefore momentarily stopping is${32}^0$ .

Using the law of conservation of angular momentum and angular velocity, the angle can be found.