Two small spheres of mass m each are attached gently to the two ends of the rod. what is the final angular frequency of the system.

There are no external torques so principle of conservation of angular momentum applies.

$l_i=I_i \omega_i=\frac{M_L^2}{1^2} \omega_.$

$l_f=I_f \omega_f$

$=\left(\frac{M L^2}{12}+m L^2+m L^2\right) \omega_f$

$\omega_f=\left(\frac{M L^2 / 12}{\frac{M L^2}{1 2}+2ML^2}\right) \omega_1$

the connecting and is hight.

taking moments about p.

• $(m \times l) \times x$

• $=4 m l(2 l-x)$

• $x=\frac{8l}{5}$

The connecting and has the same mass few unit length (m).

• $m l x=2 lm (l-x)+4 lm (2 l-x)$

• $x=\frac{10l}{7}$

A uniform rod is on a table 2m and m they stick to the bar after striking it.

(1) To determine $v_c$ velocity of cm,

Use of principle of conservation of linear momentum.

$8 m(0)+2 m(-v)+m(2 v)=p_i=0(0$

$P_f=(8 m+2m+m) v_c \quad(2)$

$v_c = 0$

There is no translational motion.

(2) To calculate the angular velocity of the cm

$l_i = l_f$

$l_i = (2m v a ) + ( m \times 2v \times 2a) = 6m v a$

$l_f = I w$

$=\left[2 m \times a^2+m \times(2a)^2 +\left(8 m \times \frac{(6 a)^2}{12}\right)\right] \omega = 30 m a^2 w$

$30 m a^2 w = 6mvx$

$\omega = \frac{v}{5a}$

(3) Kinetic energy due to rotation.

$KE_{rotation} = \frac{1}{2} Iw^2$

$= \frac{1}{2} (30ma^2)(\frac{v}{5x})^2$

$= \frac{3}{5}mv^2$

$\mu$ : the coefficient of friction.

The rod rotation about A with a constant angular acceleration $(\alpha)$

$\alpha$ is constant ; $\therefore$ then $\omega = \alpha t$

Linear acceleration of the head: a $=$ l $\alpha$

The reaction force on the bead due to the rod = N = ma = mL $\alpha$

Centripetal force on the bead : $m\left(m \theta^2\right)$

{ $\dot{\theta}=\frac{d \theta}{d t}=\omega$}

• $=m L(\alpha t)^2=m L \alpha^2 t^2$

the limiting frictional force $=\mu N=\mu m L \alpha$

for slipping $\mu ml \alpha=m l\alpha^2 t^2$

• $t=\left(\frac{\mu}{\alpha}\right)^{1 / 2}$

Cubical block resting in a rough horizontal surface. The cofficient of friction is high such that the block does not slide before tripping . Calculate the $F_{min.}$ for the block to topple.

• $\sum F_y: \quad N=M_g$

• $\sum F_x: \quad F=f$

Torque equation about c

• $\left(F \times \frac{L}{2}\right)+\left(f \times \frac{L}{2}\right)=N \times \frac{L}{2}$

• $F + f = N$

• $2 F=N \Rightarrow F=\frac{N}{2}=\frac{Mg}{2}$

Diatomic molecule, rotational frequency and quantum theory.

x : reparation between the atoms

Given oxygen atoms

The separation between the atoms, $x = 1.20 \times 10^{-10} m$

Mass of the oxygen atoms $= 2.66\times 10^{-26} kg$

Calculate the rotational frequency $\omega$?

$I=\sum m_i r_i^2=m\left(\frac{x}{2}\right)^2+m\left(\frac{x}{2}\right)^2=\frac{m x^2}{2}$

The fundamental unit of angular momentum (according to quantum theory) $=\hbar$

$\hbar=1.054 \times 10^{-34} \frac{\mathrm{kg} \mathrm{m}^2}{s}.$

$I \omega \approx \hbar$

$\omega=\frac{\hbar}{2}$

=$\frac{1.054 \times 10^{34} kgm^2/s}{(\frac{2.66 \times 10^{-26}kg}{2})(1.20 \times 10^{-10}m)^2}$

$\simeq 5.2 \times 10^{11} \mathrm{rad} / \mathrm{sec}$

A rigid rod of mass m and length L rotates in a vertical plane about a frictionless pivot through the center.

1. The moment of inertia of the system

$I=\frac{M L^2}{12}+m_1\left(\frac{L}{2}\right)^2+m_2\left(\frac{L}{2}\right)^2=\frac{L^2}{4}\left(\frac{M}{3}+M_1+M_2\right)$

2. One ' $\omega$ ' is known ' $l$ ' can be calculated

• $l=I \omega$

• $=\frac{L^2}{4}\left(\frac{m}{3}+m_1+m_2\right) \omega$

3. $\tau_1=m_1 g \frac{l}{2} \text { cos $\theta$ }(\odot \equiv \text { one of the plane })$

• $\tau_2=m_2 g \frac{1}{2} \cos \theta(\otimes \equiv \text { into the phane })$

• $\tau_{\text {Total }}=\frac{1}{2}\left(m_1-m_2\right) q l \cos \theta$

• $\odot , if m_1 > m_2$

• $\otimes if m_2 < m$

• $\text { Since } I \alpha=l \text {, we can calculate } \alpha$

• $\alpha=\frac{\tau_ {\text {Total }}}{I} =\frac{2\left(m_ 1-m_ 2\right) g \operatorname{cos \theta}}{\frac{M}{3}+m_ 1+m_2}$