COM and rotational

Centre of mass (COM) is distance or point where entire mass of the system is supposed to be concentrated. Its general formula is

$r_{\text{cm}}=\frac{\int r.dm}{\int dm}$

or, simply $r_{\text{cm}}=\frac{m_1{r}_1+m_2{r}_2+m_3{r}_3+..............+m_n{r}_n}{m_1+m_2+m_3+............+m_n}$

where $m_1,m_2,m_3$......... are generally taken as particle mass.

#If $m_1$and $m_2$are seperated by distance $\text{x}$; Position of

COM from $m_1=\frac{m_{2}x}{m_1+m_2}$

$d=\frac{m_{1}d+m_2(x-d)}{m_1+m_2}$

or, $d=\frac{m_{2}x}{m_1+m_2}$

similarly,distance from$m_2$=$\frac{m_{1}x}{m_1+m_2}$

Center of gravity: It is the point through which whole weight of the body acts irrespective of the position of the body.

Note: For all practical purposes center of mass and center of gravity are assumed to be same. It differs only when gravitational field is not uniform or not parallel. COM is important in the study of Rotational Dynamics.

Rigid body is the body whose shape and size remains same under effect of any external forces.

Moment of Inertia is a measure of an object resistance to changes to its rotation or it is the capacity of the body or cross section to resist bending.

It is given by $I=m{r}^2$where m is mass and r is the distance from axis of rotation.

i. Theorem of parallel axis

$I_{\text{xx}}=I_{\text{XX}}+m{d}^2$

ii. Theorem of perpendicular axis

$I_x=I_y+I_z$ $I_y=I_x+I_z$ $I_z=I_x+I_y$

MOI of different bodies

1.Rod

i. Passing through center and perpendicular to length uniform =$\frac{1}{12}M{L}^2$

ii. Passing through one end, perpendicular to length =$\frac{1}{3}M{L}^2$

iii. On certain angle = $\frac{1}{3}M{L}^{2}si{n}^2\theta$

2.Ring

i.$C_g$and perpendicular to plane = $M{R}^2$

ii. along diameter = $\frac{1}{2}M{R}^2$

3.Disc

i. $C_g$and perpendicular to plane = $\frac{1}{2}M{R}^2$

ii. along diameter = $\frac{1}{4}M{R}^2$

4.Solid sphere along diameter = $\frac{2}{5}M{R}^2$

5.Hollow sphere along diameter = $\frac{2}{3}M{R}^2$

6.Solid cylinder

i. along axis = $\frac{1}{2}M{R}^2$

ii. $C_g$and perpendicular to length = $M\left(\frac{L^2}{12}+\frac{R^2}{4}\right)$

7.Hollow cylinder

i. along axis =$M{R}^2$

ii. $C_g$and perpendicular to length = $M\left(\frac{L^2}{12}+\frac{R^2}{2}\right)$

8.Rectangular lamina $C_g$and perpendicular to length = $\frac{M}{12}(a^2+b^2)$

Radius of gyration is the point where entire mass of the body is concentrated (COM).

Given by $I=M{K}^2$

$K=\sqrt{\frac{I}{M}}=\sqrt{\frac{r_1^2+r_2^2+r_3^2+..........+r_n^2}{n}}$

It depends on the axis of rotation.

Torque on body: It is the measure of turning effect of force.

$\vec{\tau }=\vec{r}\times \vec{F}=Frsin\theta \hat{n}$ is a vector quantity.

Work done by torque

small work done: $dw=\tau .d\theta$

$\therefore w=\tau \theta$

Power=$\frac{dw}{dt}=\tau \frac{d\theta }{dt}=\tau \omega$ where$\omega$is the angular velocity.

Angular momentum: It is the moment of linear momentum about an axis.

It is given as $\vec{L}=\vec{r}\times \vec{P}=Prsin\theta \hat{n}$ is a vector quantity.

Relation between torque and angular momentum is

$\tau =\frac{dL}{dt}=I\alpha$

Law of conservation of angular momentum

$|L|=Prsin\theta =Pr(\theta =90)=mvr=m(wr)r=mw{r}^2=Iw$

For no torque applied,$\tau =0$so $dL=0$ $\therefore L=constant$

$\therefore Iw=constant$

eg: Ice skater use this principle, boil egg comes to rest earlier,etc.

If radius of earth is made $\frac{R}{n}$the time period =$\frac{T}{n^2}$ as,

$I_1\times \frac{2\pi }{T_1}=I_2\times \frac{2\pi }{T_2}$

or, $M{R}^2\times \frac{2\pi }{T_1}=M\times \frac{R^2}{n^2}\times \frac{2\pi }{T_2}$

$\therefore T_2=\frac{T_1}{n^2}$

Similarly, if polar ice melts the length of the day increases as radius increases.

Kinetic energy of rotating body

The rotational kinetic energy is given by $=\frac{1}{2}I{w}^2$and linear KE$=\frac{1}{2}m{v}^2$

$\therefore$$K{E}_{\text{total}}=K{E}_{\text{rotational}}+K{E}_{\text{linear}}$

$=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{v}^2\left(1+\frac{k^2{w}^2}{v^2}\right)$

$=\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)$

$\therefore$Ratio of rotational energy : Translational energy : Total energy = $K^2:R^2:(k^2+R^2)$

For a body moving on inclined plane

acceleration(a)=$\frac{gsin\theta }{\left(\frac{K^2}{R^2}+1\right)}$

velocity(v)=$\sqrt{\frac{2gh}{\left(\frac{K^2}{R^2}+1\right)}}$

Time period=$\sqrt{\frac{2h}{g}\left(1+\frac{k^2}{R^2}\right)}cosec\theta$

is obtained when it is frictionless

If friction is provided ; coefficient of friction$(\mu )$is calculated as:

$(gsin\theta -\mu gcos\theta )=\frac{gsin\theta }{\left(1+\frac{k^2}{R^2}\right)}$

Therefore, $\mu \ge \left(\frac{k^2}{k^2+R^2}\right)tan\theta$

Equation of rotational motion

i. $w=w_0+at$ ; $w^2=w_0^2+2\alpha \theta$

ii. $\theta =w_{0}t+\frac{1}{2}\alpha t^2$ ; $\theta =\frac{w+w_0}{2}\times t$

iii. $w=\frac{d\theta }{dt}$ $\theta =\frac{d^2\theta }{d{t}^2}=\frac{dw}{dt}$

iv. No. of revolution$=\frac{\theta }{2\pi }$

Change in Moment of inertia with temperature

$I=m{r}^2$

$dI=2mrdr$

$dI=2mr\times (r\alpha \Delta \theta )$

$dI=2(m{r}^2)\alpha \Delta \theta =2I\alpha \Delta \theta$

i. If two solid spheres of same material having their radius $R_1$and $R_2$respectively, then the ratio of MOI is

$I=\frac{2}{5}M{R}^2$

$I=\frac{2}{5}(V\times \rho )R^2=\frac{2}{5}\left(\frac{4}{3}\pi R^3\right)\times R^2$

$\therefore I\propto R^5$

ii. If a rod of length L is allowed to freely fall then its velocity when it reaches to ground making an angle $\theta$with the vertical is

$\sqrt{6gL}sin\frac{\theta }{2}$

iii. When two rigid bodies with angular velocities $w_1$and $w_2$are coupled and $\omega$be the angular velocity of combination. Then, in absence of external torque

$I_1{w}_1+I_2{w}_2=(I_1+I_2)\omega$

iv. The internal and external radii of circular lamina is r and R then its MOI about its own axis if mass is M is

$=\frac{M}{2}(R^2+r^2)$

v. A solid sphere is rolling without sliding with velocity V. The velocity of particle on its surface of angle $\theta$above the level of C.M. is

$v^{\prime }=2vcos(45-\frac{\theta }{2})$

as, $v^{\prime }=2vcos\varphi$ and $\varphi =\left(\frac{90+\theta }{2}\right)$