Physics System of Particles and Rotational Motion

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second

I? / I? = (MR²/2) / (mr²/4)
[ M = σπR², m = σπr² ]
= 2 (M/m) (R²/r²)
= 2 (σπR²/σπr²) (R²/r²)
= 2 (R²/r²)² = 2R? /r?

(a) I = ML²/12
(b) I = (2M)L²/3
(c) I = M (2L)²/12 = ML²/3
(d) I = (2M) (2L)²/3 = 8ML²/3

τ = Iα
I = ½MR² = ½ (10) (0.2)² = 0.2 kgm²
α = (ωf - ωi)/Δt = (0 - 600*2π/60)/10 = -2π rad/s²
τ = |Iα| = 0.2 * 2π = 0.4π = 4π * 10? ¹ Nm

$l_1=\frac{1}{2}m{r}^2$

$\begin{array}{l}{l}_2=\frac{1}{2}m{r}^2\\ l_3=\frac{1}{2}m{r}^2\\ l_4=\frac{2}{5}m{r}^2\end{array}$

mg – T = ma. (1)

T * R = l . (2)

a = R. (3)

With the help of equations (1), (2) and (3), we get

$a=\frac{mg}{m+\frac{l}{R^2}}$

$v=\sqrt[]{2ah}=\omega R$

$\Rightarrow {\omega }^2=\frac{2mgh}{l+m{R}^2}$

Let m_AB = m = 1 kg

AB = 0.4 m =  $\mathcal{l}$

d = OM = $\sqrt[]{0.16-0.04}=\sqrt[]{0.12}$

$d=2\sqrt[]{3}*10^{-1}$  

$I_{ABaboutO}=\frac{m{\mathcal{l}}^2}{12}+m{d}^2$           

$\therefore l_{hexagon,aboutO}=6\left[\frac{m{\mathcal{l}}^2}{12}+m{d}^2\right]=\frac{m{\mathcal{l}}^2}{2}+6m{d}^2$       

$=\frac{1*{\left(0.4\right)}^2}{2}+6*1*{\left(2\sqrt[]{3}*10^{-1}\right)}^2=\frac{0.16}{2}+6*4*3*10^{-2}$   

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-1 kgm²

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$=\frac{\sigma \pi a^2*a-\frac{\sigma \pi a^2}{4}*\frac{3a}{2}}{\sigma \pi a^2-\frac{\sigma \pi a^2}{4}}=\frac{\frac{5}{8}\sigma \pi a^3}{\frac{3}{4}\sigma \pi a^2}=\frac{5}{8}\sigma \pi a^2*\frac{4}{3}\sigma \pi a^2=\frac{5}{6}a$

(Material)                      (Susceptibility ( $\chi$  )

Diamagnetic                 (II) $0>\chi \ge -1$

Ferromagnetic  (III) $\chi ?1$

Paramagnetic                (IV) $0<\chi <\epsilon$

Non-magnetic               (I) $\chi =0$

$A=90^{?}$

In prism,  $r_1+c=A$

$r_1=90^{?}-c$                  …(i)

$sinc=\frac{1}{\mu }\Rightarrow cosc=\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1·sin30^{?}=\mu sin\left(r_1\right)\Rightarrow 1*\frac{1}{2}=\mu *sin\left(90^{?}-c\right)$

$\frac{1}{2}=\mu *\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

On squaring  $\frac{1}{4}={\mu }^2-1$

$\Rightarrow {\mu }^2=\frac{5}{4}\Rightarrow \mu =\frac{\sqrt[]{5}}{2}$