Question 711: Centre of Mass and Rotational Motion

Question

List I describes four systems, each with two particles $A$ and $B$ in relative motion as shown in figures. List II gives possible magnitudes of their relative velocities (in $m s^{-1}$ ) at time $t=\frac{\pi}{3} s$.
List-I
List-II
(I) $A$ and $B$ are moving on a horizontal circle of radius $1 \mathrm{~m}$ with uniform angular speed $\omega=1 \mathrm{rad} \mathrm{s}^{-1}$. The initial angular positions of $A$ and $B$ at time $t=0$ are $	heta=0$ and $	heta=\frac{\pi}{2}$, respectively.
(P) $\frac{\sqrt{3}+1}{2}$
(II) Projectiles $A$ and $B$ are fired (in the same vertical plane) at $t=0$ and $t=0.1 \mathrm{~s}$ respectively, with the same speed $v=\frac{5 \pi}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-1}$ and at $45^{\circ}$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide. $\left(g=10 \mathrm{~ms}^{-2}ight)$.
(Q) $\frac{\sqrt{3}-1}{\sqrt{2}}$
(III) Two harmonic oscillators $A$ and $B$ moving in the $x$ direction according to $x_{A}=x_{0} \sin \frac{t}{t_{0}}$ and $x_{B}=x_{0} \sin \left(\frac{t}{t_{0}}+\frac{\pi}{2}ight)$ respectively, starting from $t=0$. Take $x_{0}=1 \mathrm{~m}, t_{0}=1 \mathrm{~s}$.
(R) $\sqrt{10}$
(IV) Particle $A$ is rotating in a horizontal circular path of radius $1 \mathrm{~m}$ on the $x y$ plane, with constant angular speed $\omega=1 \mathrm{rad} \mathrm{s}^{-1}$. Particle $B$ is moving up at a constant speed $3 \mathrm{~m} \mathrm{~s}^{-1}$ in the vertical direction as shown in the figure. (Ignore gravity.)
(S) $\sqrt{2}$
(T) $\sqrt{25\pi^{2}+1}$
Which one of the following options is correct?

Accepted Answer

(I)
$${v_{net}} = \sqrt 2 \omega R = \sqrt 2 $$
I $$	o$$ S
(II) $${v_A}(t = 0.13) = {{5\pi } \over {\sqrt 2 }}\cos 45\widehat i + \left[ {{{5\pi } \over {\sqrt 2 }} 	imes {1 \over {\sqrt 2 }} - g 	imes 0.1} ight]\widehat j$$
$$ = {{5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2} - 1} ight)\widehat j$$
$${v_B}(t = 0.1\,\sec ) = {{ - 5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2}} ight)\widehat j$$
After $$t = 0.1$$, relative velocities should not change.
$${v_{rel}}(t = 0.1\,\sec ) = \left| {5\pi \widehat i - \widehat j} ight|$$
$$ = \sqrt {25{\pi ^2} + 1} $$
II $$	o$$ T
(III) $$x = {x_A} - {x_B}$$
$$ = {x_0}\sin t - {x_0}\sin \left( {t + {\pi  \over 2}} ight)$$
$$ = \sqrt 2 {x_0}\sin \left( {t - {\pi  \over 4}} ight)$$
$${v_{rel}} = {{dx} \over {dt}} = \sqrt 2 {x_0}\cos \left( {t - {\pi  \over 4}} ight)$$
$$ = \sqrt 2 \cos \left( {{\pi  \over 3} - {\pi  \over 4}} ight)$$
$$ = \sqrt 2  	imes {{\sqrt 3  + 1} \over {2\sqrt 2 }} = {{\sqrt 3  + 1} \over 2}$$
III $$	o$$ P
(IV) $${v_{rel}} = \sqrt {{3^2} + {1^2}}  = \sqrt {10} $$
IV $$	o$$ R

Answer

I $\rightarrow$ R, II $\rightarrow$ T, III $\rightarrow$ P, IV $\rightarrow$ S

Answer

I $\rightarrow$ S, II $\rightarrow$ P, III $\rightarrow$ Q, IV $\rightarrow$ R

Answer

I $\rightarrow$ S, II $\rightarrow$ T, III $\rightarrow$ P, IV $\rightarrow$ R

Answer

I $\rightarrow$ T, II $\rightarrow$ P, III $\rightarrow$ R, IV $\rightarrow$ S

Centre of Mass and Rotational Motion

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