Q: A thin circular ring of mass \$M\$ and radius \$r\$ is rotating about its axis with a constant angular velocity \$\omega\$. Two objects, each of mass \$m\$, are attached gently to the opposite ends of the diameter of the ring. The wheel now rotates with an angular velocity

Let the final angular velocity be \$w'.\$Angular momentum is conserved about center of the ring.\$L_i=L_f\$\$I w + 0 = I w' + 2 \times mr^2 w'\$\$Mr^2 w + 0 = Mr^2 w' + 2 \times mr^2 w'\$\$\implies w' =\dfrac{Mw}{M+2m}\$

Q: Find the scalar and vector products of two vectors, \$a=(3\hat {i}- 4\hat{j}+5\hat{k})\$ and \$b=(-2\hat{i}+\hat{j}-3\hat{k})\$.

\$a\cdot b=(3\hat{i}-4\hat{j}+5\hat{k})\cdot (-2\hat{i}+\hat{j}-3\hat{k})\$\$=-6-4-15\$\$=-25\$\$a\times b= \begin {vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & -4 &5\\ -2 & 1 & -3\end{vmatrix}\$ \$=7\hat{i}-\hat{j}-5\hat{k}\$Note \$b\times a=-7\hat{i}+\hat{j}+5\hat{k}\$

Q: Find the torque of a force \$7\hat{i}+3\hat{j}-5\hat{k}\$ about the origin. The force acts on a particle whose position vector is \$\hat{i}-\hat{j}+\hat{k}\$.

Here \$r=\hat{i}-\hat{j}+\hat{k}\$and \$F=7\hat{i}+3\hat{j}-5{k}\$.We shall use the determinant rule to find the torque \$\tau =r\times F\$\$\tau=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -1 & 1\\ 7 & 3 & -5\end{vmatrix}\$ \$=(5-3)\hat{i}-(-5-7)\hat{j}+(3-(-7))\hat{k}\$or \$\tau =2\hat{i}+12\hat{j}+10\hat{k}\$

Q: A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Let \$W\$ and \$W'\$ be the respective weights of the metre stick and the coin.The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.Mass of the metre stick=\$m'\$Mass of each coin=5gWhen the coins are placed \$12 cm\$ away from the end P, the centre of mass gets shifted by \$5 cm\$ from point R toward the end P. The centre of mass is located at a distance of \$45 cm\$ from point P.The net torque will be conserved for rotational equilibrium about point R.\$10g(45-12)-m'g(50-45)=0\$Thus \$m'=66g\$

Q: Three particles of mass \$1\ kg, 2\ kg\$ and \$3\ kg\$ are placed at the corners \$A, B\$ and \$C\$ respectively of an equilateral triangle \$ABC\$ of edge \$1\ m\$. Find the distance of their centre of mass from \$A\$.

Assume , Point \$A\$ is origin.From the figure, \$CP= 1\times sin(60^o)=\dfrac{\sqrt 3}{2}\$Centre of mass from Point \$A\$ in direction of \$x-axis\$, \$x_{cm}= \dfrac{m_Ax_A+m_Bx_B+ m_Cx_C}{m_A+m_B+m_C}\$\$\Rightarrow x_{cm}= \dfrac{1\times 0+ 2\times 1+ 3\times \dfrac{1}{2}}{1+2+3}=\dfrac{7}{12}\$Centre of mass from Point \$A\$ in direction of \$y-axis\$, \$y_{cm}= \dfrac{m_Ay_A+m_By_B+ m_Cy_C}{m_A+m_B+m_C}\$\$\Rightarrow y_{cm}= \dfrac{1\times 0+ 2\times 0+ 3\times \dfrac{\sqrt 3}{2}}{1+2+3}=\dfrac{\sqrt 3}{4}\$So coordinate of centre of mass \$\left ( \dfrac{7}{12} , \dfrac{\sqrt 3}{4} \right )\$, Distance from \$A\$, \$= \sqrt{\left (\dfrac{7}{12}\right )^2+ \left (\dfrac{\sqrt 3}{4}\right )^2}=\dfrac{\sqrt {19}}{6}\$

Q: The radius of gyration of a disc about its axis passing through its centre and perpendicular to its plane is

Moment of Inertia of disc about center of mass is \$I=\cfrac{mR^{2}}{2}\$Let a line parallel to axis at a distance \$d\$.Radius of Gyration\$=R\$Then, \$\cfrac{mR^{2}}{2}=mR^{2}\$\$\Rightarrow\,R^{2}=\cfrac{R^{2}}{2}\$\$\Rightarrow R=\cfrac{R}{\sqrt 2}\$

Question 1

Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when velocity of $A$ is $v$ and that of $B$ is $2v$, the velocity of centre of mass of the system

Accepted Answer

The velocity of the center of mass will be zero. The initial velocity of the particles were zero because both $A$ and $B$ were at rest so momentum will be zero. and it is given that there is no external force hence no change in momentum of center of mass and it will remain zero.By the law of conservation of momentum, the momentum is zero, and mass is not zero, velocity must be zeroHence, The option $D$ is the correct answer

Question 2

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects, each of mass $m$, are attached gently to the opposite ends of the diameter of the ring. The wheel now rotates with an angular velocity

Accepted Answer

Let the final angular  velocity be $w'.$Angular momentum is conserved about center of the ring.$L_i=L_f$$I  w + 0 = I  w' + 2 \times mr^2  w'$$Mr^2  w + 0 = Mr^2 w' + 2 \times mr^2  w'$$\implies  w' =\dfrac{Mw}{M+2m}$

Question 3

Find the scalar and vector products of two vectors, $a=(3\hat {i}- 4\hat{j}+5\hat{k})$ and $b=(-2\hat{i}+\hat{j}-3\hat{k})$.

Accepted Answer

$a\cdot b=(3\hat{i}-4\hat{j}+5\hat{k})\cdot (-2\hat{i}+\hat{j}-3\hat{k})$$=-6-4-15$$=-25$$a\times b= \begin {vmatrix} \hat{i} & \hat{j} & \hat{k}\ 3 & -4 &5\ -2 & 1 & -3\end{vmatrix}$ $=7\hat{i}-\hat{j}-5\hat{k}$Note $b\times a=-7\hat{i}+\hat{j}+5\hat{k}$

Question 4

Find the torque of a force $7\hat{i}+3\hat{j}-5\hat{k}$  about the origin. The force acts on a particle whose position vector is $\hat{i}-\hat{j}+\hat{k}$.

Accepted Answer

Here $r=\hat{i}-\hat{j}+\hat{k}$and $F=7\hat{i}+3\hat{j}-5{k}$.We shall use the determinant rule to find the torque $\tau =r\times F$$\tau=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 1 & -1 & 1\ 7 & 3 & -5\end{vmatrix}$ $=(5-3)\hat{i}-(-5-7)\hat{j}+(3-(-7))\hat{k}$or $\tau =2\hat{i}+12\hat{j}+10\hat{k}$

Question 5

Obtain the relation between linear velocity and angular velocity

Accepted Answer

Consider a particle moving with uniform circular motion in anticlock wise direction with centre $O$ and radius $r$ the particle cover an arc of length $\triangle s$ in time $\triangle t$ moving from $A$ to $B$.Hence the angular displacement is$\triangle \theta = \dfrac {\triangle s}{r}$Dividing both side by $\triangle t$$\dfrac {\triangle \theta}{\triangle t} = \dfrac {1}{r} \dfrac {\triangle s}{\triangle t}$In the time interval $\triangle t$ be infinitesimally small $(\triangle t \rightarrow 0)$, then$\displaystyle \lim_{\triangle t \rightarrow \theta} \dfrac {\triangle \theta}{\triangle t} = \dfrac {1}{r} \left [\displaystyle \lim_{\triangle t \rightarrow \theta} \dfrac {\triangle s}{\triangle t}\right ]$But $\displaystyle \lim_{\triangle t \rightarrow \theta} \dfrac {\triangle \theta}{\triangle t} = \omega$ and $\displaystyle \lim_{\triangle t \rightarrow \theta} \dfrac {\triangle s}{\triangle t} = V$$\therefore \omega = \dfrac {V}{r}$$V = r\omega$In vector form, $\vec {V} = \vec {\omega}\times \vec {r}$

Question 6

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Accepted Answer

Let $W$ and $W'$ be the respective weights of the metre stick and the coin.The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.Mass of the metre stick=$m'$Mass of each coin=5gWhen the coins are placed $12 cm$ away from the end P, the centre of mass gets shifted by $5 cm$ from point R toward the end P. The centre of mass is located at a distance of $45 cm$ from point P.The net torque will be conserved for rotational equilibrium about point R.$10g(45-12)-m'g(50-45)=0$Thus $m'=66g$

Question 7

Three particles of mass $1\ kg, 2\ kg$ and $3\ kg$ are placed at the corners $A, B$ and $C$ respectively of an equilateral triangle $ABC$ of edge $1\ m$. Find the distance of their centre of mass from $A$.

Accepted Answer

Assume , Point $A$ is origin.From the figure, $CP= 1\times sin(60^o)=\dfrac{\sqrt 3}{2}$Centre of mass from Point $A$ in direction of $x-axis$,   $x_{cm}= \dfrac{m_Ax_A+m_Bx_B+ m_Cx_C}{m_A+m_B+m_C}$$\Rightarrow x_{cm}= \dfrac{1\times 0+ 2\times 1+ 3\times \dfrac{1}{2}}{1+2+3}=\dfrac{7}{12}$Centre of mass from Point $A$ in direction of $y-axis$,   $y_{cm}= \dfrac{m_Ay_A+m_By_B+ m_Cy_C}{m_A+m_B+m_C}$$\Rightarrow y_{cm}= \dfrac{1\times 0+ 2\times 0+ 3\times \dfrac{\sqrt 3}{2}}{1+2+3}=\dfrac{\sqrt 3}{4}$So coordinate of centre of mass $\left ( \dfrac{7}{12} , \dfrac{\sqrt 3}{4} \right )$, Distance from $A$, $= \sqrt{\left (\dfrac{7}{12}\right )^2+ \left (\dfrac{\sqrt 3}{4}\right )^2}=\dfrac{\sqrt {19}}{6}$

Question 8

The radius of gyration of a disc about its axis passing through its centre and perpendicular to its plane is

Accepted Answer

Moment of Inertia of disc about center of mass is $I=\cfrac{mR^{2}}{2}$Let a line parallel to axis at a distance $d$.Radius of Gyration$=R$Then, $\cfrac{mR^{2}}{2}=mR^{2}$$\Rightarrow\,R^{2}=\cfrac{R^{2}}{2}$$\Rightarrow R=\cfrac{R}{\sqrt 2}$

Question 9

Let F be the force acting on a particle having position vector r and $\tau$ be the torque of this force about the origin. Then,

Accepted Answer

torgue is the cross product of the force F and the position vector'r'$\tau = r\times F$so the angle between$\tau$ and F is $90^0$ and between $\tau$ and r is $90^0$ for example if we take dot product between two vectors as $\theta =90^0 \implies cos\theta = cos90^0 = 0$ $\implies r.\tau = 0$$  F.\tau =0$

Question 10

State the law of conservation of momentum and derive related expression.

Accepted Answer

Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.&#10;&#10;Let,&#10;&#10;$m_A$ = Mass of ball A&#10;&#10;$m_B$= Mass of ball B&#10;&#10;$u_A$= initial velocity of ball A&#10;&#10;$u_B$= initial velocity of ball B&#10;&#10;$v_A$= Velocity after the collision of ball A&#10;&#10;$v_B$= Velocity after the collision of ball B&#10;&#10;$F_{ab}$= Force exerted by A on B&#10;&#10;$F_{ba}$= Force exerted by B on A&#10;&#10;Now,Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision&#10;&#10;= $m_A v_A - m_A u_A$&#10;&#10;Rate of change of momentum A= Change in momentum of A/ time taken&#10;&#10;= ${\dfrac{m_A v_A - m_A u_A}{t}}$&#10;&#10;Force exerted by B on A ($F_{ba}$);&#10;&#10;$F_{ba}= {\dfrac{m_A v_A - m_A u_A}{t}}$........ [i]&#10;&#10;In the same way,&#10;&#10;Rate of change of momentum of B=&#10;&#10;${\dfrac{m_b v_B - m_B u_B}{t}}$&#10;&#10;Force exerted by A on B ($F_{ab}$)=&#10;&#10;$F_{ab}= {\dfrac{m_B v_B - m_B u_B}{t}}$.......... [ii]&#10;&#10;Newton's third law of motion states that every action has an equal and opposite reaction, then,&#10;&#10;$F_ab= -F_ba$ [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision] &#10;&#10; &#10;&#10;Using [i] and [ii] , we have&#10;&#10;${\dfrac{m_B v_B - m_B u_B}{t}} = - {\dfrac{m_A v_A - m_A u_A}{t}}$&#10;&#10;$m_B v_B - m_B u_B= - m_A v_A + m_A u_A$&#10;&#10;Finally we get,&#10;&#10;$m_B v_B + m_A v_A = m_B u_B + m_A u_A$&#10;&#10;This is the derivation of conservation of linear momentum.