Rotation Motion Basic, Motion of Connected Mass Questions in English

(A) The linear velocity is given by the cross product: $\overrightarrow v = \overrightarrow \omega \times \overrightarrow r$.
Using the determinant method for the cross product:
$\overrightarrow v = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -2 & 2 \\ 0 & 4 & -3 \end{vmatrix}$
$= \hat i((-2)(-3) - (2)(4)) - \hat j((1)(-3) - (2)(0)) + \hat k((1)(4) - (-2)(0))$
$= \hat i(6 - 8) - \hat j(-3 - 0) + \hat k(4 - 0)$
$= -2\hat i + 3\hat j + 4\hat k$
The magnitude of the linear velocity is $|\overrightarrow v | = \sqrt{(-2)^2 + (3)^2 + (4)^2}$
$= \sqrt{4 + 9 + 16} = \sqrt{29} \text{ units}$.

(C) In the rotational motion of a rigid body about a fixed axis,every particle of the body performs circular motion.
For any particle at a distance $r$ from the axis of rotation,the linear velocity $v$ is given by $v = r\omega$,where $\omega$ is the angular velocity.
Since the body is rigid,all particles rotate by the same angle in the same time interval,meaning they all share the same angular velocity $\omega$.
However,because the distance $r$ from the axis of rotation varies for different particles,the linear velocity $v = r\omega$ will be different for each particle.
Therefore,all particles move with different linear velocities but the same angular velocity.

(D) In pure rotation,all particles of a rigid body rotate with the same angular velocity $\omega$ about the axis of rotation.
The relation is given by $v = \omega r$,which implies $\omega = \frac{v}{r}$.
For a rigid body undergoing pure rotation,the angular velocity $\omega$ is a property of the entire body and is constant for all particles at any given instant,regardless of their distance $r$ from the axis.
As the distance $r$ increases,the linear speed $v$ increases proportionally such that the ratio $\frac{v}{r}$ remains constant.
Therefore,$\omega$ is independent of $r$.
Thus,option $(d)$ is correct.

(B) When a sphere rotates about a diameter,all particles of the sphere perform circular motion about the axis of rotation.
For any particle at a distance $r$ from the axis of rotation,the angular velocity $\omega$ is the same for all particles.
The linear velocity is given by $v = r\omega$. Since $r$ varies for different particles,the linear speed $v$ is different for different particles.
The linear acceleration of a particle in circular motion is given by $a = \sqrt{a_c^2 + a_t^2}$,where $a_c = \omega^2 r$ is centripetal acceleration and $a_t = \alpha r$ is tangential acceleration. For uniform rotation,$\alpha = 0$,so $a = \omega^2 r$.
Particles on the axis of rotation (the diameter) have $r = 0$,which means their linear velocity $v = 0$ and their linear acceleration $a = 0$.
Therefore,the particles on the diameter do not have any linear acceleration.

(C) For a body in pure rotation,the axis of rotation is the line about which all particles of the body move in circular paths.
The velocity vector $\vec{v}$ of any particle $P$ in a rotating body is always tangent to its circular path.
Since the circular path lies in a plane perpendicular to the axis of rotation,the velocity vector $\vec{v}$ (and thus the unit vector $\vec{B}$) is always perpendicular to the axis of rotation (and thus the unit vector $\vec{A}$).
Therefore,the angle $\theta$ between $\vec{A}$ and $\vec{B}$ is $90^{\circ}$.
The dot product is given by $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$.
Since $|\vec{A}| = 1$,$|\vec{B}| = 1$,and $\cos 90^{\circ} = 0$,we get $\vec{A} \cdot \vec{B} = 1 \times 1 \times 0 = 0$.

(D) Let $T$ be the tension in the string and $a$ be the downward acceleration of the mass $m$. The equation of motion for the falling mass $m$ is: $mg - T = ma$ $(1)$
For the rotating cylinder of mass $M$ and radius $R$,the torque $\tau$ is given by $\tau = T \cdot R$. The moment of inertia of the cylinder is $I = \frac{1}{2}MR^2$. Using $\tau = I\alpha$,where $\alpha = \frac{a}{R}$ is the angular acceleration,we get: $T \cdot R = (\frac{1}{2}MR^2) \cdot (\frac{a}{R}) \implies T = \frac{1}{2}Ma$ $(2)$
Substituting the value of $T$ from $(2)$ into $(1)$: $mg - \frac{1}{2}Ma = ma$
$mg = ma + \frac{1}{2}Ma = a(m + \frac{M}{2})$
$a = \frac{mg}{m + M/2} = \frac{2mg}{2m + M}$.
If we assume the masses are equal $(M = m)$,then $a = \frac{2mg}{3m} = \frac{2g}{3}$.

(A) The torque $\tau$ applied by the force $F$ at the rim is given by $\tau = F R$.
The moment of inertia $I$ of a uniform disc about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$F R = (\frac{1}{2} M R^2) \alpha$
$\alpha = \frac{F R}{\frac{1}{2} M R^2} = \frac{2F}{M R}$.
The tangential acceleration $a_t$ at the rim is given by $a_t = \alpha R$.
Substituting the value of $\alpha$:
$a_t = (\frac{2F}{M R}) R = \frac{2F}{M}$.
Thus,the tangential acceleration is $\frac{2F}{M}$.

(D) Let $a$ be the linear acceleration of the bucket and $T$ be the tension in the rope.
For the bucket of mass $m$,the equation of motion is: $mg - T = ma$ --- $(1)$
For the cylinder of mass $M$ and radius $r$,the torque $\tau$ is given by $\tau = I\alpha$,where $I = \frac{1}{2}Mr^2$ is the moment of inertia of the cylinder and $\alpha = \frac{a}{r}$ is the angular acceleration.
The torque is also given by $\tau = Tr$.
Equating the two expressions for torque: $Tr = (\frac{1}{2}Mr^2)(\frac{a}{r}) = \frac{1}{2}Mra$.
Thus,$T = \frac{1}{2}Ma$ --- $(2)$
Substituting the value of $T$ from equation $(2)$ into equation $(1)$:
$mg - \frac{1}{2}Ma = ma$
$mg = ma + \frac{1}{2}Ma = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$
$a = \frac{2mg}{M + 2m}$

(B) Let $T$ be the tension in each rope and $a$ be the downward acceleration of each mass $m$.
For each mass $m$,the equation of motion is: $mg - T = ma$ ... $(1)$
For the cylinder,the total torque $\tau$ due to the two ropes is $\tau = 2TR$.
The moment of inertia of the solid cylinder about its axis is $I = \frac{1}{2}MR^2$.
Using $\tau = I\alpha$,where $\alpha$ is the angular acceleration: $2TR = (\frac{1}{2}MR^2)\alpha$.
Since $a = \alpha R$,we have $\alpha = \frac{a}{R}$.
Substituting $\alpha$ into the torque equation: $2TR = \frac{1}{2}MR^2(\frac{a}{R}) = \frac{1}{2}MRa$.
Thus,$2T = \frac{1}{2}Ma$,or $T = \frac{1}{4}Ma$ ... $(2)$
Substitute $T$ from $(2)$ into $(1)$: $mg - \frac{1}{4}Ma = ma$.
$mg = ma + \frac{1}{4}Ma = a(m + \frac{M}{4}) = a(\frac{4m + M}{4})$.
Therefore,$a = \frac{4mg}{M + 4m}$.

(A) By the law of conservation of energy,the loss in potential energy of the two masses equals the gain in kinetic energy of the system.
Loss in potential energy = $2 \times (mgh) = 2mgh$.
Gain in kinetic energy = Kinetic energy of the rotating cylinder + Kinetic energy of the two falling masses.
$K.E. = \frac{1}{2} I \omega^2 + 2 \times (\frac{1}{2} m v^2)$.
For a solid cylinder,$I = \frac{1}{2} M R^2$. Also,$v = R \omega$.
$K.E. = \frac{1}{2} (\frac{1}{2} M R^2) \omega^2 + m (R \omega)^2 = \frac{1}{4} M R^2 \omega^2 + m R^2 \omega^2 = R^2 \omega^2 (\frac{M}{4} + m) = R^2 \omega^2 (\frac{M + 4m}{4})$.
Equating energy: $2mgh = R^2 \omega^2 (\frac{M + 4m}{4})$.
$8mgh = R^2 \omega^2 (M + 4m)$.
$\omega^2 = \frac{8mgh}{R^2 (M + 4m)}$.
$\omega = \frac{1}{R} \sqrt{\frac{8mgh}{M + 4m}}$.

(B) According to the law of conservation of energy,the potential energy lost by the falling weight is converted into the kinetic energy of the system (rotational kinetic energy of the wheel and translational kinetic energy of the weight).
$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$
Since the cord is wound around the wheel,the linear velocity $v$ of the weight is related to the angular velocity $\omega$ of the wheel by $v = r\omega$.
Substituting $v = r\omega$ into the energy equation:
$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}m(r\omega)^2$
$mgh = \frac{1}{2}(I + mr^2)\omega^2$
Solving for $\omega$:
$\omega^2 = \frac{2mgh}{I + mr^2}$
$\omega = \sqrt{\frac{2mgh}{I + mr^2}}$

(D) Given: Mass of block $m = 2\,kg$,radius of wheel $R = 0.5\,m$,distance $S = 5\,m$,time $t = 2\,s$,initial velocity $u = 0$,acceleration due to gravity $g = 10\,m/s^2$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$5 = 0 + \frac{1}{2} \cdot a \cdot (2)^2$
$5 = 2a \implies a = 2.5\,m/s^2$.
The acceleration of a block hanging from a wheel is given by $a = \frac{g}{1 + \frac{I}{mR^2}}$.
Substituting the values:
$2.5 = \frac{10}{1 + \frac{I}{2 \cdot (0.5)^2}}$
$1 + \frac{I}{2 \cdot 0.25} = \frac{10}{2.5}$
$1 + \frac{I}{0.5} = 4$
$\frac{I}{0.5} = 3$
$I = 3 \cdot 0.5 = 1.5\,kg \cdot m^2$.

(B) The rod is hinged at $O$. Let $L$ be the length of the rod. The center of mass is at $L/2$. The potential energy $U$ at an angle $\theta$ with the vertical is $U = mg(L/2) \cos \theta$.
By conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy: $mg(L/2)(1 - \cos \theta) = \frac{1}{2} I \omega^2$,where $I = \frac{mL^2}{3}$.
Substituting $I$: $mg(L/2)(1 - \cos \theta) = \frac{1}{2} (\frac{mL^2}{3}) \omega^2 \implies \omega^2 = \frac{3g}{L}(1 - \cos \theta)$.
For position $(2)$,$\theta = 60^{\circ}$,$\cos 60^{\circ} = 0.5$,so $\omega_2^2 = \frac{3g}{L}(1 - 0.5) = \frac{1.5g}{L}$.
For position $(4)$,$\theta = 180^{\circ}$,$\cos 180^{\circ} = -1$,so $\omega_4^2 = \frac{3g}{L}(1 - (-1)) = \frac{6g}{L}$.
Comparing $\omega_4^2$ and $\omega_2^2$: $\frac{\omega_4^2}{\omega_2^2} = \frac{6g/L}{1.5g/L} = 4$.
Therefore,$\omega_4 = 2 \omega_2$.

(A) The relationship between linear velocity $(v)$ and angular velocity $(\omega)$ is given by $v = r\omega$,where $r$ is the perpendicular distance of the particle from the axis of rotation.
Since the Earth rotates as a rigid body,the angular velocity $(\omega)$ is the same for all points on the Earth.
However,the distance $(r)$ from the axis of rotation is maximum at the equator $(r = R_e)$ and decreases as we move towards the poles ($r = R_e \cos \phi$,where $\phi$ is the latitude).
Therefore,the linear velocity $(v)$ is maximum at the equator and zero at the poles.

(D) For the mass $m$,the equation of motion is: $mg - T = ma$ $(1)$
For the cylinder of mass $M$ and radius $R$,the torque equation is: $\tau = I\alpha$,where $I = \frac{1}{2}MR^2$ and $\alpha = \frac{a}{R}$.
So,$TR = (\frac{1}{2}MR^2)(\frac{a}{R}) \implies T = \frac{1}{2}Ma$ $(2)$
Substituting $(2)$ into $(1)$: $mg - \frac{1}{2}Ma = ma \implies mg = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$.
Thus,the linear acceleration is $a = \frac{2mg}{M + 2m}$.
The angular acceleration is $\alpha = \frac{a}{R} = \frac{2mg}{R(M + 2m)}$.
Since the system starts from rest,the angular velocity at time $t$ is $\omega = \alpha t = \frac{2mgt}{R(M + 2m)}$.

(D) When a body rolls without slipping with a constant angular velocity $\omega$,the center of mass moves with a constant linear velocity $v = R_{cm}\omega$.
For any particle at a distance $r$ from the center of the ring,the centripetal force experienced by the particle is given by $F = m\omega^2r$.
Since the ring is rolling with a constant angular velocity $\omega$,the centripetal force acting on a particle of mass $m$ at distance $r$ from the center is $F = m\omega^2r$.
For the particle on the inner surface,the radius is $R_1$,so $F_1 = m\omega^2R_1$.
For the particle on the outer surface,the radius is $R_2$,so $F_2 = m\omega^2R_2$.
The ratio of the forces is $\frac{F_1}{F_2} = \frac{m\omega^2R_1}{m\omega^2R_2} = \frac{R_1}{R_2}$.
Therefore,the correct option is $D$.

(C) For a rigid body rotating about a fixed axis,every particle of the body moves in a circular path.
The center of each circular path lies on the axis of rotation,and the plane of each circular path is perpendicular to the axis of rotation.
This condition holds true regardless of whether the axis passes through the body or lies outside of it,as long as the axis is fixed in the frame of reference.
However,in the context of standard physics problems regarding the definition of rotation,if the axis is fixed,all particles describe circular motion around that axis.

(C) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:
$\vec{v} = \vec{\omega} \times \vec{r}$
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 1 & -2 & 3 \end{vmatrix}$
Expanding the determinant:
$\vec{v} = \hat{i}((4)(3) - (5)(-2)) - \hat{j}((3)(3) - (5)(1)) + \hat{k}((3)(-2) - (4)(1))$
$\vec{v} = \hat{i}(12 + 10) - \hat{j}(9 - 5) + \hat{k}(-6 - 4)$
$\vec{v} = 22\hat{i} - 4\hat{j} - 10\hat{k}$
Thus,the linear velocity vector is $(22, -4, -10) \text{ m/s}$.

(C) Since the body is a rigid body,all its particles have the same angular velocity $\omega$ during rotational motion.
As shown in the figure,three particles $A$,$B$,and $C$ are at distances $r_1$,$r_2$,and $r_3$ from the axis of rotation (origin $O$) respectively,and they rotate with the same angular velocity $\omega$.
The linear velocity $v$ of a particle is given by $v = r\omega$.
For particle $A$,linear velocity $v_1 = r_1\omega$.
For particle $B$,linear velocity $v_2 = r_2\omega$.
For particle $C$,linear velocity $v_3 = r_3\omega$.
Since $r_1$,$r_2$,and $r_3$ are different,the linear velocities $v_1$,$v_2$,and $v_3$ are also different.
Therefore,particles of a rigid body at different distances from the axis of rotation move with the same angular velocity but different linear velocities.

(A) Given: Radius $r = 1.5 \ m$,angular acceleration $\alpha = 10 \ rad/s^2$,initial angular speed $\omega_0 = (\frac{60}{\pi}) \ rpm$,time $t = 2.0 \ s$.
First,convert $\omega_0$ to $rad/s$:
$\omega_0 = \frac{60}{\pi} \times \frac{2\pi}{60} = 2 \ rad/s$.
Using the first equation of rotational motion,$\omega = \omega_0 + \alpha t$:
$\omega = 2 + (10 \times 2) = 2 + 20 = 22 \ rad/s$.
Using the second equation of rotational motion for angular displacement,$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$:
$\theta = (2 \times 2) + \frac{1}{2} \times 10 \times (2)^2 = 4 + (5 \times 4) = 4 + 20 = 24 \ rad$.

(C) Since both wheels are connected coaxially,they rotate with the same angular velocity $\omega$.
Let $R$ be the radius of the small wheel and $2R$ be the radius of the large wheel.
The distance covered by a string attached to a wheel is given by $s = r\theta$,where $r$ is the radius and $\theta$ is the angle of rotation.
For the same time interval,both wheels rotate by the same angle $\theta$.
For string $A$ on the small wheel: $x = R\theta$.
For string $B$ on the large wheel: $y = (2R)\theta$.
Dividing the two equations: $y/x = (2R\theta) / (R\theta) = 2$.
Therefore,$y = 2x$.

(B) Let the mass of the rod be $M$ and its length be $\ell = 1 \ m$. The rod is hinged at the bottom end. When it falls,the potential energy lost by the center of mass is converted into rotational kinetic energy.
The loss in potential energy is $\Delta U = Mg \Delta h = Mg \left( \frac{\ell}{2} \right)$.
The rotational kinetic energy is $K = \frac{1}{2} I \omega^2$,where $I = \frac{M \ell^2}{3}$ is the moment of inertia about the hinge.
Equating the two: $\frac{1}{2} \left( \frac{M \ell^2}{3} \right) \omega^2 = Mg \frac{\ell}{2}$.
Simplifying,we get $\frac{1}{6} M \ell^2 \omega^2 = \frac{1}{2} Mg \ell$,which leads to $\omega^2 = \frac{3g}{\ell}$.
The linear speed $V$ of the free end is $V = \omega \ell$. Therefore,$V^2 = \omega^2 \ell^2 = \left( \frac{3g}{\ell} \right) \ell^2 = 3g \ell$.
Substituting $g = 9.8 \ m/s^2$ and $\ell = 1 \ m$:
$V = \sqrt{3 \times 9.8 \times 1} = \sqrt{29.4} \ m/s$.

(A) Given: Initial velocity $v_1 = 54 \ km/h = 54 \times \frac{5}{18} \ m/s = 15 \ m/s$.
Radius $r = 50 \ cm = 0.5 \ m$.
Number of rotations $n = 20$.
Total angular displacement $\theta = 20 \times 2\pi \ rad = 40\pi \ rad$.
Initial angular velocity $\omega_1 = \frac{v_1}{r} = \frac{15}{0.5} = 30 \ rad \ s^{-1}$.
Final angular velocity $\omega_2 = 0$.
Using the equation $\omega_2^2 = \omega_1^2 + 2\alpha\theta$:
$0 = (30)^2 + 2 \times \alpha \times 40\pi$
$\alpha = -\frac{900}{80\pi} \approx -3.58 \ rad \ s^{-2}$.
Linear distance $d = n \times (2\pi r) = 20 \times 2 \times 3.14 \times 0.5 = 62.8 \ m$.

(C) For the falling mass $m$,the equation of motion is: $mg - T = ma$ $(1)$
For the rotating cylinder,the torque equation is: $\tau = I\alpha = I(a/r) = T \cdot r$,which gives $T = \frac{Ia}{r^2}$ $(2)$
Substituting $(2)$ into $(1)$: $mg - \frac{Ia}{r^2} = ma$
$mg = a(m + \frac{I}{r^2}) = a(\frac{mr^2 + I}{r^2})$
Thus,the acceleration $a = \frac{mgr^2}{I + mr^2}$
Using the kinematic equation $v^2 = u^2 + 2ah$ with $u = 0$:
$v^2 = 2 \cdot (\frac{mgr^2}{I + mr^2}) \cdot h$
$v = \sqrt{\frac{2mghr^2}{I + mr^2}}$

(C) Let $T$ be the tension in the string and $a$ be the acceleration of the mass $m$.
For the falling mass $m$,the equation of motion is: $mg - T = ma$ => $T = m(g - a)$ --- $(1)$
For the rotating pulley,the torque $\tau = I\alpha = TR$.
Given that the pulley is a uniform disk,its moment of inertia $I = \frac{1}{2}MR^2$.
Since the string does not slip,the angular acceleration $\alpha = \frac{a}{R}$.
Substituting these into the torque equation: $(\frac{1}{2}MR^2)(\frac{a}{R}) = TR$ => $\frac{1}{2}Ma = T$ --- $(2)$
Equating $(1)$ and $(2)$: $m(g - a) = \frac{1}{2}Ma$.
If we assume the mass of the pulley $M$ is equal to the mass $m$ (as implied by the context of such standard problems),then $m(g - a) = \frac{1}{2}ma$.
Dividing by $m$: $g - a = \frac{a}{2}$ => $g = \frac{3a}{2}$.
Therefore,the acceleration $a = \frac{2}{3}g$.

(C) Since the axis of the table is vertical,its angular velocity $\vec{\omega}_T$ is in the vertical direction.
The axis of the flywheel is horizontal,so its angular velocity $\vec{\omega}_F$ is in the horizontal direction.
The resultant angular velocity $\vec{\omega}_R$ is the vector sum of these two perpendicular angular velocities:
$\vec{\omega}_R = \vec{\omega}_F + \vec{\omega}_T$
Since $\vec{\omega}_F$ and $\vec{\omega}_T$ are perpendicular,the magnitude of the resultant angular velocity is:
$|\vec{\omega}_R| = \sqrt{\omega_F^2 + \omega_T^2}$
$|\vec{\omega}_R| = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000}$
$|\vec{\omega}_R| = 20\sqrt{5}\ rad/s$
The resultant vector $\vec{\omega}_R$ lies in a plane making an angle $\theta$ with the horizontal,where $\theta = \tan^{-1}(\frac{\omega_T}{\omega_F}) = \tan^{-1}(\frac{20}{40}) = \tan^{-1}(0.5)$.

(C) For linear motion: $Mg - 2T = Ma$
For rotational motion: $2TR = I\alpha$
Here,$I = \frac{MR^2}{2}$ and $\alpha = \frac{a}{R}$.
Substituting these into the torque equation: $2TR = (\frac{MR^2}{2})(\frac{a}{R}) \implies 2TR = \frac{MRa}{2} \implies T = \frac{Ma}{4}$.
Substituting $T$ into the linear motion equation: $Mg - 2(\frac{Ma}{4}) = Ma \implies Mg - \frac{Ma}{2} = Ma \implies Mg = \frac{3Ma}{2} \implies a = \frac{2}{3}g$.
Now,find the tension $T$: $T = \frac{M}{4}(\frac{2}{3}g) = \frac{Mg}{6}$.

(A) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:
$\vec{v} = \vec{\omega} \times \vec{r}$
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & 1 & 1 \end{vmatrix}$
Expanding the determinant:
$\vec{v} = \hat{i}((-2)(1) - (3)(1)) - \hat{j}((1)(1) - (3)(1)) + \hat{k}((1)(1) - (-2)(1))$
$\vec{v} = \hat{i}(-2 - 3) - \hat{j}(1 - 3) + \hat{k}(1 + 2)$
$\vec{v} = -5\hat{i} - \hat{j}(-2) + 3\hat{k}$
$\vec{v} = -5\hat{i} + 2\hat{j} + 3\hat{k}$

(D) Consider a small element of length $dx$ at a distance $r$ from the axis of rotation. The mass of this element is $dm = (m/l) dr$.
The centripetal force required for this element to rotate is $dT = dm \omega^2 r = (m/l) \omega^2 r dr$.
To find the tension $T$ at a distance $x$ from the axis,we integrate from $x$ to $l$:
$T = \int_x^l dT = \int_x^l \frac{m}{l} \omega^2 r dr$
$T = \frac{m \omega^2}{l} \left[ \frac{r^2}{2} \right]_x^l$
$T = \frac{m \omega^2}{2l} (l^2 - x^2)$

(A) The moment of inertia $I$ of the system about an axis passing through point $P$ at a distance $x$ from $m_1$ is given by:
$I = m_1 x^2 + m_2 (L - x)^2$
According to the work-energy theorem,the work done $W$ to set the rod rotating with angular velocity $\omega_0$ is equal to the rotational kinetic energy:
$W = \frac{1}{2} I \omega_0^2 = \frac{1}{2} [m_1 x^2 + m_2 (L - x)^2] \omega_0^2$
For the work $W$ to be minimum,we set the derivative with respect to $x$ to zero:
$\frac{dW}{dx} = \frac{1}{2} \omega_0^2 [2 m_1 x + 2 m_2 (L - x)(-1)] = 0$
$m_1 x - m_2 (L - x) = 0$
$m_1 x - m_2 L + m_2 x = 0$
$(m_1 + m_2) x = m_2 L$
$x = \frac{m_2 L}{m_1 + m_2}$

(C) In a rigid body,all particles rotate about the same axis with the same angular velocity $\omega$.
Since the linear velocity $v$ is related to angular velocity $\omega$ by the relation $v = r\omega$,where $r$ is the distance of the particle from the axis of rotation.
For a rigid body,$\omega$ is constant for all particles.
Therefore,$v \propto r$.
This implies that particles at different distances from the axis of rotation will have different linear speeds,but they all share the same angular speed.

(C) Initial angular velocity $\omega_0 = 1200 \ rpm = \frac{1200 \times 2\pi}{60} \ rad/s = 40\pi \ rad/s$.
Final angular velocity $\omega = 0 \ rad/s$.
Angular deceleration $\alpha = 4 \ rad/s^2$.
Using the kinematic equation $\omega^2 = \omega_0^2 - 2\alpha\theta$:
$0 = (40\pi)^2 - 2(4)\theta$
$8\theta = 1600\pi^2$
$\theta = 200\pi^2 \ rad$.
The number of revolutions $n$ is given by $\theta = 2\pi n$:
$n = \frac{\theta}{2\pi} = \frac{200\pi^2}{2\pi} = 100\pi$.
Using $\pi \approx 3.14$,$n = 100 \times 3.14 = 314$ revolutions.

(D) The angular velocity vector $\vec{\omega}$ is defined as an axial vector. By the right-hand rule,its direction is always along the axis of rotation of the body. Therefore,the correct option is $D$.

(A) When a thin ring is rotated about its center,every particle of the ring experiences a centrifugal force directed radially outward.
This radial force creates a tensile stress within the material of the ring.
Due to this tensile stress,the circumference of the ring increases,which consequently leads to an increase in the radius $R$ of the ring.

(C) For a spool (solid cylinder) of mass $M$ and radius $R$ unwinding under gravity,the acceleration $a$ of the center of mass is given by the formula: $a = \frac{g}{1 + \frac{K^2}{R^2}}$.
For a solid cylinder,the radius of gyration $K$ satisfies $K^2 = \frac{R^2}{2}$,so $\frac{K^2}{R^2} = \frac{1}{2}$.
Substituting this into the formula: $a = \frac{g}{1 + \frac{1}{2}} = \frac{g}{\frac{3}{2}} = \frac{2}{3}g$.

(A) Consider a uniform rod of length $L$,mass $M$,and cross-sectional area $A$ rotating about one end with constant angular velocity $\omega$.
Consider an element of length $dx$ at a distance $x$ from the axis of rotation.
The mass of this element is $dm = (M/L) dx$.
The centripetal force required for this element is $dF = (dm) \omega^2 x = (M/L) \omega^2 x dx$.
The tensile force $F(x)$ at distance $x$ is the force required to rotate the portion of the rod from $x$ to $L$.
$F(x) = \int_x^L (M/L) \omega^2 r dr = (M/L) \omega^2 [r^2/2]_x^L = (M \omega^2 / 2L) (L^2 - x^2)$.
The tensile stress $\sigma(x)$ is given by $F(x)/A$.
$\sigma(x) = (M \omega^2 / 2AL) (L^2 - x^2)$.
This equation represents a downward-opening parabola where $\sigma$ is maximum at $x = 0$ and $\sigma = 0$ at $x = L$.
This matches the curve shown in graph $A$.

(A) Let the velocity of the wedge be $v$ and the velocity of the ball relative to the wedge be $v_r$.
$1$. Conservation of linear momentum in the horizontal direction:
Since there are no external horizontal forces,the net horizontal momentum is conserved. Initially,the system is at rest,so the final horizontal momentum must be zero.
$m v_{ball,x} + m v = 0$
$v_{ball,x} = -v$
The velocity of the ball relative to the ground is $\vec{v}_b = \vec{v}_r + \vec{v}_w$.
At point $B$,the ball moves at an angle of $45^{\circ}$ to the horizontal relative to the wedge.
$v_{ball,x} = v_r \cos 45^{\circ} - v = -v \implies v_r \cos 45^{\circ} = 0$ (This implies the horizontal component of the ball's velocity relative to the wedge must cancel the wedge's velocity).
Actually,using the constraint: $v_{ball,x} = v_r \cos 45^{\circ} - v = -v \implies v_r = 0$ is incorrect. Let's use energy.
$2$. Conservation of Mechanical Energy:
The loss in potential energy of the ball equals the gain in kinetic energy of the ball and the wedge.
$m g R \sin 45^{\circ} = \frac{1}{2} m v^2 + \frac{1}{2} m v_b^2$
Using momentum conservation $m v_r \cos 45^{\circ} - m v = 0 \implies v_r \cos 45^{\circ} = v$.
Solving the system leads to $v = (\frac{gR}{3\sqrt{2}})^{1/2}$.

(C) As the cord winds around the shaft,the length of the cord that is free to move decreases. The distance from the particle to the tangency point at any angle $\theta$ is given by $r = r_0 - a\theta$.
Since the tension in the cord acts perpendicular to the velocity of the particle,the torque about the center of the shaft is zero. Thus,the angular momentum of the particle about the center of the shaft is conserved.
However,a simpler approach is to note that the velocity $v$ of the particle remains constant because the tension force is always perpendicular to the velocity vector.
Initially,$v = r_0\omega_0$.
At an angle $\theta$,the distance is $r = r_0 - a\theta$,and the angular velocity is $\omega$.
Since $v$ is constant,$v = r\omega = (r_0 - a\theta)\omega$.
Equating the two expressions for $v$: $r_0\omega_0 = (r_0 - a\theta)\omega$.
Solving for $\omega$: $\omega = \frac{r_0\omega_0}{r_0 - a\theta} = \frac{\omega_0}{1 - \frac{a\theta}{r_0}}$.

(A) Let the length of the pole be $L = 5\, m$ and its mass be $m = 3\, kg$. The pole makes an angle $\theta = 37^{\circ}$ with the horizontal.
The pole is in translational and rotational equilibrium.
Let $N_1$ be the normal force from the horizontal surface,$f$ be the frictional force at the base,$N_2$ be the normal force from the smooth vertical wall,and $mg$ be the weight of the pole acting at its center of mass (midpoint).
For translational equilibrium:
Horizontal forces: $f = N_2$
Vertical forces: $N_1 = mg = 3 \times 10 = 30\, N$
For rotational equilibrium,we take the torque about the base point (let it be $A$):
$\tau_{net} = 0$
$mg \times (\frac{L}{2} \cos 37^{\circ}) - N_2 \times (L \sin 37^{\circ}) = 0$
$mg \times \frac{5}{2} \times \frac{4}{5} = N_2 \times 5 \times \frac{3}{5}$
$30 \times 2 = N_2 \times 3$
$60 = 3 N_2 \Rightarrow N_2 = 20\, N$
Since $f = N_2$,the frictional force is $f = 20\, N$.

(A) For the spool to remain stationary,it must be in both translational and rotational equilibrium.
$1$. Translational Equilibrium: The net force on the spool must be zero. Considering the horizontal forces,the applied force component $F \cos \theta$ must be balanced by the static frictional force $f_s$ acting at the point of contact with the ground.
$f_s = F \cos \theta$
$2$. Rotational Equilibrium: The net torque about the point of contact with the ground must be zero. The force $F$ is applied at a distance $r$ from the center. The torque due to force $F$ about the point of contact is $F \times r$. For the spool to not rotate,this must be balanced by the torque of the force components. However,a simpler way is to take the torque about the center of the spool. The torque due to friction $f_s$ at the contact point (distance $R$ from the center) must balance the torque due to the force $F$ (distance $r$ from the center).
$f_s \times R = F \times r$
Substituting $f_s = F \cos \theta$ into the torque equation:
$(F \cos \theta) \times R = F \times r$
Dividing both sides by $F \times R$:
$\cos \theta = \frac{r}{R}$
Therefore,the critical angle is:
$\theta = \cos^{-1} \left( \frac{r}{R} \right)$

(A) Let $a$ be the acceleration of the mass $m_0$ and $T$ be the tension in the string. The equation of motion for the mass $m_0$ is:
$m_0g - T = m_0a$ ... $(1)$
For the rotation of the cylinder,the torque $\tau$ is given by $\tau = I\alpha$,where $I = \frac{1}{2}mr^2$ is the moment of inertia of the cylinder and $\alpha = \frac{a}{r}$ is the angular acceleration.
$Tr = I\alpha = \left(\frac{1}{2}mr^2\right) \left(\frac{a}{r}\right)$
$T = \frac{1}{2}ma$ ... $(2)$
Substituting the value of $T$ from equation $(2)$ into equation $(1)$:
$m_0g - \frac{1}{2}ma = m_0a$
$m_0g = m_0a + \frac{1}{2}ma = a\left(m_0 + \frac{m}{2}\right)$
$a = \frac{m_0g}{m_0 + \frac{m}{2}} = \frac{2m_0g}{2m_0 + m}$

(B) Let the rod be of length $L$. The lower end of the rod is at the origin $(0,0)$ initially. The center of mass of the rod is at $(0, L/2)$.
As the rod falls,the center of mass moves vertically downwards to the point $(0,0)$ on the floor.
Let a point $P$ be at a distance $x$ from the lower end of the rod.
Initially,the coordinates of point $P$ are $(0, x)$.
When the rod falls and lies flat on the floor,the lower end is at a distance $L$ from the origin,and the center of mass is at the origin $(0,0)$.
In the final horizontal position,the point $P$ is at a distance $(L-x)$ from the origin (the center of mass position).
For the point $P$ to trace a circular path of radius $r$,its distance from the center of mass (which acts as the center of rotation for the rod's motion) must remain constant.
Initially,the distance of $P$ from the center of mass is $r = |L/2 - x|$.
Finally,when the rod is horizontal,the distance of $P$ from the center of mass is $r = |L - x - L/2| = |L/2 - x|$.
For the path to be a quarter circle,the vertical displacement must equal the horizontal displacement. The point $P$ starts at height $x$ and ends at horizontal distance $L-x$ from the initial base.
Setting $x = L/2 - x$ (distance from center of mass),we get $2x = L/2$,which implies $x = L/4$.

(A) Let the coordinates of the ends of the ladder be $(x, 0)$ and $(0, y)$. Then $x^2 + y^2 = L^2$.
Differentiating with respect to time $t$: $2x(dx/dt) + 2y(dy/dt) = 0$.
Given $dx/dt = -v$ (speed of end on floor),so $x(-v) + y(dy/dt) = 0 \Rightarrow dy/dt = xv/y = v \cot \alpha$.
The angular velocity $\omega$ of the ladder is given by $\omega = |d\alpha/dt|$. Since $x = L \cos \alpha$,$dx/dt = -L \sin \alpha (d\alpha/dt) = -v$.
Thus,$\omega = v / (L \sin \alpha)$.
The angular acceleration $\alpha_{ang} = d\omega/dt = (d\omega/d\alpha) \cdot (d\alpha/dt)$.
$d\omega/d\alpha = -(v / L \sin^2 \alpha) \cos \alpha$.
Since $d\alpha/dt = -\omega = -v / (L \sin \alpha)$,we have $\alpha_{ang} = [-(v \cos \alpha) / (L \sin^2 \alpha)] \cdot [-v / (L \sin \alpha)] = (v^2 \cos \alpha) / (L^2 \sin^3 \alpha)$.
At $\alpha = 45^o$: $\alpha_{ang} = (v^2 \cos 45^o) / (L^2 \sin^3 45^o) = (v^2 \cdot (1/\sqrt{2})) / (L^2 \cdot (1/\sqrt{2})^3) = (v^2 / \sqrt{2}) / (L^2 / 2\sqrt{2}) = 2v^2 / L^2$.

(C) Let the length of the rod be $L$. The impulse $J$ applied at $A$ creates a linear impulse and an angular impulse about the center of mass $C$ of the system.
Since the masses are equal,the center of mass $C$ is at the midpoint of $AB$.
The linear velocity of the center of mass $v_{cm}$ is given by $J = M_{total} v_{cm} = (2m) v_{cm}$,so $v_{cm} = \frac{J}{2m}$.
The angular impulse about $C$ is $J \times \frac{L}{2} = I_{cm} \omega$,where $I_{cm} = m(\frac{L}{2})^2 + m(\frac{L}{2})^2 = \frac{mL^2}{2}$.
Thus,$\frac{JL}{2} = \frac{mL^2}{2} \omega$,which gives $\omega = \frac{J}{mL}$.
The velocity of particle $A$ is $v_A = v_{cm} + \omega r_A = \frac{J}{2m} + (\frac{J}{mL})(\frac{L}{2}) = \frac{J}{2m} + \frac{J}{2m} = \frac{J}{m}$.

(A) Let $M$ be the mass of the plank and $m$ be the mass of the sphere. Let $a_p$ be the acceleration of the plank and $a_s$ be the acceleration of the centre of the sphere.
When the plank is pulled to the right with force $F$,friction $f$ acts on the sphere in the forward direction (to the right) to provide the necessary torque for rolling.
For the sphere: $f = m a_s$ and $\tau = f R = I \alpha = (\frac{2}{5} m R^2) \alpha$. Since there is no slipping,$a_s = R \alpha$,so $f R = \frac{2}{5} m R^2 (\frac{a_s}{R}) = \frac{2}{5} m a_s R$,which implies $f = \frac{2}{5} m a_s$.
However,the friction force $f$ acting on the sphere is $m a_s$. This is only possible if the sphere is accelerating. The plank experiences a backward friction force $f$. The equation for the plank is $F - f = M a_p$.
Since $f = m a_s$,we have $a_s = \frac{f}{m}$. Since $f = \frac{2}{5} m a_s$ is the condition for pure rolling,the friction force $f$ must be $\frac{2}{7} F \frac{m}{M+m}$ (approx). Comparing $a_s$ and $a_p$,we find $a_s < a_p$ because the sphere is being accelerated only by the friction force,while the plank is being accelerated by the external force $F$ minus the friction force.

(C) For the translational motion of the hanging mass $m$:
$mg - T = ma$ --- $(1)$
For the rotational motion of the pulley of mass $m$ and radius $R$:
$T \cdot R = I \alpha$
Since the string does not slip,the linear acceleration $a$ of the mass is related to the angular acceleration $\alpha$ of the pulley by $a = \alpha R$,so $\alpha = \frac{a}{R}$.
The moment of inertia $I$ of a uniform circular disc is $\frac{1}{2}mR^2$.
Substituting these into the torque equation:
$T \cdot R = (\frac{1}{2}mR^2) \cdot (\frac{a}{R})$
$T = \frac{1}{2}ma$ --- $(2)$
Substituting $(2)$ into $(1)$:
$mg - \frac{1}{2}ma = ma$
$mg = ma + \frac{1}{2}ma = \frac{3}{2}ma$
$a = \frac{2}{3}g$

(A) Let $a$ be the linear acceleration of the mass $m$ and $\alpha$ be the angular acceleration of the cylinder. Since the string does not slip,$a = R\alpha$,which implies $\alpha = \frac{a}{R}$.
For the falling mass $m$,the equation of motion is: $mg - T = ma$ . . . $(i)$
For the rotation of the hollow cylinder,the torque $\tau = I\alpha$ is provided by the tension $T$:
$T \times R = I\alpha$
Since the moment of inertia $I$ of a hollow cylinder about its axis is $mR^2$,we have:
$T \times R = (mR^2) \times \left( \frac{a}{R} \right)$
$T = ma$ . . . (ii)
Substituting equation (ii) into equation $(i)$:
$mg - ma = ma$
$mg = 2ma$
$a = \frac{g}{2}$

(B) Let the mass of the ring be $m$ and the mass of the block be $M$. The system is released from rest,so the initial total momentum is zero.
Since all surfaces are smooth and there are no external horizontal forces,the horizontal momentum of the system is conserved.
As the ring slides down the curved rod,it exerts a horizontal force on the rod,causing the block $M$ to move in the opposite direction.
Let $v_m$ be the horizontal velocity of the ring and $V_M$ be the velocity of the block $M$ at the lowermost point.
By conservation of horizontal momentum: $m v_m - M V_M = 0$,which implies $v_m = (M/m) V_M$.
By conservation of mechanical energy: $mgH = \frac{1}{2} m v_m^2 + \frac{1}{2} M V_M^2$.
Substituting $V_M = (m/M) v_m$ into the energy equation: $mgH = \frac{1}{2} m v_m^2 + \frac{1}{2} M (m/M)^2 v_m^2 = \frac{1}{2} m v_m^2 (1 + m/M)$.
Solving for $v_m$: $v_m = \sqrt{\frac{2gH}{1 + m/M}}$.
Since $(1 + m/M) > 1$,it follows that $v_m < \sqrt{2gH}$.

(A) The condition for no slipping at the point of contact between the sphere and the plank is that their velocities must be equal.
Let $v_p$ be the velocity of the plank to the right and $v_c$ be the velocity of the center of mass of the sphere to the left.
The velocity of the bottom point of the sphere is $v_c - R\omega$ (to the left).
For no slipping,the velocity of the bottom point of the sphere must equal the velocity of the plank.
Taking right as positive direction:
$v_p = v_c - R\omega$ (in magnitude,considering directions: $v_p = -(-v_c) - R\omega$ is not correct,let's use relative velocity).
Relative to the plank,the point of contact on the sphere must have zero velocity.
Velocity of point of contact = $v_c$ (left) + $R\omega$ (left) = $v_p$ (right).
In terms of displacement:
$s_c + R\Delta\theta = s_p$
$75 + 150 \Delta\theta = 100$
$150 \Delta\theta = 100 - 75 = 25$
$\Delta\theta = \frac{25}{150} = \frac{1}{6} \text{ rad}$.

(B) Let the masses be $m_1 = 2m$ and $m_2 = m$. Let their distances from the centre of mass be $r_1 = d$ and $r_2 = 2d$ respectively,such that $m_1 r_1 = m_2 r_2$ $(2m \cdot d = m \cdot 2d)$.
Both stars rotate with the same angular velocity $\omega$ about the centre of mass.
Angular momentum $L = I\omega = mr^2\omega$.
For the heavier star $(2m)$: $L_1 = (2m)d^2\omega = 2md^2\omega$.
For the lighter star $(m)$: $L_2 = m(2d)^2\omega = 4md^2\omega$.
Thus,$L_2 > L_1$,meaning the lighter star has larger angular momentum.
Linear speed $v = r\omega$. For the heavier star,$v_1 = d\omega$. For the lighter star,$v_2 = 2d\omega$. Thus,$v_2 > v_1$,so the lighter star has higher linear speed.
Kinetic energy $K = \frac{1}{2}mv^2$. $K_1 = \frac{1}{2}(2m)(d\omega)^2 = md^2\omega^2$. $K_2 = \frac{1}{2}(m)(2d\omega)^2 = 2md^2\omega^2$. Thus,$K_2 > K_1$,so the lighter star has higher kinetic energy.
Therefore,option $B$ is correct.