Mix Example - System of Particles and Rotational Motion Questions in English

(B) Initial state: The first disc has moment of inertia $I$ and angular velocity $\omega$. The second disc is at rest $(I_2 = I, \omega_2 = 0)$.
By the principle of conservation of angular momentum: $L_i = L_f$.
$I \omega + I(0) = (I + I) \omega'$,where $\omega'$ is the common angular velocity.
$I \omega = 2I \omega' \implies \omega' = \frac{\omega}{2}$.
Initial rotational kinetic energy: $K_i = \frac{1}{2} I \omega^2$.
Final rotational kinetic energy: $K_f = \frac{1}{2} (2I) (\omega')^2 = I (\frac{\omega}{2})^2 = \frac{I \omega^2}{4}$.
Loss in rotational kinetic energy: $\Delta K = K_i - K_f = \frac{1}{2} I \omega^2 - \frac{1}{4} I \omega^2 = \frac{1}{4} I \omega^2$.

(B) Given: Radius $r = 0.5 \ m$,Moment of inertia $I = 10 \ kg \cdot m^2$,Initial angular speed $\omega_0 = 70 \ rev/min = 70 \times \frac{2\pi}{60} \ rad/s = \frac{7\pi}{3} \ rad/s$,Time $t = 5.0 \ s$,Normal force $N = 88 \ N$.
Final angular speed $\omega = 0 \ rad/s$.
Angular acceleration $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 7\pi/3}{5} = -\frac{7\pi}{15} \ rad/s^2$.
Magnitude of torque $\tau = I|\alpha| = 10 \times \frac{7\pi}{15} = \frac{14\pi}{3} \ N \cdot m$.
Also,torque $\tau = f_k \cdot r = \mu_k N \cdot r$.
Equating the two: $\mu_k N r = \frac{14\pi}{3} \implies \mu_k (88)(0.5) = \frac{14\pi}{3}$.
$\mu_k (44) = \frac{14 \times 3.14159}{3} \approx 14.66$.
$\mu_k = \frac{14.66}{44} \approx 0.333$.
Thus,the coefficient of kinetic friction is approximately $0.33$.

(A) The rotational kinetic energy $(KE)$ of a solid sphere is given by $KE = \frac{1}{2} I \omega^2$.
For a solid sphere,the moment of inertia $I = \frac{2}{5} m r^2$ and angular velocity $\omega = 2 \pi n$.
Substituting these values,$KE = \frac{1}{2} \times (\frac{2}{5} m r^2) \times (2 \pi n)^2 = \frac{1}{5} m r^2 \times 4 \pi^2 n^2 = \frac{4}{5} m r^2 \pi^2 n^2$.
Given that $50 \%$ of this energy is converted into heat,the heat produced $\Delta Q = \frac{1}{2} KE = \frac{1}{2} \times (\frac{4}{5} m r^2 \pi^2 n^2) = \frac{2}{5} m r^2 \pi^2 n^2$.
Using the relation $\Delta Q = m S \Delta t$,where $\Delta t$ is the rise in temperature,we get $\Delta t = \frac{\Delta Q}{m S}$.
Substituting $\Delta Q$,$\Delta t = \frac{2/5 m r^2 \pi^2 n^2}{m S} = \frac{2 \pi^2 n^2 r^2}{5 S}$.

(B, C) Let the rod make an angle $\theta$ with the $OX$ axis. The velocity of end $A$ is $v_A = 3 \ m/s$ along $OX$ and the velocity of end $B$ is $v_B = 4 \ m/s$ along $OY$ (downwards).
Using the condition that the length of the rod is constant,the component of velocity along the rod must be the same for both ends:
$v_A \cos \theta = v_B \sin \theta \implies 3 \cos \theta = 4 \sin \theta \implies \tan \theta = \frac{3}{4}$.
Thus,$\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The angular velocity $\omega$ is given by $\omega = \frac{v_A \sin \theta + v_B \cos \theta}{\ell} = \frac{3(3/5) + 4(4/5)}{1} = \frac{9/5 + 16/5}{1} = \frac{25}{5} = 5 \ rad/s$.
Since the rod is rotating such that $A$ moves right and $B$ moves down,the rotation is clockwise.
The rotational kinetic energy is $(KE)_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{12} m \ell^2) \omega^2 = \frac{1}{2} \times \frac{1}{12} \times 4 \times (1)^2 \times (5)^2 = \frac{1}{24} \times 4 \times 25 = \frac{25}{6} \ J$.
Both statements $B$ and $C$ are correct.

(A) For a simple pendulum of length $\ell$,the angular velocity $\omega_1$ at an angle $\theta$ below the horizontal is derived from energy conservation: $mg(\ell \sin \theta) = \frac{1}{2} m v^2 = \frac{1}{2} m (\omega_1 \ell)^2$. Thus,$\omega_1 = \sqrt{\frac{2g \sin \theta}{\ell}}$.
For a uniform bar of length $L$ pivoted at one end,the moment of inertia is $I = \frac{1}{3} m L^2$. The potential energy lost when it rotates by angle $\theta$ is $mg(\frac{L}{2} \sin \theta)$. By energy conservation: $mg(\frac{L}{2} \sin \theta) = \frac{1}{2} I \omega_2^2 = \frac{1}{2} (\frac{1}{3} m L^2) \omega_2^2$. Simplifying gives $\omega_2 = \sqrt{\frac{3g \sin \theta}{L}}$.
Since the motions are synchronous,$\omega_1 = \omega_2$,so $\frac{2}{\ell} = \frac{3}{L}$.
Therefore,$L = \frac{3}{2} \ell$.

(A) The radius of the circular path of the mass is $R = L/2$. The linear velocity of the mass at the moment of detachment is $v = \omega R = \frac{\omega L}{2}$.
Since the mass is moving vertically upwards,the time of flight $T$ is given by $T = \frac{2v}{g} = \frac{\omega L}{g}$.
For the mass to reach the bar at the same point,the bar must complete an integer number of rotations $n$ in time $T$. The time for $n$ rotations is $T = n \cdot \frac{2\pi}{\omega}$.
Equating the two expressions for $T$: $\frac{\omega L}{g} = n \frac{2\pi}{\omega} \implies n = \frac{\omega^{2} L}{2\pi g}$. Thus,$n$ must be an integer.
For option $C$,the distance travelled by the mass in air is $2h = 2 \cdot \frac{v^2}{2g} = \frac{v^2}{g} = \frac{(\omega L/2)^2}{g} = \frac{\omega^2 L^2}{4g}$,which is proportional to $\omega^2$.

(A, C) Let the rod have length $L$ and mass $m$. The center of mass is at $L/2$ from $A$.
When the rod makes an angle $\theta$ with the horizontal,the center of mass has descended by $h = (L/2) \sin \theta$.
By conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mg(L/2) \sin \theta = \frac{1}{2} I \omega^2$,where $I = mL^2/3$.
$mg(L/2) \sin \theta = \frac{1}{2} (mL^2/3) \omega^2 \Rightarrow \omega^2 \propto \sin \theta \Rightarrow \omega \propto \sqrt{\sin \theta}$.
Since the speed of end $B$ is $v = \omega L$,we have $v \propto \sqrt{\sin \theta}$. Thus,option $A$ is correct.
For angular acceleration $\alpha$,the torque about $A$ is $\tau = mg(L/2) \cos \theta$.
Using $\tau = I \alpha$,we get $mg(L/2) \cos \theta = (mL^2/3) \alpha$.
This implies $\alpha \propto \cos \theta$. Thus,option $C$ is correct.
Note: Both $A$ and $C$ are correct statements.

(A) Let the center of mass of the rod be the origin. The moment of inertia of the rod about its center is $I_{rod} = \frac{(20m)(12)^2}{12} = 240m \text{ cm}^2$.
Using the conservation of angular momentum about the center of mass of the rod:
$L_i = L_f$
Initial angular momentum $L_i = (2m)(v)(2) + (m)(v)(4) = 4mv + 4mv = 8mv$.
The final moment of inertia of the system is $I_f = I_{rod} + I_{mass1} + I_{mass2} = 240m + (2m)(2)^2 + (m)(4)^2 = 240m + 8m + 16m = 264m \text{ cm}^2$.
Since $L_f = I_f \omega$,we have $8mv = 264m \omega$.
Therefore,$\frac{v}{\omega} = \frac{264}{8} = 33$.

(B) Density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.
Mass of smaller part $m_1 = \frac{M}{8}$. Since $\rho = \frac{m_1}{V_1} = \frac{m_1}{\frac{4}{3}\pi r^3}$,we have $\frac{M}{\frac{4}{3}\pi R^3} = \frac{M/8}{\frac{4}{3}\pi r^3} \Rightarrow r^3 = \frac{R^3}{8} \Rightarrow r = \frac{R}{2}$.
$I_1$ (moment of inertia of sphere about its centre) $= \frac{2}{5}m_1r^2 = \frac{2}{5}(\frac{M}{8})(\frac{R}{2})^2 = \frac{2}{5} \times \frac{M}{8} \times \frac{R^2}{4} = \frac{M R^2}{80}$.
Mass of larger part $m_2 = M - \frac{M}{8} = \frac{7M}{8}$.
$I_2$ (moment of inertia of disc about its diameter) $= \frac{m_2 R_D^2}{4}$,where $R_D = 2R$.
$I_2 = \frac{(\frac{7M}{8})(2R)^2}{4} = \frac{(\frac{7M}{8})(4R^2)}{4} = \frac{7M R^2}{8}$.
Ratio $\frac{I_2}{I_1} = \frac{7M R^2 / 8}{M R^2 / 80} = \frac{7}{8} \times 80 = 70$.

(A) For a body of mass $M$ and radius $R$ with moment of inertia $I = kMR^2$,a tangential force $F$ applied at the highest point (at distance $2R$ from the contact point) creates a torque $\tau = F(2R) - f(R) = I\alpha$,where $f$ is the friction force and $\alpha = a/R$ is the angular acceleration.
Thus,$2FR - fR = (kMR^2)(a/R) \Rightarrow 2F - f = kMa$.
The linear force equation is $F + f = Ma$.
Adding these two equations: $(2F - f) + (F + f) = kMa + Ma \Rightarrow 3F = (k+1)Ma \Rightarrow a = \frac{3F}{(k+1)M}$.
For a solid sphere $(A)$,$k = 2/5$ and $M = 5m$: $a_A = \frac{3F}{(2/5 + 1)5m} = \frac{3F}{(7/5)5m} = \frac{3F}{7m}$.
For a spherical shell $(B)$,$k = 2/3$ and $M = m$: $a_B = \frac{3F}{(2/3 + 1)m} = \frac{3F}{(5/3)m} = \frac{9F}{5m}$.
The ratio $a_A/a_B = (3F/7m) / (9F/5m) = (3/7) \times (5/9) = 15/63 = 5/21$.

(D) Given position vector $\vec{r} = 10t^2\hat{i} + 5t^3\hat{j}$.
Velocity $\vec{v} = \frac{d\vec{r}}{dt} = 20t\hat{i} + 15t^2\hat{j}$.
At $t = 1 \text{ s}$,$\vec{v} = 20\hat{i} + 15\hat{j} \text{ m/s}$.
Linear momentum $\vec{p} = m\vec{v} = 0.1(20\hat{i} + 15\hat{j}) = (2\hat{i} + 1.5\hat{j}) \text{ kg} \cdot \text{m/s}$. Thus,statement $A$ is correct.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 20\hat{i} + 30t\hat{j}$.
At $t = 1 \text{ s}$,$\vec{a} = 20\hat{i} + 30\hat{j} \text{ m/s}^2$.
Force $\vec{F} = m\vec{a} = 0.1(20\hat{i} + 30\hat{j}) = (2\hat{i} + 3\hat{j}) \text{ N}$. Thus,statement $B$ is correct.
At $t = 1 \text{ s}$,$\vec{r} = 10(1)^2\hat{i} + 5(1)^3\hat{j} = 10\hat{i} + 5\hat{j}$.
Angular momentum $\vec{L} = \vec{r} \times \vec{p} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 1.5\hat{j}) = (15\hat{k} - 10\hat{k}) = 5\hat{k} \text{ J} \cdot \text{s}$. Thus,statement $C$ is incorrect.
Torque $\vec{\tau} = \vec{r} \times \vec{F} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 3\hat{j}) = (30\hat{k} - 10\hat{k}) = 20\hat{k} \text{ N} \cdot \text{m}$. Thus,statement $D$ is correct.
Therefore,statements $A, B$ and $D$ are correct.