Conservation of angular momentum (combined translation and rotational motion) Questions in English

(C) Before impact,the bullet is moving with velocity $v$. The initial angular momentum of the system about the hinge point $O$ is $J = m v L$.
After the bullet gets embedded in the rod,suppose the system attains angular velocity $\omega$. The moment of inertia of the bullet-rod system about the axis through $O$ is:
$I = I_{\text{bullet}} + I_{\text{rod}} = m L^2 + \frac{1}{3} M L^2 = \left( \frac{M + 3m}{3} \right) L^2$.
By the principle of conservation of angular momentum,the initial angular momentum equals the final angular momentum:
$J = J' \implies m v L = I \omega$
$m v L = \left( \frac{M + 3m}{3} \right) L^2 \omega$
Solving for $\omega$:
$\omega = \frac{3 m v}{(M + 3m) L}$.

(A) The initial moment of inertia of the ring about the transverse axis is $I = M R^2$. The initial angular momentum is $L_i = I \omega = M R^2 \omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia becomes $I' = M R^2 + m R^2 + m R^2 = (M + 2m) R^2$.
Since no external torque acts on the system,the angular momentum is conserved,so $L_i = L_f$.
$M R^2 \omega = (M + 2m) R^2 \omega'$
$\omega' = \frac{M R^2 \omega}{(M + 2m) R^2} = \frac{M \omega}{M + 2m}$.

(A) Initial linear velocity $= v$
Initial radius $= r$
Angular momentum $L = mvr$
When the string is halved,the new radius $r' = \frac{r}{2}$.
Since the angular momentum $L$ remains constant:
$mvr = mv'r'$
$mvr = mv' \left(\frac{r}{2}\right)$
$v = \frac{v'}{2}$
$v' = 2v$
Therefore,the new linear velocity is $2v$.

(A) Since there is no external torque about the pivot $C$,the angular momentum of the system is conserved.
Initial angular momentum $L_i = (2m)(v)(L) + (m)(2v)(2L) = 2mvL + 4mvL = 6mvL$.
Final moment of inertia $I_f = I_{\text{rod}} + I_{2m} + I_{m} = \frac{(8m)(6L)^2}{12} + (2m)(L)^2 + (m)(2L)^2$.
$I_f = \frac{8m \cdot 36L^2}{12} + 2mL^2 + 4mL^2 = 24mL^2 + 2mL^2 + 4mL^2 = 30mL^2$.
Using $L_i = I_f \omega$:
$6mvL = (30mL^2) \omega$.
$\omega = \frac{6mvL}{30mL^2} = \frac{v}{5L}$.

(C) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. Since the mass remains constant, $V \propto R^3$.
If the new volume $V_2 = \frac{1}{8} V_1$, then $\frac{V_2}{V_1} = \frac{R_2^3}{R_1^3} = \frac{1}{8}$.
Taking the cube root, we get $\frac{R_2}{R_1} = \frac{1}{2}$, so $R_2 = \frac{1}{2} R_1$.
By the law of conservation of angular momentum, $L = I \omega = \text{constant}$.
Since $I = \frac{2}{5} M R^2$ and $\omega = \frac{2 \pi}{T}$, we have $I_1 \omega_1 = I_2 \omega_2$.
Substituting the values, $\frac{2}{5} M R_1^2 \cdot \frac{2 \pi}{T_1} = \frac{2}{5} M R_2^2 \cdot \frac{2 \pi}{T_2}$.
This simplifies to $\frac{R_1^2}{T_1} = \frac{R_2^2}{T_2}$, which gives $T_2 = T_1 \left( \frac{R_2}{R_1} \right)^2$.
Given $T_1 = 24$ hours, $T_2 = 24 \cdot (1/2)^2 = 24 \cdot (1/4) = 6$ hours.

(D) The earth is an isolated system with respect to external torques, so its angular momentum $L = I\omega$ remains constant.
When ice at the poles melts and water flows to the equator, the mass distribution of the earth changes such that more mass is concentrated further from the axis of rotation.
This increases the moment of inertia $I$ of the earth $(I = \sum mr^2)$.
Since $L = I\omega$ is constant, an increase in $I$ must result in a decrease in angular velocity $\omega$.
As $\omega = 2\pi / T$, where $T$ is the duration of the day, a decrease in $\omega$ leads to an increase in the duration of the day $T$.

(C) The spinning of the 'ground chakkar' is a rotational motion where no external torque acts on the system to change its state of rotation.
According to Newton's second law for rotation,the torque $\tau$ is equal to the rate of change of angular momentum $L$,given by $\tau = \frac{dL}{dt}$.
Since there is no external torque acting on the 'ground chakkar' $(\tau = 0)$,the rate of change of angular momentum is zero,which implies $\frac{dL}{dt} = 0$.
Therefore,the angular momentum $L$ remains constant.
This phenomenon is based on the law of conservation of angular momentum.

(A) Since there is no external torque acting on the system,the angular momentum of the system is conserved.
$L_i = L_f$
$I_i \omega_i = I_f \omega_f$
Initially,the boy is at the center,so his moment of inertia is zero. The moment of inertia of the disc is $I_{\text{disc}} = \frac{1}{2} M R^2$.
$I_i = \frac{1}{2} \times 100 \times 2^2 = 200 \ kg \cdot m^2$.
Finally,the boy is at a distance $r = 1 \ m$ from the center. His moment of inertia is $I_{\text{boy}} = m r^2 = 60 \times 1^2 = 60 \ kg \cdot m^2$.
The final moment of inertia of the system is $I_f = I_{\text{disc}} + I_{\text{boy}} = 200 + 60 = 260 \ kg \cdot m^2$.
Using conservation of angular momentum:
$200 \times 1 = 260 \times \omega_f$
$\omega_f = \frac{200}{260} = \frac{20}{26} \approx 0.77 \ rad/s$.

(A) Given: Mass of the first disc $m_1 = M$,radius $r_1 = R$,and initial angular velocity $\omega_1 = \omega$.
The moment of inertia of the first disc is $I_1 = \frac{1}{2} M R^2$.
The second disc has mass $m_2 = \frac{1}{8} M$ and radius $r_2 = R$.
The moment of inertia of the second disc is $I_2 = \frac{1}{2} (\frac{1}{8} M) R^2 = \frac{1}{16} M R^2$.
When the second disc is placed on the first,the total moment of inertia of the system becomes $I_{total} = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{16} M R^2 = \frac{8+1}{16} M R^2 = \frac{9}{16} M R^2$.
By the principle of conservation of angular momentum,$L_{initial} = L_{final}$.
$I_1 \omega = I_{total} \omega_{final}$
$\frac{1}{2} M R^2 \omega = \frac{9}{16} M R^2 \omega_{final}$
$\frac{1}{2} \omega = \frac{9}{16} \omega_{final}$
$\omega_{final} = \frac{16}{18} \omega = \frac{8}{9} \omega$.

(D) Applying the principle of conservation of angular momentum about the center of mass (midpoint $O$) of the rod:
Initial angular momentum $L_i = m(2v)(a) + (2m)(v)(2a) = 2mav + 4mav = 6mav$.
Final angular momentum $L_f = I_{total} \omega$,where $I_{total}$ is the moment of inertia of the system about the center of mass after the particles stick to the rod.
$I_{total} = I_{rod} + I_{m} + I_{2m} = \frac{(6m)(8a)^2}{12} + m(a)^2 + (2m)(2a)^2$.
$I_{total} = \frac{6m(64a^2)}{12} + ma^2 + 8ma^2 = 32ma^2 + ma^2 + 8ma^2 = 41ma^2$.
Equating $L_i = L_f$:
$6mav = (41ma^2) \omega$.
Therefore,$\omega = \frac{6mav}{41ma^2} = \frac{6v}{41a}$.

(C) The angular momentum of the Earth remains conserved because no external torque acts on it.
$L = I \omega$
Since $I = \frac{2}{5} MR^2$ and $\omega = \frac{2 \pi}{T}$,we have:
$I_1 \omega_1 = I_2 \omega_2$
$\frac{2}{5} MR^2 \times \frac{2 \pi}{T_1} = \frac{2}{5} M(xR)^2 \times \frac{2 \pi}{T_2}$
Canceling common terms:
$\frac{R^2}{T_1} = \frac{x^2 R^2}{T_2}$
$T_2 = T_1 x^2$
Given the present period of rotation $T_1 = 24 \text{ hours}$,the new period is:
$T_2 = 24 x^2 \text{ hours}$.

(A) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the final angular momentum.
Initial angular momentum $L_i = I \omega_0$,where $\omega_0$ is the initial angular velocity.
When the mouse of mass $m$ lands on the edge at radius $R$,the new moment of inertia of the system becomes $I' = I + m R^2$.
Let the new angular velocity be $\omega$. Then,$L_f = (I + m R^2) \omega$.
Equating $L_i = L_f$,we get $I \omega_0 = (I + m R^2) \omega$.
Thus,$\omega = \frac{I \omega_0}{I + m R^2}$.
The fractional loss of angular velocity is given by $\frac{\omega_0 - \omega}{\omega_0} = 1 - \frac{\omega}{\omega_0}$.
Substituting the value of $\omega$,we get $1 - \frac{I}{I + m R^2} = \frac{I + m R^2 - I}{I + m R^2} = \frac{m R^2}{I + m R^2}$.

(C) According to the law of conservation of angular momentum,since no external torque acts on the Earth,the angular momentum remains constant: $L_1 = L_2$.
Since $L = I\omega$,we have $I_1 \omega_1 = I_2 \omega_2$.
The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.
Let $R_1 = R$ and $R_2 = \frac{R}{n}$.
Substituting these into the equation: $\frac{2}{5}MR^2 \left(\frac{2\pi}{T_1}\right) = \frac{2}{5}M\left(\frac{R}{n}\right)^2 \left(\frac{2\pi}{T_2}\right)$.
Simplifying the equation: $R^2 \left(\frac{1}{T_1}\right) = \frac{R^2}{n^2} \left(\frac{1}{T_2}\right)$.
Therefore,$T_2 = \frac{T_1}{n^2}$.
Given the initial duration of the day $T_1 = 24 \text{ hr}$,the new duration is $T_2 = \frac{24}{n^2} \text{ hr}$.