Rotational Motion MCQs with Answers – Part 3 (NEET Physics)

Explanation: Linear momentum of the particle-plus-disc system is not the convenient conserved quantity because the axle can exert an external impulse. However, the impulsive force of the axle has zero moment about the rotation axis. Hence axial angular momentum is conserved. \(\textbf{Angular momentum before impact:}\) The particle moves tangentially at radius \(R\), so \[ L_i=mvR \] The disc is initially stationary. \(\textbf{Rotational inertia after sticking:}\) \[ I_f=2mR^2+mR^2=3mR^2 \] Angular-momentum conservation gives \[ mvR=(3mR^2)\omega_f \] Cancelling \(mR\), \[ v=3R\omega_f \] Therefore, \[ \omega_f=\frac{v}{3R} \] The sticking collision is inelastic, but angular momentum about the frictionless axis is still conserved. \(\textbf{Final answer:}\) \[ \boxed{\omega_f=\frac{v}{3R}} \]

Explanation: The particle's initial angular momentum is determined by the perpendicular distance of its line of motion from the axis. It is not necessary for the particle's velocity to be tangential to the rim before impact. \(\textbf{Initial angular momentum:}\) \[ L_i=mvd_\perp \] Here, \[ d_\perp=\frac R2 \] Therefore, \[ L_i=\frac12mvR \] \(\textbf{Moment of inertia after sticking:}\) The particle finally sticks at radius \(R\), so \[ I_f=2mR^2+mR^2=3mR^2 \] With negligible external torque about the axis, \[ L_i=I_f\omega_f \] Thus, \[ \frac12mvR=3mR^2\omega_f \] Hence, \[ \omega_f=\frac{v}{6R} \] The impact parameter controls the incoming angular momentum, while the final radius controls the particle's contribution to the rotational inertia after sticking. \(\textbf{Final answer:}\) \[ \boxed{\omega_f=\frac{v}{6R}} \]

Explanation: Internal forces can exchange angular momentum between different parts of a system. Therefore the angular momentum of an individual constituent may change even when the total remains constant. Statement I is correct. The fundamental condition for conservation about a specified origin is \[ \vec\tau_{\rm ext}=0. \] Then, \[ \frac{d\vec L_{\rm total}}{dt}=0, \] so Statement II is correct. Statement III is false because conservation applies to the sum, not necessarily to each part separately. Statement IV is also false. A system can have zero resultant force but still experience a non-zero external couple. Such a couple produces non-zero torque and changes angular momentum. Therefore only Statements I and II are correct.

Explanation: Friction between the discs is internal to the two-disc system. With negligible external torque, \[ L_i=L_f \] \(\textbf{Initial angular momentum:}\) \[ L_i=I(4\omega)+3I(-2\omega) \] Thus, \[ L_i=4I\omega-6I\omega=-2I\omega \] After contact, the combined moment of inertia is \[ I_f=I+3I=4I \] Therefore, \[ -2I\omega=(4I)\Omega \] Hence, \[ \Omega=-\frac{\omega}{2} \] \(\textbf{Initial rotational kinetic energy:}\) \[ K_i=\frac12I(4\omega)^2+\frac12(3I)(2\omega)^2 \] Thus, \[ K_i=8I\omega^2+6I\omega^2=14I\omega^2 \] \(\textbf{Final rotational kinetic energy:}\) \[ K_f=\frac12(4I)\left(\frac{\omega}{2}\right)^2 \] Therefore, \[ K_f=\frac12I\omega^2 \] Hence, \[ \frac{K_f}{K_i} =\frac{\frac12}{14} =\frac1{28} \] Angular momentum is conserved, but substantial mechanical energy is dissipated while the discs acquire a common angular speed. \(\textbf{Final answer:}\) \[ \boxed{\Omega=-\frac{\omega}{2},\qquad \frac{K_f}{K_i}=\frac1{28}} \]

Explanation: Angular momentum about the chosen axis is conserved when the net external torque about that axis is zero. In P, the person and held masses redistribute internally while the turntable bearing is frictionless. Thus no appreciable external axial torque acts, so angular momentum is conserved. In Q, the external tangential force has a non-zero moment arm and therefore produces a non-zero axial torque. Angular momentum changes. In R, the force acts radially through the axis. Its moment arm about that axis is zero, so it produces no axial torque. Angular momentum about the axis can therefore remain conserved. In S, a couple applied about the axis is itself a non-zero external torque. Angular momentum cannot remain constant while that couple acts. Hence the appropriate situations are P and R.

Explanation: Angular momentum must be evaluated for the complete system consisting of the person, turntable and package during the throw. \(\textbf{Before release:}\) The package contributes \[ mR^2 \] to the moment of inertia. Thus, \[ I_i=2mR^2+mR^2=3mR^2 \] The initial angular momentum is \[ L_i=3mR^2\omega \] \(\textbf{Angular momentum carried by the thrown package:}\) Its final tangential velocity is opposite to the original rotation and has magnitude \(R\omega\). Therefore, \[ L_{\rm pkg}=-mR(R\omega) =-mR^2\omega \] Let the turntable-person angular speed after release be \(\Omega\). Their angular momentum is \[ L_{\rm tp}=2mR^2\Omega \] Conservation gives \[ 3mR^2\omega =2mR^2\Omega-mR^2\omega \] Therefore, \[ 4\omega=2\Omega \] Hence, \[ \Omega=2\omega \] \(\textbf{Final answer:}\) \[ \boxed{\Omega=2\omega} \]

Explanation: Initially the person is on the axis and therefore contributes no \(mr^2\) term. \(\textbf{Initial angular momentum:}\) \[ I_i=3mR^2 \] Thus, \[ L_i=3mR^2\omega_0 \] \(\textbf{After the person reaches the rim:}\) The total moment of inertia becomes \[ I_f=3mR^2+mR^2=4mR^2 \] With zero external torque, \[ 3mR^2\omega_0=4mR^2\omega_f \] Therefore, \[ \omega_f=\frac34\omega_0 \] At the rim, the person's speed relative to the ground is \[ v=R\omega_f \] Hence, \[ v=\frac34R\omega_0 \] The person's outward motion increases the system's rotational inertia, so the angular speed decreases. \(\textbf{Final answer:}\) \[ \boxed{\omega_f=\frac34\omega_0,\qquad v=\frac34R\omega_0} \]

Explanation: Because the system is isolated with respect to axial torque, \[ I_i\omega_0=I_f\left(\frac32\omega_0\right) \] \(\textbf{Initial moment of inertia:}\) \[ I_i=2ma^2+m(a)^2+m(3a)^2 \] Thus, \[ I_i=(2+1+9)ma^2=12ma^2 \] Therefore, \[ 12ma^2=\frac32I_f \] Hence, \[ I_f=8ma^2 \] \(\textbf{Changed configuration:}\) \[ I_f=2ma^2+m(2a)^2+mx^2 \] Therefore, \[ 8ma^2=2ma^2+4ma^2+mx^2 \] Thus, \[ mx^2=2ma^2 \] Hence, \[ x^2=2a^2, \] so \[ x=\sqrt2\,a \] The angular-speed information therefore constrains the final mass distribution through angular-momentum conservation. \(\textbf{Final answer:}\) \[ \boxed{x=\sqrt2\,a} \]

Explanation: Zero net external torque implies conservation of angular momentum: \[ L=I\omega=\text{constant}. \] Therefore, \[ \omega=\frac{L}{I}. \] For the same non-zero conserved \(L\), \[ \omega\propto\frac1I. \] Thus increasing the moment of inertia decreases the angular speed, while decreasing the moment of inertia increases it. The relation is not linear. A plot of \(\omega\) against \(I\) therefore has the form of a decreasing rectangular-hyperbola type curve over positive values of \(I\). Zero torque does not by itself require constant \(\omega\) if the moment of inertia is being changed internally. The directly conserved quantity is angular momentum. Hence option C gives the correct graph description.

Explanation: Friction between the two rotors transfers angular momentum internally from one rotor to the other. It does not provide an external torque about the common frictionless axis. Therefore, \[ L_i=L_f \] \(\textbf{Initial total angular momentum:}\) For rotor A, \[ L_A=(2I)(3\omega)=6I\omega \] For rotor B, \[ L_B=I(-\omega)=-I\omega \] Hence, \[ L_i=5I\omega \] \(\textbf{After the rotors lock together:}\) Their combined moment of inertia is \[ I_f=2I+I=3I \] If the common angular velocity is \(\Omega\), \[ L_f=3I\Omega \] Thus, \[ 3I\Omega=5I\omega \] Therefore, \[ \Omega=\frac53\omega \] The positive sign means that the final rotation follows the original sense of rotor A because its initial positive angular momentum dominates. Internal friction may dissipate mechanical energy, but it does not destroy the total angular momentum of the isolated rotational system. \(\textbf{Final answer:}\) \[ \Omega=\frac53\omega \]

Explanation: Because an external torque acts, angular momentum is not conserved during the redistribution. \(\textbf{Initial angular momentum:}\) \[ L_i=I_i\omega_i \] Thus, \[ L_i=(4)(5)=20\,\text{kg m}^2\text{s}^{-1} \] \(\textbf{Angular impulse of the external torque:}\) \[ \Delta L=\tau\Delta t \] Therefore, \[ \Delta L=(-3)(2)=-6\,\text{kg m}^2\text{s}^{-1} \] Hence, \[ L_f=20-6=14\,\text{kg m}^2\text{s}^{-1} \] The final moment of inertia is \(2\,\text{kg m}^2\), so \[ L_f=I_f\omega_f \] Thus, \[ 14=2\omega_f \] Therefore, \[ \omega_f=+7\,\text{rad s}^{-1} \] Although the external torque reduces the angular momentum, the simultaneous decrease in moment of inertia is large enough for the angular speed to increase. \(\textbf{Final answer:}\) \[ \boxed{\omega_f=+7\,\text{rad s}^{-1}} \]

Explanation: The required torque must be obtained from the change in angular momentum, not merely from the change in angular velocity because the moment of inertia also changes. \(\textbf{Initial state:}\) \[ L_i=I_i\omega_i=(3)(4)=12\,\text{kg m}^2\text{s}^{-1} \] \(\textbf{Final state:}\) \[ L_f=I_f\omega_f=(5)(1)=5\,\text{kg m}^2\text{s}^{-1} \] Therefore, \[ \Delta L=L_f-L_i=5-12=-7\,\text{kg m}^2\text{s}^{-1} \] The average external torque is \[ \tau_{\rm av}=\frac{\Delta L}{\Delta t} \] Substituting, \[ \tau_{\rm av}=\frac{-7}{2} \] Hence, \[ \tau_{\rm av}=-3.5\,\text{N m} \] The negative sign indicates that the net angular impulse acts opposite to the chosen positive axial direction. \(\textbf{Final answer:}\) \[ \boxed{\tau_{\rm av}=-3.5\,\text{N m}} \]

Explanation: The axle can exert an impulse on the system, but its torque about the rotation axis is negligible. Therefore axial angular momentum is conserved during the collision. \(\textbf{Angular momentum before impact:}\) \[ L_i=mvR \] After rebounding, the particle carries angular momentum in the opposite sense: \[ L_{\rm p,f}=-m\left(\frac v4\right)R \] Let the disc angular speed be \(\Omega\). Conservation gives \[ mvR=(2mR^2)\Omega-\frac14mvR \] Hence, \[ \frac54mvR=2mR^2\Omega \] Therefore, \[ \Omega=\frac{5v}{8R} \] \(\textbf{Initial kinetic energy:}\) \[ K_i=\frac12mv^2 \] \(\textbf{Final kinetic energy of the disc:}\) \[ K_{\rm d} =\frac12(2mR^2)\left(\frac{5v}{8R}\right)^2 =\frac{25}{64}mv^2 \] The rebounding particle has \[ K_{\rm p} =\frac12m\left(\frac v4\right)^2 =\frac1{32}mv^2 \] Thus, \[ K_f=\frac{27}{64}mv^2 \] Therefore, \[ \frac{K_f}{K_i} =\frac{27/64}{1/2} =\frac{27}{32} \] Angular momentum is conserved, whereas kinetic energy decreases during the collision. \(\textbf{Final answer:}\) \[ \boxed{\Omega=\frac{5v}{8R},\qquad \frac{K_f}{K_i}=\frac{27}{32}} \]

Explanation: The initial angular momentum of the disc is \[ L_{\rm d}=2mR^2\omega_0 \] \(\textbf{Incoming particle angular momentum:}\) For a straight-line trajectory, \[ L_{\rm p}=mvd_\perp \] Its magnitude is \[ |L_{\rm p}| =m(2R\omega_0)\left(\frac R2\right) =mR^2\omega_0 \] Because it is opposite to the disc angular momentum, \[ L_{\rm p}=-mR^2\omega_0 \] Hence the total initial angular momentum is \[ L_i=mR^2\omega_0 \] After sticking at the rim, the total moment of inertia is \[ I_f=2mR^2+mR^2=3mR^2 \] Conservation of axial angular momentum gives \[ mR^2\omega_0=(3mR^2)\omega_f \] Therefore, \[ \omega_f=\frac{\omega_0}{3} \] The final sign remains positive because the disc's initial positive angular momentum exceeds the incoming negative angular momentum. \(\textbf{Final answer:}\) \[ \boxed{\omega_f=+\frac{\omega_0}{3}} \]

Explanation: Before release, the total angular momentum is \[ L_i=(I+mR^2)\omega \] A simple release without a push produces no impulsive tangential interaction that would suddenly change the platform's angular velocity. The small mass therefore leaves with its instantaneous tangential speed \[ v=R\omega \] Its angular momentum about the axis immediately after release is \[ L_m=mRv=mR^2\omega \] If the platform continues with angular speed \(\omega\), its angular momentum is \[ L_{\rm p}=I\omega \] The total after separation is then \[ L_f=I\omega+mR^2\omega \] Thus, \[ L_f=(I+mR^2)\omega=L_i \] Angular momentum is conserved without requiring any sudden change in the platform's angular speed. An active throw would be different because an internal impulse could redistribute angular momentum between the mass and platform.

Explanation: Let the final angular speed of the person-turntable be \(\Omega\). The package is thrown backward with relative tangential speed \(3R\omega\). Therefore its ground-frame tangential velocity is \[ v_{\rm pkg}=R\Omega-3R\omega \] \(\textbf{Initial angular momentum:}\) Initially the package also rotates with the turntable, so \[ L_i=(2mR^2+mR^2)\omega =3mR^2\omega \] \(\textbf{Final angular momentum:}\) The person-turntable contributes \[ 2mR^2\Omega \] The package contributes \[ mR\,v_{\rm pkg} =mR^2(\Omega-3\omega) \] Conservation gives \[ 3mR^2\omega =2mR^2\Omega+mR^2(\Omega-3\omega) \] Cancelling \(mR^2\), \[ 3\omega=3\Omega-3\omega \] Thus, \[ \Omega=2\omega \] The package ground velocity is then \[ v_{\rm pkg} =R(2\omega)-3R\omega =-R\omega \] \(\textbf{Final answer:}\) \[ \boxed{\Omega=2\omega,\qquad v_{\rm pkg}=-R\omega} \]

Explanation: The force is always parallel or antiparallel to the particle's position vector from \(O\). Therefore, \[ \vec\tau_O=\vec r\times\vec F=\vec0. \] Using \[ \frac{d\vec L_O}{dt}=\vec\tau_O, \] we obtain \[ \frac{d\vec L_O}{dt}=\vec0. \] Thus the angular momentum about \(O\) remains constant. However, the force itself need not be zero. Hence, \[ \frac{d\vec p}{dt}=\vec F \] can be non-zero, so the linear momentum may change in both magnitude and direction. The zero-torque conclusion is specifically about the chosen point \(O\). A different reference point need not have zero torque. Therefore angular-momentum conservation here does not imply conservation of linear momentum.

Explanation: Statement I is the conservation condition: \[ \tau_{\rm ext}=0 \quad\Rightarrow\quad L=\text{constant}. \] Statement II is also correct. Internal friction can transfer angular momentum between interacting parts, as when two coaxial rotors come into contact. The individual angular momenta change, while their total can remain constant if external torque is negligible. Statement III is false. Processes such as sticking collisions or frictional coupling can conserve angular momentum while dissipating mechanical energy. Statement IV is correct because torque depends on moment arm, not merely on whether an external force exists. If the force line passes through the axis, \[ d_\perp=0 \] and its torque about that axis is zero. Therefore Statements I, II and IV are correct.

Explanation: The complete process has two distinct stages, but angular momentum about the axis is conserved throughout. \(\textbf{Angular momentum before impact:}\) The disc contributes \[ L_{\rm d}=2mR^2\omega_0 \] The incoming particle contributes \[ L_{\rm p}=m(2R\omega_0)R =2mR^2\omega_0 \] Hence, \[ L_{\rm total}=4mR^2\omega_0 \] \(\textbf{Immediately after sticking:}\) The moment of inertia is \[ I_1=2mR^2+mR^2=3mR^2 \] Thus, \[ \omega_1=\frac{4}{3}\omega_0 \] \(\textbf{After moving the particle inward:}\) The final moment of inertia becomes \[ I_f=2mR^2+m\left(\frac R2\right)^2 \] Therefore, \[ I_f=\frac94mR^2 \] Since the same total angular momentum is still conserved, \[ 4mR^2\omega_0 =\frac94mR^2\omega_f \] Hence, \[ \omega_f=\frac{16}{9}\omega_0 \] \(\textbf{Final answer:}\) \[ \boxed{\omega_f=\frac{16}{9}\omega_0} \]

Explanation: With zero external torque, \[ L=I\omega \] must remain constant. For state P, \[ L_P=(2)(3)=6\,\text{kg m}^2\text{s}^{-1}. \] For Q, \[ L_Q=(3)(2)=6, \] so Q is compatible with angular-momentum conservation. For R, \[ L_R=(1.5)(4)=6, \] so R is also compatible. For S, \[ L_S=(4)(1)=4. \] This differs from the initial value. Therefore state S would require a net external angular impulse or some exchange of angular momentum with the surroundings. Internal rearrangement alone cannot change the total angular momentum of the isolated system. Hence S is the only incompatible listed state.