Apply Rigid Body Mechanics - IB Physics

1. A spinning cloud of interstellar gas collapses under its own gravity to form a more compact protostar. Assuming the cloud is an isolated system, which quantities are conserved during the collapse?
   Answer: Angular momentum and total energy Since the system is isolated, there are no external torques, so angular momentum (L) is conserved. Also, since there are no external non-conservative forces doing work, the total mechanical energy (gravitational potential energy + kinetic energy) is conserved. Gravitational potential energy is converted into rotational and translational kinetic energy (and thermal energy), so rotational kinetic energy alone is not conserved (it increases). Angular velocity (ω) is not conserved; as the cloud's radius decreases, its moment of inertia (I) decreases, and since L=Iω is constant, ω must increase.
2. Two point masses, $m_1 = 3.0 \text{ kg}$ and $m_2 = 1.0 \text{ kg}$, are fixed to the ends of a massless rod of length 2.0 m. What is the moment of inertia of the system about an axis of rotation perpendicular to the rod and located 0.50 m from mass $m_2$?
   Answer: 7.0 kg m² The moment of inertia of a system of point masses is $I = \sum m_i r_i^2$. The distance of $m_2$ from the axis is $r_2 = 0.50 \text{ m}$. The distance of $m_1$ from the axis is $r_1 = 2.0 \text{ m} - 0.50 \text{ m} = 1.5 \text{ m}$. Therefore, $I = m_1 r_1^2 + m_2 r_2^2 = (3.0)(1.5)^2 + (1.0)(0.50)^2 = (3.0)(2.25) + (1.0)(0.25) = 6.75 + 0.25 = 7.0 \text{ kg m}^2$.
3. A child of mass 40 kg stands at the edge of a merry-go-round of radius 2.0 m and moment of inertia 160 kg m², which is rotating at 1.0 rad s⁻¹. The child then walks to the center of the merry-go-round. What is the final angular velocity of the system? Treat the child as a point mass.
   Answer: 2.0 rad s⁻¹ Angular momentum is conserved. Initially, the total moment of inertia is $I_i = I_{mgr} + I_{child} = 160 + mr^2 = 160 + (40)(2.0)^2 = 160 + 160 = 320 \text{ kg m}^2$. The initial angular momentum is $L_i = I_i \omega_i = (320)(1.0) = 320 \text{ kg m}^2 \text{s}^{-1}$. When the child is at the center, their distance from the axis is r=0, so their moment of inertia is zero. The final moment of inertia is just that of the merry-go-round, $I_f = 160 \text{ kg m}^2$. By conservation of angular momentum, $L_f = L_i$, so $I_f \omega_f = 320$. The final angular velocity is $\omega_f = \frac{320}{160} = 2.0 \text{ rad s}^{-1}$.
4. A uniform rod of mass $M$ and length $L$ lies on a frictionless horizontal surface. A bullet of mass $m$ moving with velocity $v$ strikes the rod at a distance $\frac{L}{3}$ from one end, perpendicular to the rod's length, and embeds in it. What is the velocity of the center of mass of the rod immediately after the collision?
   Answer: $\frac{mv}{M + m}$ When you see a collision problem involving embedding (a perfectly inelastic collision), the key principle is conservation of momentum. However, you need to carefully distinguish between linear momentum of the entire system and the motion of specific parts. Since the bullet embeds in the rod, this is a perfectly inelastic collision where linear momentum is conserved for the entire system. Before collision, only the bullet has momentum: $mv$. After collision, both the bullet and rod move together as a single system with total mass $(M + m)$. Using conservation of linear momentum: $mv = (M + m)v_{cm}$, where $v_{cm}$ is the velocity of the center of mass of the entire system. Solving: $v_{cm} = \frac{mv}{M + m}$. This is answer C. The wrong answers represent common misconceptions about this type of problem. Answer A ($\frac{3mv}{2(M + m)}$) incorrectly attempts to account for the collision location by multiplying by a factor, but the center of mass velocity doesn't depend on where the collision occurs. Answer B ($\frac{mv}{M + m} \cdot \frac{L}{3}$) and Answer D ($\frac{mv}{M + m} \cdot \frac{2L}{3}$) make the dimensional error of multiplying a velocity by a length, which gives units of $\frac{length^2}{time}$, not velocity. These also incorrectly assume the collision location affects the center of mass velocity. Remember: In collision problems, the velocity of the center of mass depends only on total momentum and total mass, never on where the collision occurs on the object. The collision location affects rotation, not translation.
5. A thin rod of length $L$ and mass $M$ is pivoted at a point $\frac{L}{4}$ from one end. The rod is released from rest at an angle of $60°$ from the vertical. When the rod passes through the vertical position, what is the angular acceleration at the instant it reaches the vertical?
   Answer: $\frac{9g}{14L}$ The moment of inertia about the pivot is $I = \frac{ML^2}{12} + M\left(\frac{L}{4} - \frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{7ML^2}{48}$. Using energy conservation from $60°$ to vertical: $Mg \cdot \frac{L}{4}(1-\cos 60°) = \frac{1}{2}I\omega^2$, giving $\omega^2 = \frac{18g}{7L}$. At the vertical position, the net torque is $\tau = Mg \cdot \frac{L}{4} = \frac{MgL}{4}$ (since the center of mass is still $\frac{L}{4}$ below the pivot). Therefore, $\alpha = \frac{\tau}{I} = \frac{MgL/4}{7ML^2/48} = \frac{12g}{7L} \cdot \frac{1}{4} = \frac{9g}{14L}$.
6. A uniform disk of mass $M$ and radius $R$ is mounted on a frictionless axle through its center. A massless string is wound around the disk's rim, and a block of mass $m$ hangs from the string. The system is released from rest. What is the tension in the string?
   Answer: $\frac{Mmg}{M + 2m}$ Let $T$ be the tension and $a$ be the acceleration of the block. For the block: $mg - T = ma$. For the disk: $TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa$. So $T = \frac{Ma}{2}$. Substituting into the first equation: $mg - \frac{Ma}{2} = ma$, which gives $a = \frac{2mg}{M + 2m}$. Therefore, $T = \frac{Ma}{2} = \frac{M}{2} \cdot \frac{2mg}{M + 2m} = \frac{Mmg}{M + 2m}$.
7. A solid sphere starts from rest at the top of an incline of vertical height $h$ and rolls without slipping to the bottom. What is its translational speed at the bottom? The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$.
   Answer: $\sqrt{\frac{10}{7}gh}$ By conservation of energy, the initial potential energy $mgh$ is converted into translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For rolling without slipping, $\omega = v/r$. Substituting this and $I = \frac{2}{5}mr^2$: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. Solving for $v$, we get $v^2 = \frac{10}{7}gh$, so $v = \sqrt{\frac{10}{7}gh}$.
8. A disk is rotating with angular velocity $\omega$. A constant frictional torque $\tau$ is applied, bringing the disk to rest. Which expression represents the total angular displacement $\Delta\theta$ of the disk before it stops? Let $I$ be the moment of inertia of the disk.
   Answer: $\frac{I\omega^2}{2\tau}$ We can use the work-energy theorem for rotation: $W = \Delta KE_{rot}$. The work done by the frictional torque is $W = -\tau \Delta\theta$ (negative because it opposes the motion). The change in kinetic energy is $\Delta KE_{rot} = KE_f - KE_i = 0 - \frac{1}{2}I\omega^2$. Setting them equal: $-\tau \Delta\theta = -\frac{1}{2}I\omega^2$. Solving for $\Delta\theta$ gives $\Delta\theta = \frac{I\omega^2}{2\tau}$.
9. A block of mass $m$ hangs from a light string wrapped around a solid cylindrical pulley of mass $M$ and radius $R$. The block is released from rest and accelerates downwards. What is the tension $T$ in the string? The moment of inertia of the pulley is $I = \frac{1}{2}MR^2$.
   Answer: $mg \left( \frac{M}{2m+M} \right)$ For the hanging mass: $mg - T = ma$. For the pulley: $\tau = TR = I\alpha$. With $a = \alpha R$ and $I = \frac{1}{2}MR^2$, we get $TR = (\frac{1}{2}MR^2)(\frac{a}{R})$, which simplifies to $T = \frac{1}{2}Ma$. Substituting this expression for T into the first equation: $mg - \frac{1}{2}Ma = ma$. Solving for acceleration $a$: $mg = a(m + \frac{M}{2}) \implies a = \frac{mg}{m + M/2}$. Now substitute this $a$ back into the equation for tension: $T = \frac{1}{2}Ma = \frac{1}{2}M \left( \frac{mg}{m+M/2} \right) = mg \frac{M/2}{m+M/2} = mg \left( \frac{M}{2m+M} \right)$.
10. A constant torque of 5.0 N m is applied to a grinding wheel with a moment of inertia of 0.20 kg m², initially at rest. What is the rotational kinetic energy of the wheel after it has completed 10 revolutions?
    Answer: 314 J The work-energy theorem for rotation states that the work done by the net torque equals the change in rotational kinetic energy, $W = \Delta KE_{rot}$. The work done is $W = \tau \theta$. First, the angular displacement $\theta$ must be in radians: $\theta = 10 \text{ rev} \times 2\pi \text{ rad/rev} = 20\pi \text{ rad}$. Then, $W = (5.0 \text{ N m})(20\pi \text{ rad}) = 100\pi \text{ J}$. Since the wheel starts from rest, its change in kinetic energy is equal to its final kinetic energy. $KE_{final} = 100\pi \approx 314 \text{ J}$.
11. A solid sphere and a hollow sphere have the same mass $M$ and radius $R$. Both are initially at rest. The same constant net torque is applied to each, causing them to rotate through the same total angular displacement. Which statement correctly compares their final states?
    Answer: The solid sphere has a greater final angular velocity. The work done by the torque is $W = \tau \theta$. Since both $\tau$ and $\theta$ are the same for both spheres, the work done on each is identical. By the work-energy theorem, they have the same final rotational kinetic energy ($KE_{rot}$). The formula for rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2$. The moment of inertia of a solid sphere ($I_s = \frac{2}{5}MR^2$) is less than that of a hollow sphere ($I_h = \frac{2}{3}MR^2$). Since $KE_{rot}$ is the same, the sphere with the smaller moment of inertia (the solid sphere) must have the greater angular velocity $\omega$.
12. A uniform rod of length 2.0 m is pivoted at its center. A force of 10 N is applied 0.50 m from the pivot at an angle of 30° to the rod, tending to cause a clockwise rotation. Another force of 5.0 N is applied at one end of the rod, perpendicular to it, tending to cause a counter-clockwise rotation. What is the magnitude of the net torque about the pivot?
    Answer: 2.5 N m Torque is calculated as $\tau = Fr \sin \theta$. The counter-clockwise torque from the 5.0 N force is $\tau_{ccw} = (5.0 \text{ N})(1.0 \text{ m})\sin(90^\circ) = 5.0 \text{ N m}$. The clockwise torque from the 10 N force is $\tau_{cw} = (10 \text{ N})(0.50 \text{ m})\sin(30^\circ) = (10)(0.50)(0.5) = 2.5 \text{ N m}$. The net torque is the difference between the two, as they act in opposite directions: $\tau_{net} = \tau_{ccw} - \tau_{cw} = 5.0 - 2.5 = 2.5 \text{ N m}$.
13. A uniform solid sphere of mass $m$ and radius $r$ is released from rest at the top of a hemispherical bowl of radius $R$ (where $R >> r$). The sphere rolls without slipping inside the bowl. At what angle $\theta$ from the vertical will the sphere lose contact with the bowl surface?
    Answer: $\cos^{-1}\left(\frac{10}{17}\right)$ When you see a rolling object losing contact with a curved surface, you're dealing with a combination of energy conservation and circular motion dynamics. The key insight is that contact is lost when the normal force becomes zero. Let's work through this systematically. Using energy conservation from the top to angle $\theta$, the sphere drops a vertical distance $R(1-\cos\theta)$. For a rolling sphere, the kinetic energy splits between translational and rotational motion: $mgR(1-\cos\theta) = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ For a solid sphere, $I = \frac{2}{5}mr^2$, and the rolling condition gives $v = \omega r$. Substituting: $mgR(1-\cos\theta) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$ This yields $v^2 = \frac{10gR(1-\cos\theta)}{7}$ At the point of losing contact, the centripetal force equals the component of weight toward the center: $\frac{mv^2}{R} = mg\cos\theta$ Combining these equations: $mg\cos\theta = \frac{m \cdot 10gR(1-\cos\theta)}{7R} = \frac{10mg(1-\cos\theta)}{7}$ Simplifying: $7\cos\theta = 10(1-\cos\theta) = 10 - 10\cos\theta$ Therefore: $17\cos\theta = 10$, so $\cos\theta = \frac{10}{17}$ Answer C is correct: $\cos^{-1}\left(\frac{10}{17}\right)$ The other answers represent common errors: A assumes a sliding sphere ($\cos^{-1}\frac{2}{3}$), B uses incorrect energy distribution, and D misapplies the centripetal force condition. Strategy tip: Always remember that rolling objects have both translational and rotational kinetic energy, and the moment of inertia depends on the object's shape. The factor $\frac{7}{10}$ is characteristic of solid spheres.
14. A solid cylinder of mass $M$ and radius $R$ has a moment of inertia $I = \frac{1}{2}MR^2$. It is rotating with angular velocity $\omega$. What is the ratio of its rotational kinetic energy to its translational kinetic energy if it were moving with a translational speed $v = \omega R$?
    Answer: 1/2 The rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = \frac{1}{4}MR^2\omega^2$. The translational kinetic energy is $KE_{trans} = \frac{1}{2}Mv^2$. Substituting $v = \omega R$, we get $KE_{trans} = \frac{1}{2}M(\omega R)^2 = \frac{1}{2}MR^2\omega^2$. The ratio is $\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{4}MR^2\omega^2}{\frac{1}{2}MR^2\omega^2} = \frac{1/4}{1/2} = 1/2$.
15. A yo-yo consists of two uniform disks of mass $M$ and radius $R$ connected by a thin axle of mass $m$ and radius $r$. The yo-yo is released from rest with the string wound around the axle. As it falls and unwinds, what is the ratio of its rotational kinetic energy to its translational kinetic energy?
    Answer: $\frac{2MR^2 + mr^2}{2(M + m)r^2}$ The total moment of inertia about the center is $I = 2 \cdot \frac{1}{2}MR^2 + \frac{1}{2}mr^2 = MR^2 + \frac{mr^2}{2}$. For a yo-yo, the constraint is $v = r\omega$ (the string unwinds at radius $r$). The rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(MR^2 + \frac{mr^2}{2}\right)\omega^2$. The translational kinetic energy is $KE_{trans} = \frac{1}{2}(M + m)v^2 = \frac{1}{2}(M + m)r^2\omega^2$. The ratio is $\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{2}\left(MR^2 + \frac{mr^2}{2}\right)\omega^2}{\frac{1}{2}(M + m)r^2\omega^2} = \frac{MR^2 + \frac{mr^2}{2}}{(M + m)r^2} = \frac{2MR^2 + mr^2}{2(M + m)r^2}$.
16. A constant torque of 12 N m is applied to a flywheel, causing its angular velocity to increase from 20 rad s⁻¹ to 50 rad s⁻¹. The flywheel has a moment of inertia of 6.0 kg m². What is the duration of the applied torque?
    Answer: 15 s The relationship between torque, moment of inertia, and angular acceleration is $\tau = I\alpha$. The angular acceleration can be found from $\alpha = \frac{\Delta\omega}{\Delta t}$. Substituting this into the torque equation gives $\tau = I \frac{\Delta\omega}{\Delta t}$. This is the angular impulse equation $\tau \Delta t = I \Delta \omega = \Delta L$. Rearranging to solve for $\Delta t$: $\Delta t = \frac{I \Delta \omega}{\tau} = \frac{(6.0 \text{ kg m}^2)(50 \text{ rad s}^{-1} - 20 \text{ rad s}^{-1})}{12 \text{ N m}} = \frac{(6.0)(30)}{12} = \frac{180}{12} = 15 \text{ s}$.
17. A torque $\tau$ applied to an object for a time $\Delta t$ causes its angular momentum to change by $\Delta L$. If the same torque is applied for a time $2\Delta t$, what is the change in the object's rotational kinetic energy? Assume the object starts from rest.
    Answer: It quadruples. The angular impulse is $\tau \Delta t = \Delta L$. If the time is doubled to $2\Delta t$, the new change in angular momentum is $\Delta L' = \tau (2\Delta t) = 2(\tau \Delta t) = 2\Delta L$. Since the object starts from rest, its final angular momentum is $L' = 2L$. Rotational kinetic energy is related to angular momentum by $KE = \frac{L^2}{2I}$. The initial KE was $KE = \frac{L^2}{2I}$. The new KE is $KE' = \frac{(L')^2}{2I} = \frac{(2L)^2}{2I} = \frac{4L^2}{2I} = 4 \times KE$. Therefore, the kinetic energy quadruples.
18. A rigid disk of radius $R$ rotates with constant angular velocity $\omega$. A point P is at a distance $R/2$ from the center, and a point Q is on the rim at distance $R$. What is the ratio of the centripetal acceleration of Q to that of P ($a_Q / a_P$)?
    Answer: 2 Centripetal acceleration is given by $a = \omega^2 r$. Since the disk is rigid, the angular velocity $\omega$ is the same for all points. For point Q, $a_Q = \omega^2 R$. For point P, $a_P = \omega^2 (R/2)$. The ratio is $\frac{a_Q}{a_P} = \frac{\omega^2 R}{\omega^2 (R/2)} = 2$.
19. A solid cylinder of mass $M$ and radius $R$ is initially spinning with angular velocity $\omega_0$ about its central axis while moving with linear velocity $v_0$ along a rough horizontal surface. The coefficient of kinetic friction is $\mu$. After what time will the cylinder begin to roll without slipping?
    Answer: $\frac{v_0 - \omega_0 R}{3\mu g}$ The friction force $f = \mu Mg$ opposes the relative motion at the contact point. Initially, the contact point moves with velocity $v_0 - \omega_0 R$. If this is positive, friction acts backward on translation and forward on rotation. The linear acceleration is $a = -\mu g$ and angular acceleration is $\alpha = \frac{\mu MgR}{I} = \frac{\mu MgR}{\frac{1}{2}MR^2} = \frac{2\mu g}{R}$. The velocities become: $v(t) = v_0 - \mu gt$ and $\omega(t) = \omega_0 + \frac{2\mu gt}{R}$. Rolling without slipping occurs when $v(t) = \omega(t)R$: $v_0 - \mu gt = (\omega_0 + \frac{2\mu gt}{R})R = \omega_0 R + 2\mu gt$. Solving: $v_0 - \omega_0 R = 3\mu gt$, so $t = \frac{v_0 - \omega_0 R}{3\mu g}$.
20. A figure skater is spinning with an initial angular velocity $\omega_0$ and moment of inertia $I_0$. She pulls her arms in, reducing her moment of inertia to $I_0/3$. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?
    Answer: 3 By conservation of angular momentum, $L_i = L_f$, so $I_0\omega_0 = I_f\omega_f$. With $I_f = I_0/3$, her final angular velocity is $\omega_f = \frac{I_0\omega_0}{I_0/3} = 3\omega_0$. The initial kinetic energy is $KE_i = \frac{1}{2}I_0\omega_0^2$. The final kinetic energy is $KE_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(\frac{I_0}{3})(3\omega_0)^2 = \frac{1}{2}\frac{I_0}{3}(9\omega_0^2) = 3(\frac{1}{2}I_0\omega_0^2) = 3KE_i$. The ratio $KE_f/KE_i$ is 3. The extra energy comes from the work the skater does to pull her arms in.

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