Understand Rigid Body Mechanics - IB Physics

1. A student sits at rest on a rotating stool holding a spinning bicycle wheel with its axis vertical. The wheel is spinning clockwise when viewed from above. The student then inverts the wheel so its axis is again vertical, but it spins counter-clockwise when viewed from above. What is the subsequent motion of the student and the stool?
   Answer: They rotate clockwise. The system consists of the student, the stool, and the wheel. The action of flipping the wheel is an internal torque, so the total angular momentum of the system must be conserved. Let the upward direction be positive. Initially, the wheel's angular momentum is negative ($-L_{wheel}$) and the student is at rest ($L_{student}=0$), so the total initial angular momentum is $L_{initial} = -L_{wheel}$. After inversion, the wheel's angular momentum is positive ($+L_{wheel}$). To conserve total angular momentum, the student and stool must acquire angular momentum $L_{final, student}$ such that $L_{final, student} + (+L_{wheel}) = -L_{wheel}$. This gives $L_{final, student} = -2L_{wheel}$. The negative sign indicates that the student and stool will rotate in the original direction of the wheel's spin, which was clockwise.
2. A constant torque of 5.0 N m is applied to a grinding wheel, initially at rest, with a moment of inertia of 2.0 kg m². What is the work done by the torque after the wheel has completed 10 revolutions?
   Answer: 314 J The work done by a constant torque is given by $W = \tau \theta$, where $\theta$ must be in radians. First, convert the number of revolutions to radians: $\theta = 10 \text{ rev} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 20\pi \text{ rad}$. Now, calculate the work done: $W = (5.0 \text{ N m}) \times (20\pi \text{ rad}) = 100\pi \text{ J}$. Using $\pi \approx 3.14$, the work done is approximately $100 \times 3.14 = 314 \text{ J}$. The moment of inertia is extra information not needed for this method, but could be used to find the final kinetic energy, which by the work-energy theorem equals the work done.
3. A thin uniform rod of mass $m$ and length $L$ is hinged at one end and hangs vertically. It is struck at its center by a horizontal impulse $J$. What is the angular velocity of the rod immediately after the impulse?
   Answer: $\frac{3J}{2mL}$ The impulse $J$ is applied at the center of the rod (distance $\frac{L}{2}$ from the hinge). The angular impulse about the hinge is $J \cdot \frac{L}{2}$. The moment of inertia of the rod about the hinge (at one end) is $I = \frac{1}{3}mL^2$. Using the angular impulse-momentum theorem: $\Delta L = J \cdot \frac{L}{2}$, where $\Delta L = I\omega - 0 = \frac{1}{3}mL^2\omega$. Therefore: $\frac{1}{3}mL^2\omega = \frac{JL}{2}$, which gives $\omega = \frac{JL}{2} \cdot \frac{3}{mL^2} = \frac{3J}{2mL}$.
4. A solid disk of mass $M$ and radius $R$ is free to rotate about its fixed central axis. The moment of inertia of the disk is $I = \frac{1}{2}MR^2$. A string is wrapped around its rim and pulled with a constant tangential force $F$. What is the angular acceleration of the disk?
   Answer: $\frac{2F}{MR}$ The torque applied to the disk is $\tau = F \times R$ since the force is tangential. According to Newton's second law for rotation, $\tau = I\alpha$. Substituting the expressions for torque and moment of inertia, we get $FR = (\frac{1}{2}MR^2)\alpha$. Solving for the angular acceleration $\alpha$, we find $\alpha = \frac{FR}{\frac{1}{2}MR^2} = \frac{2F}{MR}$.
5. Four identical point masses $m$ are placed at the corners of a square with side length $a$. What is the moment of inertia of the system about an axis perpendicular to the square and passing through its centre?
   Answer: $2ma^2$ The moment of inertia of a system of point masses is $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of each mass from the axis of rotation. The centre of the square is the origin (0,0). The corners are at distances $x=\pm a/2$ and $y=\pm a/2$. The distance $r$ from the centre to any corner is the same. Using the Pythagorean theorem, $r^2 = (a/2)^2 + (a/2)^2 = a^2/4 + a^2/4 = a^2/2$. Since all four masses are identical and at the same distance from the axis, the total moment of inertia is $I = 4 \times mr^2 = 4m(a^2/2) = 2ma^2$.
6. A horizontal turntable with moment of inertia $I$ rotates freely with angular velocity $\omega$. A piece of clay of mass $m$ is dropped from rest and sticks to the turntable at a distance $r$ from the axis. After the clay sticks, a brake applies a constant external torque to stop the rotation. How does the angular momentum of the turntable-clay system change, first during the collision with the clay, and then during the application of the brake?
   Answer: Is conserved during collision; Decreases during braking During the collision, the forces between the clay and the turntable are internal to the system. Assuming no external torques act on the system during this brief period, the total angular momentum of the turntable-clay system is conserved. After the collision, the brake applies an external torque. An external torque causes a change in the system's angular momentum, according to $\Delta L = \tau_{ext} \Delta t$. Since the brake opposes the motion, the torque is retarding, and the angular momentum of the system decreases until it becomes zero.
7. A uniform rod of length $L$ and mass $M$ is pivoted at its centre. A small object of mass $m$ is placed on the rod at a distance $x$ from the pivot. The rod is released from a horizontal position. What is the initial angular acceleration of the rod?
   Answer: $\frac{mgx}{\frac{1}{12}ML^2 + mx^2}$ The initial torque on the system about the pivot is caused by the weight of the small object $m$, so $\tau = mgx$. The weight of the rod $Mg$ acts at the pivot, so it produces no torque. The total moment of inertia of the system is the sum of the moment of inertia of the rod about its centre ($I_{rod} = \frac{1}{12}ML^2$) and the moment of inertia of the point mass $m$ ($I_{mass} = mx^2$). So, $I_{total} = \frac{1}{12}ML^2 + mx^2$. Using Newton's second law for rotation, $\tau = I\alpha$, the initial angular acceleration is $\alpha = \frac{\tau}{I_{total}} = \frac{mgx}{\frac{1}{12}ML^2 + mx^2}$.
8. A disk is spinning about a horizontal axle oriented along an east-west axis. Viewed from the east, the disk spins clockwise. The axle is supported at both ends. The eastern support is then removed. What is the initial motion of the free eastern end of the axle due to precession?
   Answer: It moves horizontally towards the north. This is a problem of gyroscopic precession. First, determine the direction of the angular momentum vector $\vec{L}$. Using the right-hand rule for a clockwise spin viewed from the east, the $\vec{L}$ vector points to the west. When the eastern support is removed, the force of gravity acts downwards at the center of mass, creating a torque $\vec{\tau}$ about the western support (the pivot). The torque vector is given by $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ points east (from pivot to center of mass) and $\vec{F}$ points down. Using the right-hand rule for the cross product, $\vec{\tau}$ points horizontally to the north. The torque causes a change in angular momentum, $\Delta \vec{L}$, in the same direction as the torque. So, $\Delta \vec{L}$ points north. The new angular momentum is $\vec{L}_{new} = \vec{L}_{old} + \Delta \vec{L}$. The axle will precess in the direction of the torque, so the eastern end moves horizontally to the north.
9. A uniform solid cylinder rolls down a ramp and then moves across a horizontal surface with coefficient of kinetic friction $\mu$. If the cylinder has initial translational velocity $v_0$ when it reaches the horizontal surface, what distance will it travel before coming to rest?
   Answer: $\frac{3v_0^2}{4\mu g}$ For a rolling cylinder, the total kinetic energy is $KE = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2 = \frac{1}{2}mv_0^2 + \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v_0^2}{R^2} = \frac{1}{2}mv_0^2 + \frac{1}{4}mv_0^2 = \frac{3}{4}mv_0^2$. The work done by friction is $W = \mu mg \cdot d$, where $d$ is the distance traveled. By the work-energy theorem: $\mu mgd = \frac{3}{4}mv_0^2$. Solving for $d$: $d = \frac{3v_0^2}{4\mu g}$.
10. A uniform block of height H and square base of side length W is placed on a plane. The plane is slowly tilted by an angle θ. Assuming the block does not slide, what is tan(θ) when the block is on the verge of toppling?
    Answer: $\frac{W}{H}$ The block is on the verge of toppling when its centre of mass is vertically directly above the pivot point, which is the lower edge of its base of support. The centre of mass of a uniform block is at its geometric centre, at a height of $H/2$ from the base and a horizontal distance of $W/2$ from the edge. When the plane is tilted by an angle $\theta$, the angle between the vertical line through the centre of mass and the line perpendicular to the base is also $\theta$. A right-angled triangle is formed with the adjacent side $H/2$ and the opposite side $W/2$. From trigonometry, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{W/2}{H/2} = \frac{W}{H}$.
11. A uniform ladder of mass $M$ and length $L$ leans against a frictionless vertical wall at an angle $\theta$ to the horizontal ground. The ground is rough. Which equation represents the condition for rotational equilibrium about the point where the ladder touches the ground?
    Answer: $Mg\frac{L}{2}\cos\theta = N_{wall}L\sin\theta$ For rotational equilibrium, the net torque about any point must be zero. Choosing the pivot at the base of the ladder simplifies the calculation, as the normal force from the ground and the friction force exert no torque. The weight of the ladder $Mg$ acts downwards at its centre of mass, at a distance $L/2$ from the base. The horizontal distance from the pivot to the line of action of the weight is the lever arm, which is $(\frac{L}{2})\cos\theta$. This creates a clockwise torque of $\tau_{cw} = Mg\frac{L}{2}\cos\theta$. The normal force from the wall, $N_{wall}$, acts horizontally at the top of the ladder. Its lever arm is the vertical distance from the pivot, which is $L\sin\theta$. This creates a counter-clockwise torque of $\tau_{ccw} = N_{wall}L\sin\theta$. Setting clockwise torques equal to counter-clockwise torques gives $Mg\frac{L}{2}\cos\theta = N_{wall}L\sin\theta$.
12. Two identical uniform rods, each of mass $m$ and length $L$, are connected at their ends to form an L-shape. The system rotates about an axis through the corner, perpendicular to the plane containing both rods. If the system starts from rest and a constant torque $\tau_0$ is applied, what is the angular velocity after the system has rotated through angle $\theta$?
    Answer: $\sqrt{\frac{3\tau_0\theta}{mL^2}}$ The moment of inertia of each rod about the corner is $I_{rod} = \frac{1}{3}mL^2$ (about one end). The total moment of inertia is $I_{total} = 2 \times \frac{1}{3}mL^2 = \frac{2}{3}mL^2$. Using the work-energy theorem: $W = \Delta KE$, where $W = \tau_0 \theta$ and $\Delta KE = \frac{1}{2}I_{total}\omega^2 - 0 = \frac{1}{2} \cdot \frac{2}{3}mL^2 \cdot \omega^2 = \frac{1}{3}mL^2\omega^2$. Therefore: $\tau_0\theta = \frac{1}{3}mL^2\omega^2$, which gives $\omega = \sqrt{\frac{3\tau_0\theta}{mL^2}}$.
13. A solid sphere of mass $m$ and radius $r$ rolls on a horizontal surface with a linear speed $v$. It then rolls up an incline without slipping. What is the maximum vertical height $h$ it reaches? The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$.
    Answer: $\frac{7v^2}{10g}$ By conservation of mechanical energy, the initial total kinetic energy on the horizontal surface will be equal to the final gravitational potential energy at the maximum height. The initial total kinetic energy is the sum of translational and rotational kinetic energy: $E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For rolling without slipping, $\omega = v/r$. Substituting for $I$ and $\omega$: $E_k = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. The final potential energy is $E_p = mgh$. Setting $E_k = E_p$, we have $\frac{7}{10}mv^2 = mgh$. Solving for $h$ gives $h = \frac{7v^2}{10g}$.
14. A thin hoop and a solid disk have the same mass $M$ and outer radius $R$. They are rotated about an axis passing through their centres and perpendicular to their planes. Which statement correctly compares their moments of inertia ($I_{hoop}$ and $I_{disk}$) and the torques ($\tau_{hoop}$ and $\tau_{disk}$) required to produce the same non-zero angular acceleration $\alpha$?
    Answer: $I_{hoop} > I_{disk}$ and $\tau_{hoop} > \tau_{disk}$ The moment of inertia depends on the distribution of mass relative to the axis of rotation. For a hoop, all the mass is at radius $R$, so $I_{hoop} = MR^2$. For a solid disk, the mass is distributed from the centre to the radius $R$, so its moment of inertia is smaller: $I_{disk} = \frac{1}{2}MR^2$. Therefore, $I_{hoop} > I_{disk}$. From Newton's second law for rotation, $\tau = I\alpha$. To produce the same angular acceleration $\alpha$, the object with the greater moment of inertia requires the greater torque. Since $I_{hoop} > I_{disk}$, it follows that $\tau_{hoop} > \tau_{disk}$.
15. A flywheel with moment of inertia $I$ is accelerated from rest to a final angular velocity $\omega$ in a time $t$ by a constant torque. What is the total angle in radians through which the flywheel turns during this time?
    Answer: $\frac{1}{2}\omega t$ We can use the rotational kinematic equation $\theta = \frac{1}{2}(\omega_0 + \omega)t$. Since the flywheel starts from rest, the initial angular velocity $\omega_0 = 0$. Therefore, the angle turned is $\theta = \frac{1}{2}(0 + \omega)t = \frac{1}{2}\omega t$. Alternatively, one could find the constant angular acceleration $\alpha = \frac{\omega - \omega_0}{t} = \frac{\omega}{t}$ and then use $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(\frac{\omega}{t})t^2 = \frac{1}{2}\omega t$.
16. A figure skater is spinning with angular velocity $\omega$ and has a moment of inertia $I$. Their rotational kinetic energy is $E_k$. They pull their arms in, reducing their moment of inertia to $I/3$. What is the new rotational kinetic energy in terms of $E_k$?
    Answer: $3E_k$ Since there is no external torque, angular momentum is conserved. Let the initial state be 1 and final state be 2. $L_1 = L_2$, so $I_1\omega_1 = I_2\omega_2$. We are given $I_1 = I$, $\omega_1 = \omega$, and $I_2 = I/3$. Therefore, $I\omega = (I/3)\omega_2$, which gives $\omega_2 = 3\omega$. The initial kinetic energy is $E_k = E_{k1} = \frac{1}{2}I\omega^2$. The final kinetic energy is $E_{k2} = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(\frac{I}{3})(3\omega)^2 = \frac{1}{2}(\frac{I}{3})(9\omega^2) = 3(\frac{1}{2}I\omega^2) = 3E_k$.
17. A force of 50 N is applied to the end of a wrench of length 20 cm. The force is applied at an angle of 60° to the handle of the wrench. What is the magnitude of the torque on the bolt?
    Answer: 8.7 N m The magnitude of the torque is given by the formula $\tau = Fr\sin\theta$, where $r$ is the length of the lever arm and $\theta$ is the angle between the force vector and the lever arm vector. First, convert the length to SI units: $r = 20 \text{ cm} = 0.20 \text{ m}$. Now, calculate the torque: $\tau = (50 \text{ N})(0.20 \text{ m})\sin(60°)$. Since $\sin(60°) = \frac{\sqrt{3}}{2} \approx 0.866$, the torque is $\tau = 10 \times 0.866 = 8.66 \text{ N m}$, which is approximately 8.7 N m.
18. A solid ball of mass $M$ rolls without slipping along a horizontal surface with speed $v$. It makes a perfectly inelastic collision with a stationary block of mass $M$. After the collision, the combined object slides without rotating. What is the speed of the combined object?
    Answer: $v/2$ In this collision, we should consider the conservation of linear momentum for the system (ball + block) in the horizontal direction, as there are no net external horizontal forces. Mechanical energy is not conserved because the collision is perfectly inelastic. The fact that the ball was initially rolling is relevant for its energy, but not for its linear momentum. The initial linear momentum of the system is just that of the ball, $P_i = Mv$. The block is stationary. After the collision, the two objects stick together, forming a combined mass of (2M) moving with a final speed $v_f$. The final linear momentum is $P_f = (2M)v_f$. By conservation of linear momentum, $P_i = P_f$, so $Mv = (2M)v_f$. Solving for $v_f$ gives $v_f = \frac{Mv}{2M} = v/2$.
19. A wheel of radius $R$ rolls without slipping on a horizontal surface. The centre of the wheel moves with a constant linear speed $v$. What is the instantaneous speed, relative to the surface, of a point on the very top of the wheel?
    Answer: $2v$ The motion of any point on the wheel is the superposition of the translational motion of the centre of mass and the rotational motion about the centre of mass. The translational speed of every point is $v$. The tangential speed of a point on the rim due to rotation is $v_{tan} = \omega R$. The condition for rolling without slipping is $v = \omega R$, so $v_{tan} = v$. At the very top of the wheel, the translational velocity vector and the tangential velocity vector both point in the same direction (forward). Therefore, the total speed relative to the surface is the sum of these speeds: $v_{total} = v + v_{tan} = v + v = 2v$.
20. A centrifuge rotor accelerates uniformly from rest to a final angular speed of 400π rad s⁻¹ in 30 s. How many revolutions does the rotor make during this time?
    Answer: 3000 First, calculate the total angular displacement $\theta$ using the rotational kinematic equation $\theta = \frac{1}{2}(\omega_0 + \omega)t$. Given $\omega_0 = 0$, $\omega = 400\pi$ rad s⁻¹, and $t = 30$ s. $\theta = \frac{1}{2}(0 + 400\pi)(30) = (200\pi)(30) = 6000\pi$ radians. The question asks for the number of revolutions. Since one revolution is $2\pi$ radians, we convert radians to revolutions: Number of revolutions = $\frac{\text{total angle in radians}}{2\pi} = \frac{6000\pi}{2\pi} = 3000$.

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