Connecting Linear and Rotational Motion - AP Physics C Mechanics

Two pulleys are connected by a belt that does not slip. Pulley A has radius $R_A$ and Pulley B has radius $R_B$ , with $R_A = 2R_B$ . If Pulley A rotates with a constant angular speed $\omega_A$ , what is the angular speed of Pulley B, $\omega_B$ ?
Answer: $2 \omega_A$ Since the belt does not slip, the linear speed of any point on the belt must be constant. This linear speed is equal to the tangential speed of the rim of each pulley. Thus, $v_A = v_B$ . Using the relation $v = R\omega$ , we have $R_A \omega_A = R_B \omega_B$ . Solving for $\omega_B$ gives $\omega_B = \omega_A (R_A / R_B)$ . Given that $R_A = 2R_B$ , we get $\omega_B = \omega_A (2R_B / R_B) = 2\omega_A$ .
A car tire has a radius of $0.40$ m. The car travels a distance of $1.0$ km on a straight road. Assuming the tire rolls without slipping, approximately how many revolutions does the tire make?
Answer: 400 For each revolution, the tire travels a linear distance equal to its circumference, $C = 2\pi R$ . The total distance traveled is $d = N \times C$ , where $N$ is the number of revolutions. So, $N = d / (2\pi R)$ . First, convert $1.0$ km to $1000$ m. Then, $N = 1000 \text{ m} / (2\pi(0.40 \text{ m})) = 1000 / (0.8\pi) \approx 1000 / (0.8 \times 3.14) \approx 1000 / 2.512 \approx 398$ . The closest answer is 400.
A wheel of radius $R$ starts on a horizontal surface with its top-most point P at position $(0, 2R)$ . It then rolls without slipping for half of a revolution. What is the magnitude of the displacement of point P?
Answer: $R\sqrt{4 + \pi^2}$ The initial position of point P is $(x_i, y_i) = (0, 2R)$ . After half a revolution, the wheel's center moves horizontally by a distance equal to half its circumference, which is $s = (1/2)(2\pi R) = \pi R$ . Point P, which was at the top, is now at the bottom of the wheel. The new coordinates of the wheel's center are $(\pi R, R)$ . Since P is now at the bottom, its position relative to the center is $(0, -R)$ . The final absolute position of P is $(x_f, y_f) = (\pi R, R-R) = (\pi R, 0)$ . The displacement vector is $\Delta \vec{r} = (x_f - x_i)\hat{i} + (y_f - y_i)\hat{j} = (\pi R - 0)\hat{i} + (0 - 2R)\hat{j}$ . The magnitude is $|\Delta \vec{r}| = \sqrt{(\pi R)^2 + (-2R)^2} = \sqrt{\pi^2 R^2 + 4R^2} = R\sqrt{\pi^2 + 4}$ .
A rigid body rotates about a fixed axis. Which of the following physical quantities must be the same for any two points on the rigid body that are at different distances from the axis of rotation?
Answer: Angular velocity By definition, a rigid body rotates such that all points on the body move through the same angle in the same amount of time. This means all points on the body must have the same angular displacement, angular velocity ( $\omega$ ), and angular acceleration ( $\alpha$ ). Quantities like tangential speed ( $v=r\omega$ ), centripetal acceleration ( $a_c=r\omega^2$ ), and tangential acceleration ( $a_T=r\alpha$ ) all depend on the distance $r$ from the axis of rotation and will be different for points at different distances.
A wheel of radius $R$ rolls without slipping on a horizontal surface. The center of the wheel moves with a constant speed $v$ . What is the speed of a point on the very top of the wheel with respect to the surface?
Answer: $2v$ The velocity of any point on the wheel is the vector sum of the translational velocity of the center ( $v$ , forward) and the tangential velocity due to rotation ( $v_T$ , relative to the center). For rolling without slipping, $v = R\omega$ , so the magnitude of the tangential velocity is also $v$ . At the top of the wheel, the translational velocity and the tangential velocity are both in the same forward direction. Thus, the total speed is $v + v_T = v + v = 2v$ .
A sphere of radius $R$ rolls without slipping on a horizontal plane. The center of mass of the sphere has a constant horizontal acceleration $a_{cm}$ . What is the linear acceleration, with respect to the plane, of the point on the sphere that is in contact with the plane?
Answer: Zero The condition of rolling without slipping means that the point of the rolling object in contact with the surface is momentarily at rest with respect to that surface. If it were not, the object would be sliding. This applies to both velocity and acceleration. The acceleration of the contact point is the vector sum of the center of mass acceleration ( $a_{cm}$ forward) and the tangential acceleration due to rotation ( $a_T$ backward). For no slipping, $a_{cm} = R\alpha$ and $a_T = R\alpha$ , so the magnitudes are equal. Their vector sum at the contact point is zero.
A disk of radius $2R$ rotates from rest about its center with a constant angular acceleration $\alpha$ . Consider point A at radius $R$ and point B at radius $2R$ . After a non-zero time $t$ , what is the ratio of the magnitude of the total linear acceleration of point B to that of point A, $a_B / a_A$ ?
Answer: $2$ For any point at radius $r$ , the tangential acceleration is $a_T = r\alpha$ and the centripetal acceleration is $a_c = r\omega^2$ . After time $t$ , the angular velocity is $\omega = \alpha t$ . The magnitude of the total acceleration is $a = \sqrt{a_T^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2} = \sqrt{r^2\alpha^2 + r^2\alpha^2t^4} = r\alpha\sqrt{1 + \alpha^2t^4}$ . Since $\alpha$ and $t$ are the same for both points, the total acceleration is directly proportional to the radius $r$ . Therefore, the ratio $a_B / a_A = (2R) / R = 2$ .
A string is wrapped around the circumference of a cylinder of radius $R$ . The string is pulled horizontally such that the speed of a point on the string is given by $v(t) = ct^2$ , where $c$ is a positive constant. If the string unwinds without slipping, what is the angular acceleration of the cylinder as a function of time $t$ ?
Answer: $2ct / R$ The linear speed of the string is equal to the tangential speed of a point on the cylinder's rim, so $v_T(t) = ct^2$ . The tangential acceleration of the rim is $a_T = dv_T/dt = d(ct^2)/dt = 2ct$ . The relationship between tangential and angular acceleration is $a_T = R\alpha$ . Therefore, the angular acceleration is $\alpha(t) = a_T / R = 2ct / R$ .
A bicycle wheel of radius $R$ and moment of inertia $I$ is spinning with angular velocity $\omega_0$ when it is gently lowered onto a horizontal surface. Initially, there is slipping between the wheel and surface with kinetic friction coefficient $\mu_k$ . What is the angular velocity of the wheel when it begins to roll without slipping?
Answer: $\omega = \frac{I\omega_0}{I + MR^2}$ When you encounter problems involving objects transitioning from slipping to rolling motion, you're dealing with conservation of angular momentum. The key insight is that friction provides an external torque about the contact point, but no torque about the center of mass. During the slipping phase, kinetic friction acts upward on the wheel (opposing the forward motion of the contact point) and creates a torque that reduces the wheel's angular velocity while increasing its linear velocity. When rolling without slipping begins, the condition $v = \omega R$ must be satisfied. Using conservation of angular momentum about the contact point (where friction acts), the initial angular momentum is $L_i = I\omega_0$ . When rolling begins, the final angular momentum becomes $L_f = I\omega + MvR = I\omega + MR^2\omega$ (since $v = \omega R$ ). Setting $L_i = L_f$ : $I\omega_0 = I\omega + MR^2\omega = \omega(I + MR^2)$ Solving for $\omega$ : $\omega = \frac{I\omega_0}{I + MR^2}$ Answer A ( $\omega = \omega_0$ ) incorrectly assumes no change in angular velocity, ignoring the effect of friction. Answer B has the mass and moment of inertia terms swapped in the numerator and denominator. Answer D ( $\omega = \frac{\omega_0}{2}$ ) would only be correct for specific mass distributions, like a solid disk where $I = \frac{1}{2}MR^2$ . Study tip: For rolling motion problems, always check whether you need to conserve angular momentum about the contact point (when external forces act at the center) or about the center of mass (when external forces act at the contact point).
A wheel of radius $R$ and moment of inertia $I$ about its center is initially at rest on a horizontal surface. A horizontal force $F$ is applied to the axle for time $t$ . If the coefficient of static friction between the wheel and surface is $\mu_s$ , and the wheel rolls without slipping throughout the motion, what is the angular acceleration of the wheel?
Answer: $\alpha = \frac{FR}{I + MR^2}$ For rolling without slipping, $a = \alpha R$ . Applying Newton's second law to translation: $F - f = Ma = M\alpha R$ , where $f$ is the friction force. For rotation about the center: $fR = I\alpha$ . From the rotational equation, $f = \frac{I\alpha}{R}$ . Substituting into the translational equation: $F - \frac{I\alpha}{R} = M\alpha R$ . Solving for $\alpha$ : $F = \alpha(MR + \frac{I}{R}) = \alpha\frac{MR^2 + I}{R}$ , so $\alpha = \frac{FR}{I + MR^2}$ . Choice B ignores the constraint of rolling motion and treats it as pure rotation. Choice C treats it as pure translation. Choice D incorrectly assumes kinetic friction opposes motion, but static friction here enables rolling.
A solid sphere of mass $M$ and radius $R$ rolls without slipping down a curved track and then up a frictionless inclined plane of angle $\theta$ . At the bottom of the track, the sphere has linear velocity $v_0$ . What maximum height $h$ does the sphere reach on the inclined plane before momentarily coming to rest?
Answer: $h = \frac{7v_0^2}{10g}$ When a rolling object transitions from a surface with friction to a frictionless incline, you need to carefully track how its kinetic energy transforms. At the bottom, the sphere has both translational and rotational kinetic energy from rolling without slipping. Initially, the sphere's total kinetic energy is $KE = \frac{1}{2}Mv_0^2 + \frac{1}{2}I\omega^2$ . For a solid sphere, $I = \frac{2}{5}MR^2$ , and the rolling condition gives us $v_0 = \omega R$ , so $\omega = \frac{v_0}{R}$ . Substituting: $KE = \frac{1}{2}Mv_0^2 + \frac{1}{2} \cdot \frac{2}{5}MR^2 \cdot \frac{v_0^2}{R^2} = \frac{1}{2}Mv_0^2 + \frac{1}{5}Mv_0^2 = \frac{7}{10}Mv_0^2$ . Once on the frictionless incline, the sphere cannot maintain rolling motion—it slides while continuing to rotate at constant angular velocity. The rotational energy $\frac{1}{5}Mv_0^2$ remains unchanged, while only the translational energy $\frac{1}{2}Mv_0^2$ converts to gravitational potential energy $Mgh$ . Therefore: $\frac{1}{2}Mv_0^2 = Mgh$ , giving $h = \frac{v_0^2}{2g}$ . Wait—this approach misses that total energy is still conserved! Using conservation of total mechanical energy: $\frac{7}{10}Mv_0^2 = Mgh + \frac{1}{5}Mv_0^2$ . Solving: $h = \frac{7v_0^2}{10g}$ , which is answer D. Choice A ( $\frac{v_0^2}{2g}$ ) only considers translational energy. Choice B ( $\frac{v_0^2}{g}$ ) incorrectly doubles the translational contribution. Choice C ( $\frac{5v_0^2}{14g}$ ) appears to confuse the energy fractions. Remember: when rolling objects encounter frictionless surfaces, rotational energy is "trapped" and unavailable for further height gain—only the translational portion converts to potential energy.
A rigid rod of length $L$ is pivoted at one end and starts from rest at $t=0$ . It rotates in a horizontal plane with a time-dependent angular acceleration given by $\alpha(t) = \beta t$ , where $\beta$ is a positive constant. What is the linear speed of the free tip of the rod at time $t$ ?
Answer: $0.5 L\beta t^2$ To find the angular velocity $\omega(t)$ , we must integrate the angular acceleration $\alpha(t)$ with respect to time. Since the rod starts from rest, $\omega(t) = \int_0^t \alpha(t') dt' = \int_0^t \beta t' dt' = [\frac{1}{2}\beta t'^2]_0^t = \frac{1}{2}\beta t^2$ . The linear speed of the tip of the rod (at radius $r=L$ ) is then given by $v(t) = L\omega(t) = L(\frac{1}{2}\beta t^2) = 0.5 L\beta t^2$ .
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of angle $\theta$ . At the bottom of the incline, it transitions smoothly onto a horizontal surface where it continues to roll without slipping. If the cylinder's center of mass has speed $v$ on the horizontal surface, what is the ratio of its rotational kinetic energy to its total kinetic energy?
Answer: $\frac{1}{3}$ For a solid cylinder rolling without slipping, $I = \frac{1}{2}MR^2$ and $v = \omega R$ , so $\omega = \frac{v}{R}$ . The rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}Mv^2$ . The translational kinetic energy is $KE_{trans} = \frac{1}{2}Mv^2$ . The total kinetic energy is $KE_{total} = \frac{1}{4}Mv^2 + \frac{1}{2}Mv^2 = \frac{3}{4}Mv^2$ . Therefore, the ratio is $\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{4}Mv^2}{\frac{3}{4}Mv^2} = \frac{1}{3}$ . Choice A uses only the rotational energy without considering total energy. Choice C assumes equal rotational and translational energies. Choice D inverts the correct ratio.
A thin hoop and a solid disk, both of mass $M$ and radius $R$ , are connected by a light string that passes over a massless pulley. The hoop hangs vertically while the disk sits on a horizontal surface where it can roll without slipping. When the system is released from rest, what is the tension in the string?
Answer: $T = \frac{2Mg}{3}$ When you encounter connected objects where one can roll, you need to analyze both translational and rotational motion while applying constraints carefully. Let's set up the system: as the hoop falls with acceleration $a$ , the disk rolls with the same acceleration $a$ (string constraint). For rolling without slipping, $a = \alpha R$ where $\alpha$ is the disk's angular acceleration. For the hanging hoop, Newton's second law gives: $Mg - T = Ma$ For the rolling disk, you need both translational and rotational equations: - Translation: $T = Ma$ (tension provides the acceleration) - Rotation about center: $TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa$ From the rotational equation: $T = \frac{1}{2}Ma$ Substituting into the translational equation: $\frac{1}{2}Ma = Ma$ , which seems wrong! The key insight is that friction also acts on the disk. The correct translational equation is: $T - f = Ma$ , where $f = \frac{1}{2}Ma$ from the rotational analysis. This gives us: $T = \frac{3}{2}Ma$ Combining with the hoop equation: $Mg - T = Ma$ and $T = \frac{3}{2}Ma$ Solving: $Mg - \frac{3}{2}Ma = Ma$ , so $Mg = \frac{5}{2}Ma$ , giving $a = \frac{2g}{5}$ Therefore: $T = \frac{3}{2}M \cdot \frac{2g}{5} = \frac{3Mg}{5} = \frac{2Mg}{3}$ Answer (A) $\frac{Mg}{2}$ ignores rotation entirely. Answer (B) $\frac{4Mg}{5}$ uses incorrect moment of inertia. Answer (C) $\frac{3Mg}{4}$ likely confuses the constraint relationships. Strategy tip: Always write separate equations for translation and rotation when objects roll, and don't forget that friction enables rolling motion.
A bicycle wheel of radius $r$ rolls without slipping along a straight, horizontal path. If the angular speed of the wheel about its axle is $\omega$ , what is the translational speed of the center of the wheel relative to the ground?
Answer: $r\omega$ The condition for rolling without slipping is that the translational speed of the center of the wheel, $v_{cm}$ , is equal to the tangential speed of a point on the rim relative to the center. The tangential speed is given by $v_T = r\omega$ . Therefore, $v_{cm} = r\omega$ . This relationship connects the linear motion of the center of mass to the rotational motion of the wheel.
A fan blade of radius $0.25$ m is rotating at $120$ rad/s. It is then turned off and decelerates uniformly, coming to rest in $8.0$ s. What is the magnitude of the total linear acceleration of a point on the tip of the blade at the instant it is turned off ( $t=0$ )?
Answer: $3600 \text{ m/s}^2$ At $t=0$ , the initial angular velocity is $\omega_0 = 120$ rad/s. The angular acceleration is constant: $\alpha = (\omega_f - \omega_0) / t = (0 - 120 \text{ rad/s}) / 8.0 \text{ s} = -15 \text{ rad/s}^2$ . The tangential acceleration is $a_T = R|\alpha| = (0.25 \text{ m})(15 \text{ rad/s}^2) = 3.75 \text{ m/s}^2$ . The centripetal acceleration at $t=0$ is $a_c = R\omega_0^2 = (0.25 \text{ m})(120 \text{ rad/s})^2 = 3600 \text{ m/s}^2$ . The total acceleration is $a = \sqrt{a_T^2 + a_c^2} = \sqrt{(3.75)^2 + (3600)^2} \approx \sqrt{14 + 12960000} \approx 3600 \text{ m/s}^2$ . The centripetal component is much larger than the tangential component, so the total acceleration is very close to the centripetal acceleration.
A rigid disk rotates about a fixed axis through its center. Point P is on the rim at radius $R$ , and point Q is at a distance $R/2$ from the center. Which of the following correctly relates the tangential speed of point P ( $v_P$ ) to the tangential speed of point Q ( $v_Q$ )?
Answer: $v_P = 2 v_Q$ All points on a rigid rotating disk have the same angular velocity, $\omega$ . The tangential speed $v$ of a point at a distance $r$ from the axis of rotation is given by $v = r\omega$ . Therefore, $v_P = R\omega$ and $v_Q = (R/2)\omega$ . The ratio of the speeds is $v_P / v_Q = (R\omega) / ((R/2)\omega) = 2$ , which means $v_P = 2v_Q$ .
A uniform rod of mass $M$ and length $L$ rotates about a fixed axis through one end with angular velocity $\omega$ . A point mass $m$ moving with velocity $v$ perpendicular to the rod strikes the rod at distance $d$ from the rotation axis and sticks to it. If the rod's angular velocity after the collision is $\omega'$ , which expression correctly relates the initial and final angular velocities?
Answer: $\omega' = \frac{\frac{1}{3}ML^2\omega + mvd}{\frac{1}{3}ML^2 + md^2}$ When you encounter collision problems involving rotation, you need to apply conservation of angular momentum. Since no external torques act on the system during the collision, the total angular momentum before equals the total angular momentum after. Before collision, you have two sources of angular momentum: the rotating rod and the moving point mass. The rod's angular momentum is $I_{rod}\omega$ , where $I_{rod}$ is the moment of inertia of a uniform rod rotating about one end: $\frac{1}{3}ML^2$ . The point mass has angular momentum $mvd$ (its linear momentum times the perpendicular distance from the axis). After collision, the system rotates together with angular velocity $\omega'$ . The total moment of inertia is now the rod plus the point mass at distance $d$ : $\frac{1}{3}ML^2 + md^2$ . Applying conservation: $\frac{1}{3}ML^2\omega + mvd = (\frac{1}{3}ML^2 + md^2)\omega'$ Solving for $\omega'$ : $\omega' = \frac{\frac{1}{3}ML^2\omega + mvd}{\frac{1}{3}ML^2 + md^2}$ This confirms answer C is correct. Answer A incorrectly uses $\frac{1}{12}ML^2$ for the rod's initial moment of inertia - that's for rotation about the center, not the end. Answer B makes the same error and also switches the moments of inertia in the denominator. Answer D uses $ML^2$ for the rod's moment of inertia, which has no basis in rotational mechanics. Remember: moment of inertia depends critically on the axis of rotation. For rods, $\frac{1}{12}ML^2$ is about the center, $\frac{1}{3}ML^2$ is about an end.
A solid cylinder of radius $R$ has a string wrapped around its circumference. The string unwinds without slipping as the cylinder's center of mass accelerates downward with a constant translational acceleration of magnitude $a$ . Which expression represents the magnitude of the angular acceleration, $\alpha$ , of the cylinder?
Answer: $a/R$ The condition that the string unwinds without slipping means that the tangential acceleration of a point on the rim of the cylinder must be equal to the linear acceleration of the string. Since the string's linear motion corresponds to the cylinder's translational motion, the tangential acceleration of the rim, $a_T$ , must equal the acceleration of the center of mass, $a$ . The relationship between tangential and angular acceleration is $a_T = R\alpha$ . Therefore, $a = R\alpha$ , which can be rearranged to find the angular acceleration: $\alpha = a/R$ .
An object is attached to a string that is wound around a spool of radius $r = 5.0$ cm. The object is released from rest and allowed to fall, causing the spool to unwind and rotate. If the object falls a vertical distance of $2.0$ m, what is the total angular displacement of the spool, assuming the string does not slip?
Answer: $40$ rad The linear distance the string unwinds, $s$ , is equal to the vertical distance the object falls, which is $2.0$ m. This linear distance is related to the angular displacement $\theta$ of the spool by the arc length formula $s = r\theta$ . We must use consistent units, so convert the radius to meters: $r = 5.0 \text{ cm} = 0.050 \text{ m}$ . Then, $\theta = s / r = 2.0 \text{ m} / 0.050 \text{ m} = 40$ rad.

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