Rolling - AP Physics C Mechanics

1. A uniform solid cylinder of mass $m$ and radius $R$ is initially at rest on a rough horizontal surface with coefficient of static friction $\mu_s$ and kinetic friction $\mu_k$. A horizontal force $F$ is applied to the cylinder's center of mass. For what range of applied force will the cylinder roll without slipping?
   Answer: $0 \leq F \leq 3\mu_s mg$ For rolling without slipping, apply Newton's laws with constraint $a = R\alpha$. Translation: $F - f = ma$. Rotation about center: $fR = I\alpha = \frac{1}{2}mR^2 \alpha = \frac{1}{2}mR^2 \cdot \frac{a}{R} = \frac{1}{2}mRa$, so $f = \frac{1}{2}ma$. Substituting into translation equation: $F - \frac{1}{2}ma = ma$, giving $F = \frac{3}{2}ma$ and $f = \frac{1}{2}ma = \frac{F}{3}$. For no slipping, friction force cannot exceed maximum static friction: $|f| \leq \mu_s mg$. So $\frac{F}{3} \leq \mu_s mg$, giving $F \leq 3\mu_s mg$. Combined with $F \geq 0$, the range is $0 \leq F \leq 3\mu_s mg$. Choice A uses wrong coefficient relationship. Choice B uses wrong moment calculation. Choice D neglects rotational effects.
2. A bowling ball is initially projected with a center of mass speed $v_0$ but zero angular velocity along a rough horizontal lane. The ball eventually begins to roll without slipping. During the initial slipping phase:
   Answer: the translational speed decreases, and the angular velocity increases. Initially, the ball is sliding, so a kinetic friction force acts opposite to the direction of motion. This force causes a linear deceleration, so the translational speed decreases. The friction force also exerts a torque about the center of mass, causing an angular acceleration, so the angular velocity increases from zero. This process continues until the condition for rolling without slipping, $v_{cm} = R\omega$, is met.
3. A solid block slides down a frictionless incline of height $h$. In a second experiment, a solid sphere of the same mass rolls without slipping down a separate, but identical, incline. Let $v_{slide}$ be the final speed of the block and $v_{roll}$ be the final speed of the sphere. Which of the following correctly compares the two speeds?
   Answer: $v_{slide} > v_{roll}$ For the sliding block, all initial potential energy $Mgh$ becomes translational kinetic energy: $Mgh = \frac{1}{2}Mv_{slide}^2$, so $v_{slide} = \sqrt{2gh}$. For the rolling sphere, the potential energy becomes both translational and rotational kinetic energy: $Mgh = \frac{7}{10}Mv_{roll}^2$, so $v_{roll} = \sqrt{\frac{10}{7}gh}$. Since $2 > 10/7$, it is clear that $v_{slide} > v_{roll}$. Less energy is available for translation in the rolling case.
4. A small solid sphere is to roll without slipping from rest at a height $H$ to complete a circular loop-the-loop of radius $R$. To just maintain contact with the track at the top of the loop, what is the minimum required height $H$? Assume the sphere's radius is negligible compared to $R$.
   Answer: $2.7 R$ At the top of the loop, the centripetal force is provided by gravity: $Mg = Mv_{top}^2/R$, so $v_{top}^2 = gR$. By conservation of energy from the start to the top of the loop: $MgH = Mg(2R) + K_{total,top}$. The total kinetic energy for a rolling sphere is $K_{total} = \frac{7}{10}Mv^2$. So, $MgH = 2MgR + \frac{7}{10}Mv_{top}^2 = 2MgR + \frac{7}{10}M(gR) = (2 + 0.7)MgR = 2.7MgR$. Therefore, $H = 2.7R$.
5. A ball is sliding on a rough horizontal surface such that its center of mass speed $v_{cm}$ is greater than $R\omega$, where $R$ is the ball's radius and $\omega$ is its angular speed. Which statement describes the effect of the kinetic friction force on the ball?
   Answer: The friction force is in the backward direction, decreasing $v_{cm}$ and increasing $\omega$. Since the ball is sliding, the point of contact is moving forward relative to the surface. The kinetic friction force opposes this relative motion, so it acts in the backward direction. This backward force creates a net force that decreases the translational speed $v_{cm}$. This force also creates a torque about the center of mass that increases the angular speed $\omega$. This continues until $v_{cm} = R\omega$ and the ball rolls without slipping.
6. A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping down an inclined plane of angle $\theta$. At the bottom of the incline, it encounters a horizontal surface with coefficient of kinetic friction $\mu_k$. If the sphere was released from rest at height $h$ above the horizontal surface, what is the distance it travels on the horizontal surface before coming to rest?
   Answer: $\frac{5h}{7\mu_k}$ Using energy conservation on the incline: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = \frac{7}{10}Mv^2$. So $v = \sqrt{\frac{10gh}{7}}$. On the horizontal surface, friction does work: $\mu_k Mgd = \frac{7}{10}Mv^2 = \frac{7}{10}M(\frac{10gh}{7}) = Mgh$. Therefore $d = \frac{h}{\mu_k} \cdot \frac{5}{7} = \frac{5h}{7\mu_k}$. Choice B neglects rotational energy. Choice C uses wrong moment of inertia. Choice D inverts the fraction.
7. A uniform solid sphere rolls down a curved track and then moves along a horizontal loop-the-loop of radius $R$. If the sphere is released from rest at height $h$ above the bottom of the track, what is the minimum value of $h$ required for the sphere to maintain contact with the track at the top of the loop?
   Answer: $\frac{27R}{10}$ This problem combines rotational motion with circular motion dynamics. When you see a rolling object going through a loop, you need to consider both translational and rotational kinetic energy, plus the condition for maintaining contact. At the top of the loop, the minimum condition for contact occurs when the normal force equals zero, meaning gravity alone provides the centripetal force: $mg = \frac{mv_{top}^2}{R}$, so $v_{top} = \sqrt{gR}$. Using energy conservation from release point to loop top: The initial potential energy $mgh$ converts to kinetic energy (both translational and rotational) plus potential energy at height $2R$: $mgh = \frac{1}{2}mv_{top}^2 + \frac{1}{2}I\omega^2 + mg(2R)$ For a solid sphere, $I = \frac{2}{5}mr^2$ and $v = \omega r$, so $\frac{1}{2}I\omega^2 = \frac{1}{5}mv^2$. Therefore: $mgh = \frac{1}{2}mv_{top}^2 + \frac{1}{5}mv_{top}^2 + 2mgR = \frac{7}{10}mv_{top}^2 + 2mgR$ Substituting $v_{top}^2 = gR$: $h = \frac{7gR}{10g} + 2R = \frac{7R}{10} + 2R = \frac{27R}{10}$ Choice A ($3R$) ignores rotational energy entirely. Choice B ($\frac{5R}{2}$) incorrectly uses the condition for a point mass. Choice C ($\frac{7R}{2}$) applies the rolling energy ratio incorrectly. Remember: For rolling objects in loops, always account for rotational kinetic energy using $\frac{1}{2}I\omega^2$, and know the moment of inertia for common shapes—solid spheres have $I = \frac{2}{5}mr^2$.
8. A uniform disk of mass $M$ and radius $R$ rolls without slipping with a center-of-mass speed $v$. Which of the following expressions represents the total kinetic energy of the disk? The rotational inertia of a uniform disk is $I = \frac{1}{2}MR^2$.
   Answer: $\frac{3}{4}Mv^2$ The total kinetic energy is the sum of the translational and rotational kinetic energies: $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For a disk, $I = \frac{1}{2}MR^2$, and for rolling without slipping, $\omega = v/R$. Substituting these gives $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$.
9. A hollow cylinder and a solid cylinder, both with the same mass $M$ and radius $R$, are released simultaneously from rest at the top of an inclined plane. Both roll without slipping. When the solid cylinder has traveled a distance $d$ down the incline, what distance has the hollow cylinder traveled?
   Answer: $\frac{3d}{4}$ For rolling without slipping, $a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}$. For solid cylinder: $I = \frac{1}{2}MR^2$, so $a_s = \frac{2g\sin\theta}{3}$. For hollow cylinder: $I = MR^2$, so $a_h = \frac{g\sin\theta}{2}$. Using $s = \frac{1}{2}at^2$, when solid travels distance $d$: $t = \sqrt{\frac{3d}{g\sin\theta}}$. In this time, hollow travels: $s_h = \frac{1}{2} \cdot \frac{g\sin\theta}{2} \cdot \frac{3d}{g\sin\theta} = \frac{3d}{4}$. Choice A uses wrong moment ratios. Choice C confuses with sphere values. Choice D assumes equal accelerations.
10. A solid cylinder of mass $M$ and radius $R$ rolls without slipping down a ramp inclined at an angle $\theta$ to the horizontal. What is the magnitude of the static friction force acting on the cylinder? The rotational inertia of a solid cylinder is $I = \frac{1}{2}MR^2$.
    Answer: $\frac{1}{3} Mg \sin\theta$ The linear acceleration of a rolling cylinder down an incline is $a = \frac{g\sin\theta}{1+I/MR^2} = \frac{g\sin\theta}{1+1/2} = \frac{2}{3}g\sin\theta$. The friction force $f$ provides the torque $fR = I\alpha$. Since $a = R\alpha$, we have $f = I a / R^2$. Substituting for $I$ and $a$: $f = (\frac{1}{2}MR^2)(\frac{2}{3}g\sin\theta) / R^2 = \frac{1}{3}Mg\sin\theta$.
11. A solid ball with mass $M$ and radius $R$ rolls without slipping along a horizontal surface with a center of mass speed $v$. It then encounters a ramp and rolls up to a maximum vertical height $h$. What is the value of $h$? The rotational inertia of a solid ball is $I = \frac{2}{5}MR^2$.
    Answer: $\frac{7v^2}{10g}$ The initial total kinetic energy is converted to potential energy. $K_{total} = K_{trans} + K_{rot} = Mgh$. $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. Setting this equal to $Mgh$ and solving for $h$ gives $h = \frac{7v^2}{10g}$.
12. A solid cylinder is placed on a rough incline. The coefficient of static friction is $\mu_s$. What is the maximum angle of inclination $\theta_{max}$ for which the cylinder will roll without slipping? The rotational inertia of a solid cylinder is $I=\frac{1}{2}MR^2$.
    Answer: $\arctan(3\mu_s)$ For rolling without slipping, the required static friction force is $f = \frac{1}{3}Mg\sin\theta$. The maximum available static friction force is $f_{max} = \mu_s N = \mu_s Mg\cos\theta$. To prevent slipping, we need $f \le f_{max}$. Therefore, $\frac{1}{3}Mg\sin\theta \le \mu_s Mg\cos\theta$. This simplifies to $\tan\theta \le 3\mu_s$. The maximum angle is thus $\theta_{max} = \arctan(3\mu_s)$.
13. A uniform solid disk of mass 4.0 kg and radius 0.2 m rolls without slipping along a horizontal surface. The speed of its center of mass is 2.0 m/s. What is the total kinetic energy of the disk? The rotational inertia of a solid disk is $I = \frac{1}{2}MR^2$.
    Answer: 12.0 J The total kinetic energy is the sum of translational and rotational kinetic energies. $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. Given $M=4.0$ kg, $R=0.2$ m, $v=2.0$ m/s. $I = \frac{1}{2}(4.0)(0.2)^2 = 0.08$ kg m$^2$. Since it rolls without slipping, $\omega = v/R = 2.0/0.2 = 10$ rad/s. $K_{total} = \frac{1}{2}(4.0)(2.0)^2 + \frac{1}{2}(0.08)(10)^2 = 8.0 + 4.0 = 12.0$ J.
14. A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping down an incline that makes an angle $\theta$ with the horizontal. What is the magnitude of the linear acceleration of the sphere's center of mass? The rotational inertia of a solid sphere is $I = \frac{2}{5}MR^2$.
    Answer: $\frac{5}{7} g \sin\theta$ The net force down the incline is $Mg\sin\theta - f = Ma$. The torque causing rotation is $\tau = fR = I\alpha$. For rolling without slipping, $a = R\alpha$. Substituting for $f$ and $\alpha$: $f = I\alpha/R = (\frac{2}{5}MR^2)(a/R)/R = \frac{2}{5}Ma$. Now substitute this into the force equation: $Mg\sin\theta - \frac{2}{5}Ma = Ma$. This simplifies to $Mg\sin\theta = \frac{7}{5}Ma$, so $a = \frac{5}{7}g\sin\theta$.
15. A sphere rolls without slipping down a rough inclined plane. What is the net work done by the force of static friction on the sphere as it rolls a distance $L$ down the incline?
    Answer: Zero, because the point of application of the force is instantaneously at rest. Work is defined as the integral of force dotted with displacement, $W = \int \vec{F} \cdot d\vec{s}$. For the force of static friction on an object rolling without slipping, the point of application of the force (the point of contact) has zero instantaneous velocity relative to the surface. Therefore, the displacement of the point of application of the force is zero, and the work done by the static friction force is zero.
16. A yo-yo consists of a disk of mass $M$ and radius $R$ with a string wound around its edge. The yo-yo is released from rest and falls under gravity while the string unwinds. If the string does not slip on the disk's edge, what is the acceleration of the yo-yo's center of mass?
    Answer: $\frac{2g}{3}$ Apply Newton's second law and rotational dynamics. Forces on yo-yo: weight $Mg$ downward, tension $T$ upward. For translation: $Mg - T = Ma$. For rotation about center: $TR = I\alpha = \frac{1}{2}MR^2 \alpha$. No-slip condition: $a = R\alpha$, so $\alpha = \frac{a}{R}$. Substituting: $TR = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa$, giving $T = \frac{1}{2}Ma$. From translation equation: $Mg - \frac{1}{2}Ma = Ma$, so $Mg = \frac{3}{2}Ma$, yielding $a = \frac{2g}{3}$. Choice A uses wrong moment calculation. Choice C ignores rotational inertia. Choice D assumes free fall.
17. A solid cylinder rolls without slipping at a constant velocity across a frictionless horizontal surface. Which of the following statements about the forces acting on the cylinder is true?
    Answer: The net force and the net torque on the cylinder are both zero. According to Newton's first law for translational and rotational motion, an object will maintain a constant linear velocity and a constant angular velocity if the net external force and net external torque are both zero. Since the cylinder's velocity is constant, its linear and angular accelerations are zero. Therefore, both the net force and the net torque must be zero. No friction is needed to maintain this state of motion on a frictionless surface.
18. A thin hoop of mass $M$ and radius $R$ is released from rest at the top of an incline of vertical height $h$. It rolls without slipping to the bottom. What is the speed of its center of mass at the bottom of the incline? The rotational inertia of a hoop is $I = MR^2$.
    Answer: $\sqrt{gh}$ 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 a hoop, $I=MR^2$, and for rolling without slipping, $\omega = v/R$. So, $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$. Solving for $v$ gives $v = \sqrt{gh}$.
19. A wheel is projected along a rough horizontal surface with an initial angular speed but zero initial translational speed. It slips on the surface until it begins to roll without slipping. During the slipping phase, the work done by the kinetic friction force on the wheel:
    Answer: is positive, because it causes the center of mass to accelerate from rest. The bottom of the wheel is moving backward relative to the surface due to the initial spin. The kinetic friction force opposes this motion, so it acts in the forward direction. Since the friction force is in the same direction as the displacement of the center of mass, the work done by friction is positive. This positive work increases the translational kinetic energy of the wheel. The torque from friction does negative work on the rotation, decreasing rotational kinetic energy. The total mechanical energy decreases, but the work done by the friction force on the center of mass is positive.
20. A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. What is the ratio of its rotational kinetic energy to its translational kinetic energy?
    Answer: $2/5$ Translational kinetic energy is $K_t = \frac{1}{2}Mv_{cm}^2$. Rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2$. For a solid sphere, $I = \frac{2}{5}MR^2$. For rolling without slipping, $v_{cm} = R\omega$. Substituting these into the rotational energy equation gives $K_r = \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v_{cm}}{R})^2 = \frac{1}{5}Mv_{cm}^2$. The ratio $K_r/K_t$ is $(\frac{1}{5}Mv_{cm}^2) / (\frac{1}{2}Mv_{cm}^2) = 2/5$.

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