Rolling
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AP Physics C: Mechanics › Rolling
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 $$I = \frac{2}{5}MR^2$$.
$$g \sin\theta$$
$$\frac{2}{3} g \sin\theta$$
$$\frac{1}{2} g \sin\theta$$
$$\frac{5}{7} g \sin\theta$$
Explanation
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 $$a = \frac{5}{7}g\sin\theta$$.
A small solid sphere of radius $$r$$ rolls without slipping inside a large fixed hemispherical bowl of radius $$R$$, starting from rest at the same height as the center of the bowl. What is the speed of the sphere's center of mass at the bottom of the bowl?
$$\sqrt{\frac{5}{7}gR}$$
$$\sqrt{\frac{10}{7}g(R-r)}$$
$$\sqrt{gR}$$
$$\sqrt{2gR}$$
Explanation
The center of mass of the small sphere falls a vertical distance of $$h = R-r$$. The initial potential energy is $$Mg(R-r)$$. This is converted into translational and rotational kinetic energy. For a solid sphere rolling, $$K_{total} = \frac{7}{10}Mv^2$$. By conservation of energy, $$Mg(R-r) = \frac{7}{10}Mv^2$$. Solving for $$v$$ gives $$v = \sqrt{\frac{10}{7}g(R-r)}$$.
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 $$I=\frac{1}{2}MR^2$$.
$$\arctan(\frac{2}{3}\mu_s)$$
$$\arctan(\mu_s)$$
$$\arctan(3\mu_s)$$
$$\arctan(2\mu_s)$$
Explanation
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 $$\theta_{max} = \arctan(3\mu_s)$$.
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 $$I = \frac{1}{2}MR^2$$.
16.0 J
4.0 J
8.0 J
12.0 J
Explanation
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.
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 $$I = \frac{1}{2}MR^2$$.
$$\frac{1}{4}Mv^2$$
$$Mv^2$$
$$\frac{1}{2}Mv^2$$
$$\frac{3}{4}Mv^2$$
Explanation
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 $$K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$$.
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 $$I = MR^2$$.
$$\sqrt{\frac{1}{2}gh}$$
$$\sqrt{gh}$$
$$\sqrt{2gh}$$
$$\sqrt{\frac{4}{3}gh}$$
Explanation
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 $$v = \sqrt{gh}$$.
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?
$$1/2$$
$$2/3$$
$$1/5$$
$$2/5$$
Explanation
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 $$(\frac{1}{5}Mv_{cm}^2) / (\frac{1}{2}Mv_{cm}^2) = 2/5$$.
A disk rolls without slipping along a straight, horizontal path. Which of the following statements about the motion of the disk is correct?
The force of static friction does positive work on the disk to maintain its rolling motion.
The speed of the center of mass is greater than the tangential speed of a point on the rim.
The point on the disk in contact with the path is instantaneously at rest with respect to the path.
Every point on the disk has the same linear velocity at any given instant.
Explanation
The condition for rolling without slipping is that the point of contact between the rolling object and the surface is instantaneously at rest. This means that the velocity of the center of mass $$v_{cm}$$ is exactly balanced by the tangential velocity $$R\omega$$ at that point, relative to the ground. The speeds are related by $$v_{cm} = R\omega$$. Different points have different velocities. The static friction force does no work because its point of application does not move.
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?
Zero, because the point of application of the force is instantaneously at rest.
Positive, equal to the rotational kinetic energy gained.
Negative, equal in magnitude to the energy lost from the system.
It depends on the coefficient of static friction and cannot be determined.
Explanation
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.
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?
The friction force is in the forward direction, increasing both $$v_{cm}$$ and $$\omega$$.
The friction force is in the backward direction, decreasing $$v_{cm}$$ and increasing $$\omega$$.
The friction force is in the backward direction, decreasing both $$v_{cm}$$ and $$\omega$$.
The friction force is in the forward direction, increasing $$v_{cm}$$ and decreasing $$\omega$$.
Explanation
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.