## Chapter 12 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Spinning Grinding Wheel At time a grinding wheel has an angular velocity of 26.0 . It has a constant angular acceleration of 33.0 until a circuit breaker trips at time = 1.80 . From then on, the wheel turns through an angle of 432 as it coasts to a stop at constant angular deceleration. Part A Through what total angle did the wheel turn between and the time it stopped? Express your answer in radians. You did not open hints for this part. ANSWER: Part B At what time does the wheel stop? Express your answer in seconds. You did not open hints for this part. ANSWER: t = 0 rad/s rad/s2 t s rad t = 0 rad Part C What was the wheel’s angular acceleration as it slowed down? Express your answer in radians per second per second. You did not open hints for this part. ANSWER: An Exhausted Bicyclist An exhausted bicyclist pedals somewhat erratically when exercising on a static bicycle. The angular velocity of the wheels follows the equation , where represents time (measured in seconds), = 0.500 , = 0.250 and = 2.00 . Part A There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time to be at angle radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel’s rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures. s rad/s2 (t) = at − bsin(ct) for t 0 t a rad/s2 b rad/s c rad/s t = 0 = 0 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Express the angular displacement undergone by the spot of paint at seconds in degrees. Remember to use the unrounded value from Part A, should you need it. Express your answer in degrees using three significant figures. You did not open hints for this part. ANSWER: Part C What distance has the spot of paint moved in 2 seconds if the radius of the wheel is 50 centimeters? Express your answer in centimeters using three significant figures. You did not open hints for this part. ANSWER: = rad t = 2 = d Typesetting math: 29% Part D Which one of the following statements describes the motion of the spot of paint at seconds? You did not open hints for this part. ANSWER: Flywheel Kinematics A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration . The flywheel is assumed to be at rest at time in Parts A and B of this problem. Part A Find the time it takes to accelerate the flywheel to if the angular acceleration is . Express your answer in terms of and . d = cm t = 2.0 The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing. t = 0 t1 1 1 Typesetting math: 29% You did not open hints for this part. ANSWER: Part B Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity . Express your answer in terms of some or all of the following: , , and . You did not open hints for this part. ANSWER: Part C Assume that the motor has accelerated the wheel up to an angular velocity with angular acceleration in time . At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of . Find , the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Express your answer in terms of some or all of the following: , \texttip{\alpha }{alpha}, and \texttip{t_{\rm 1}}{t_1}. You did not open hints for this part. t1 = 1 1 1 t1 1 = 1 t1 −5 t2 1 Typesetting math: 29% ANSWER: Surprising Exploding Firework A mortar fires a shell of mass \texttip{m}{m} at speed \texttip{v_{\rm 0}}{v_0}. The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass \large{\frac15m} and a larger piece of mass \large{\frac45m}. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance \texttip{d}{d} from the mortar. If there had been no explosion, the shell would have landed a distance \texttip{r}{r} from the mortar. Assume that air resistance and the mass of the shell’s explosive charge are negligible. Part A Find the distance \texttip{d}{d} from the mortar at which the larger piece of the shell lands. Express \texttip{d}{d} in terms of \texttip{r}{r}. You did not open hints for this part. \texttip{t_{\rm 2}}{t_2} = s Typesetting math: 29% ANSWER: Kinetic Energy of a Dumbbell This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy \texttip{K_{\rm total}}{K_total} of a dumbbell of mass \texttip{m}{m} when it is rotating with angular speed \texttip{\omega }{omega} and its center of mass is moving translationally with speed \texttip{v}{v}. Denote the dumbbell’s moment of inertia about its center of mass by \texttip{I_{\rm cm}}{I_cm}. Note that if you approximate the spheres as point masses of mass m/2 each located a distance \texttip{r}{r} from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by I_{\rm cm} = mr^2, but this fact will not be necessary for this problem. Part A Find the total kinetic energy \texttip{K_{\rm tot}}{K_tot} of the dumbbell. Express your answer in terms of \texttip{m}{m}, \texttip{v}{v}, \texttip{I_{\rm cm}}{I_cm}, and \texttip{\omega }{omega}. You did not open hints for this part. \texttip{d}{d} = Typesetting math: 29% ANSWER: Part B This question will be shown after you complete previous question(s). Unwinding Cylinder A cylinder with moment of inertia \texttip{I}{I} about its center of mass, mass \texttip{m}{m}, and radius \texttip{r}{r} has a string wrapped around it which is tied to the ceiling . The cylinder’s vertical position as a function of time is y(t). At time t = 0 the cylinder is released from rest at a height \texttip{h}{h} above the ground. Part A The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate \texttip{\omega }{omega} and \texttip{v}{v}, the velocity of the center of mass of the cylinder? \texttip{K_{\rm tot}}{K_tot} = Typesetting math: 29% Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity. Express \texttip{\omega }{omega} in terms of \texttip{v}{v} and other given quantities. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C Suppose that at a certain instant the velocity of the cylinder is \texttip{v}{v}. What is its total kinetic energy, \texttip{K_{\rm total}}{K_total}, at that instant? Express \texttip{K_{\rm total}}{K_total} in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{I}{I}, and \texttip{v}{v}. You did not open hints for this part. ANSWER: Part D \texttip{\omega }{omega} = \texttip{K_{\rm total}}{K_total} = Typesetting math: 29% Find \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, the cylinder’s vertical velocity when it hits the ground. Express \texttip{v_{\rm f \hspace{1 pt}}}{v_f}, in terms of \texttip{g}{g}, \texttip{h}{h}, \texttip{I}{I}, \texttip{m}{m}, and \texttip{r}{r}. You did not open hints for this part. ANSWER: Kinetic Energy of a Rotating Wheel A simple wheel has the form of a solid cylinder of radius \texttip{r}{r} with a mass \texttip{m}{m} uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about the axle at a constant angular speed. The wheel rotates \texttip{n}{n} full revolutions in a time interval \texttip{t}{t}. Part A What is the kinetic energy \texttip{K}{K} of the rotating wheel? Express your answer in terms of \texttip{m}{m}, \texttip{r}{r}, \texttip{n}{n}, \texttip{t}{t} and, \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Finding Torque \texttip{v_{\rm f \hspace{1 pt}}}{v_f} = \texttip{K}{K} = Typesetting math: 29% A force \texttip{\vec{F}}{F_vec} of magnitude \texttip{F}{F} making an angle \texttip{\theta }{theta} with the x axis is applied to a particle located along axis of rotation A, at Cartesian coordinates (0, 0) in the figure. The vector \texttip{\vec{F}}{F_vec} lies in the xy plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the xy plane. A particle is located at a vector position \texttip{\vec{r}}{r_vec} with respect to an axis of rotation (thus \texttip{\vec{r}}{r_vec} points from the axis to the point at which the particle is located). The magnitude of the torque \texttip{\tau }{tau} about this axis due to a force \texttip{\vec{F}}{F_vec} acting on the particle is given by \tau = r F \sin(\alpha) = ({\rm moment \; arm}) \cdot F = rF_{\perp}, where \texttip{\alpha }{alpha} is the angle between \texttip{\vec{r}}{r_vec} and \texttip{\vec{F}}{F_vec}, \texttip{r}{r} is the magnitude of \texttip{\vec{r}}{r_vec}, \texttip{F}{F} is the magnitude of \texttip{\vec{F}}{F_vec}, the component of \texttip{\vec{r}}{r_vec} that is perpendicualr to \texttip{\vec{F}}{F_vec} is the moment arm, and \texttip{F_{\rm \perp}}{F_\perp} is the component of the force that is perpendicular to \texttip{\vec{r}}{r_vec}. Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation. In this problem, you must express the angle \texttip{\alpha }{alpha} in the above equation in terms of \texttip{\theta }{theta}, \texttip{\phi }{phi}, and/or \texttip{\pi }{pi} when entering your answers. Keep in mind that \pi = 180\;\rm degrees and (\pi/2) = 90\;\rm degrees . Part A What is the torque \texttip{\tau_{\rm A}}{tau_A} about axis A due to the force \texttip{\vec{F}}{F_vec}? Express the torque about axis A at Cartesian coordinates (0, 0). You did not open hints for this part. Typesetting math: 29% ANSWER: Part B What is the torque \texttip{\tau_{\rm B}}{tau_B} about axis B due to the force \texttip{\vec{F}}{F_vec}? (B is the point at Cartesian coordinates (0, b), located a distance \texttip{b}{b} from the origin along the y axis.) Express the torque about axis B in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: Part C What is the torque \texttip{\tau_{\rm C}}{tau_C} about axis C due to \texttip{\vec{F}}{F_vec}? (C is the point at Cartesian coordinates (c, 0), a distance \texttip{c}{c} along the x axis.) Express the torque about axis C in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. You did not open hints for this part. ANSWER: \texttip{\tau_{\rm A}}{tau_A} = \texttip{\tau_{\rm B}}{tau_B} = Typesetting math: 29% Part D What is the torque \texttip{\tau_{\rm D}}{tau_D} about axis D due to \texttip{\vec{F}}{F_vec}? (D is the point located at a distance \texttip{d}{d} from the origin and making an angle \texttip{\phi }{phi} with the x axis.) Express the torque about axis D in terms of \texttip{F}{F}, \texttip{\theta }{theta}, \texttip{\phi }{phi}, \texttip{\pi }{pi}, and/or other given coordinate data. ANSWER: Torque Magnitude Ranking Task The wrench in the figure has six forces of equal magnitude acting on it. \texttip{\tau_{\rm C}}{tau_C} = \texttip{\tau_{\rm D}}{tau_D} = Typesetting math: 29% Part A Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: The Parallel-Axis Theorem Typesetting math: 29% Learning Goal: To understand the parallel-axis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition (I=\sum m_ir_i^2), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as I=I_{\rm cm}+md^2, where \texttip{I_{\rm cm}}{I_cm} is the moment of inertia about an axis passing through the center of mass, \texttip{m}{m} is the total mass of the object, and \texttip{I}{I} is the moment of inertia about another axis, parallel to the one for which \texttip{I_{\rm cm}}{I_cm} is calculated and located a distance \texttip{d}{d} from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length \texttip{2r}{2r} made of two small spheres of mass \texttip{m}{m} each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page. Part A Using the definition of moment of inertia, calculate I_{\rm cm}, the moment of inertia about the center of mass, for this object. Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Using the definition of moment of inertia, calculate I_{\rm B}, the moment of inertia about an axis through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part C Now calculate I_{\rm B} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: I_{\rm cm} = I_{\rm B} = I_{\rm B} = Typesetting math: 29% Part D Using the definition of moment of inertia, calculate I_{\rm C}, the moment of inertia about an axis through point C, for this object. Point C is located a distance \texttip{r}{r} from the center of mass (see the figure). Express your answer in terms of \texttip{m}{m} and \texttip{r}{r}. You did not open hints for this part. ANSWER: Part E Now calculate I_{\rm C} for this object using the parallel-axis theorem. Express your answer in terms of \texttip{I_{\rm cm}}{I_cm}, \texttip{m}{m}, and \texttip{r}{r}. ANSWER: Consider an irregular object of mass \texttip{m}{m}. Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which passes through the center of mass (see the second diagram), is given by I_{\rm A}=0.64mr^2. Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram. In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, I_{\rm C} is the moment of inertia about axis C. I_{\rm C} = I_{\rm C} = Typesetting math: 29% Part F Which moment of inertia is the smallest? ANSWER: Part G Which moment of inertia is the largest? ANSWER: I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} I_{\rm A} I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} Typesetting math: 29% Part H Which moments of inertia are equal? ANSWER: Part I Which moment of inertia equals 4.64mr^2? ANSWER: Part J Axis X, not shown in the diagram, is parallel to the axes shown. It is known that I_{\rm X}=6mr^2. Which of the following is a possible location for axis X? ANSWER: I_{\rm A} and I_{\rm D} I_{\rm B} and I_{\rm C} I_{\rm C} and I_{\rm E} No two moments of inertia are equal. I_{\rm B} I_{\rm C} I_{\rm D} I_{\rm E} between axes A and C between axes C and D between axes D and E to the right of axis E Typesetting math: 29% Torque and Angular Acceleration Learning Goal: To understand and apply the formula \tau=I\alpha to rigid objects rotating about a fixed axis. To find the acceleration \texttip{a}{a} of a particle of mass \texttip{m}{m}, we use Newton’s second law: \vec {F}_{\rm net}=m\vec{a}, where \texttip{\vec{F}_{\rm net}}{F_vec_net} is the net force acting on the particle. To find the angular acceleration \texttip{\alpha }{alpha} of a rigid object rotating about a fixed axis, we can use a similar formula: \tau_{\rm net}=I\alpha, where \tau_{\rm net}=\sum \tau is the net torque acting on the object and \texttip{I}{I} is its moment of inertia. In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} are attached to a seesaw. The seesaw is made of a bar that has length \texttip{l}{l} and is pivoted so that it is free to rotate in the vertical plane without friction. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m_1>m_2, and that counterclockwise is considered the positive rotational direction. Part A The seesaw is pivoted in the middle, and the mass of the swing bar is negligible. Find the angular acceleration \texttip{\alpha }{alpha} of the seesaw. Express your answer in terms of some or all of the quantities \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{l}{l}, as well as the acceleration due to gravity \texttip{g}{g}. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B In what direction will the seesaw rotate, and what will the sign of the angular acceleration be? ANSWER: Part C This question will be shown after you complete previous question(s). \texttip{\alpha }{alpha} = The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part D In what direction will the seesaw rotate and what will the sign of the angular acceleration be? ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Pivoted Rod with Unequal Masses The figure shows a simple model of a seesaw. These consist of a plank/rod of mass \texttip{m_{\rm r}}{m_r} and length 2x allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass \texttip{m_{\rm 1}}{m_1} is attached to the left end of the rod, and a small sphere of mass \texttip{m_{\rm 2}}{m_2} is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to \texttip{g}{g}. The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Typesetting math: 29% Part A What is the moment of inertia \texttip{I}{I} of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of \texttip{m_{\rm r}}{m_r}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. Keep in mind that the length of the rod is 2x, not \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Weight and Wheel Consider a bicycle wheel that initially is not rotating. A block of mass \texttip{m}{m} is attached to the wheel and is allowed to fall a distance \texttip{h}{h}. Assume that the wheel has a moment of inertia \texttip{I}{I} about its rotation axis. Part A Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius \texttip{r_{\mit A}}{r_A} \texttip{I}{I} = Typesetting math: 29% . Find \texttip{\omega _{\mit A}}{omega_A}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit A}}{omega_A} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit A}}{r_A}, and \texttip{I}{I}. You did not open hints for this part. ANSWER: Part B Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius \texttip{r_{\mit B}}{r_B} . Find \texttip{\omega _{\mit B}}{omega_B}, the angular speed of the wheel after the block has fallen a distance \texttip{h}{h}, for this case. Express \texttip{\omega _{\mit B}}{omega_B} in terms of \texttip{m}{m}, \texttip{g}{g}, \texttip{h}{h}, \texttip{r_{\mit B}}{r_B}, and \texttip{I}{I}. \texttip{\omega _{\mit A}}{omega_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C Which of the following describes the relationship between \texttip{\omega _{\mit A}}{omega_A} and \texttip{\omega _{\mit B}}{omega_B}? You did not open hints for this part. ANSWER: A Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass \texttip{m_{\rm 1}}{m_1} and length \texttip{L}{L} is supported by two identical massless strings. Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to \texttip{\omega _{\mit B}}{omega_B} = \omega_A > \omega_B \omega_B > \omega_A \omega_A = \omega_B Typesetting math: 29% the left end of the bar and is connected to the floor. A small block of mass \texttip{m_{\rm 2}}{m_2} is supported against gravity by the bar at a distance \texttip{x}{x} from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use \texttip{g}{g} for the magnitude of the acceleration due to gravity. Part A Find \texttip{T_{\mit A}}{T_A}, the tension in string A. Express the tension in string A in terms of \texttip{g}{g}, \texttip{m_{\rm 1}}{m_1}, \texttip{L}{L}, \texttip{d}{d}, \texttip{m_{\rm 2}}{m_2}, and \texttip{x}{x}. You did not open hints for this part. ANSWER: Part B Find \texttip{T_{\mit B}}{T_B}, the magnitude of the tension in string B. Express the magnitude of the tension in string B in terms of \texttip{T_{\mit A}}{T_A}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. \texttip{T_{\mit A}}{T_A} = Typesetting math: 29% You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Part D If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of \texttip{x}{x} such that the bar remains stable (call it \texttip{x_{\rm critical}}{x_critical})? Express your answer for \texttip{x_{\rm critical}}{x_critical} in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{d}{d}, and \texttip{L}{L}. You did not open hints for this part. ANSWER: Part E \texttip{T_{\mit B}}{T_B} = \texttip{x_{\rm critical}}{x_critical} = Typesetting math: 29% This question will be shown after you complete previous question(s). A Tale of Two Nutcrackers This problem explores the ways that torque can be used in everyday life. Case 1 To crack a nut a force of magnitude \texttip{F_{\rm n}}{F_n} (or greater) must be applied on both sides, as shown in the figure. One can see that a nutcracker only applies this force at the point in which it contacts the nut (at a distance \texttip{d}{d} from the nutcracker pivot). In this problem the nut is placed in a nutcracker and equal forces of magnitude \texttip{F}{F} are applied to each end, directed perpendicular to the handle, at a distance \texttip{D}{D} from the pivot. The frictional forces between the nut and the nutcracker are equal and large enough that the nut doesn’t shoot out of the nutcracker. Part A Find \texttip{F}{F}, the magnitude of the force applied to each side of the nutcracker required to crack the nut. Express the force in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: Typesetting math: 29% Case 2 The nut is now placed in a nutcracker with only one lever, as shown, and again friction keeps the nut from slipping. The top “jaw” (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that \texttip{F_{\rm 2}}{F_2} is directed perpendicular to the handle. Part B Find the magnitude of the force \texttip{F_{\rm 2}}{F_2} required to crack the nut. Express your answer in terms of \texttip{F_{\rm n}}{F_n}, \texttip{d}{d}, and \texttip{D}{D}. You did not open hints for this part. ANSWER: \texttip{F}{F} = \texttip{F_{\rm 2}}{F_2} = Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Precarious Lunch A uniform steel beam of length \texttip{L}{L} and mass \texttip{m_{\rm 1}}{m_1} is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle \texttip{\theta }{theta}, as shown. Through the hinge, the wall exerts an unknown force, \texttip{F}{F}, on the beam. A workman of mass \texttip{m_{\rm 2}}{m_2} sits eating lunch a distance \texttip{d}{d} from the building. Part A Find \texttip{T}{T}, the tension in the cable. Remember to account for all the forces in the problem. Express your answer in terms of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{L}{L}, \texttip{d}{d}, \texttip{\theta }{theta}, and \texttip{g}{g}, the magnitude of the acceleration due to gravity. You did not open hints for this part. Typesetting math: 29% ANSWER: Part B Find \texttip{F_{\mit x}}{F_x}, the \texttip{x}{x}-component of the force exerted by the wall on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T} and other given quantities. You did not open hints for this part. ANSWER: Part C Find \texttip{F_{\mit y}}{F_y}, the y-component of force that the wall exerts on the beam ( \texttip{F}{F}), using the axis shown. Remember to pay attention to the direction that the wall exerts the force. Express your answer in terms of \texttip{T}{T}, \texttip{\theta }{theta}, \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}. ANSWER: \texttip{T}{T} = \texttip{F_{\mit x}}{F_x} = \texttip{F_{\mit y}}{F_y} = Typesetting math: 29% Pulling Out a Nail A nail is hammered into a board so that it would take a force \texttip{F_{\rm nail}}{F_nail}, applied straight upward on the head of the nail, to pull it out. (Take an upward force to be positive.) A carpenter uses a crowbar to try to pry it out. The length of the handle of the crowbar is \texttip{L_{\rm h}}{L_h}, and the length of the forked portion of the crowbar (which fits around the nail) is \texttip{L_{\rm n}}{L_n}. Assume that the forked portion of the crowbar is perfectly horizontal. The handle of the crowbar makes an angle \texttip{\theta }{theta} with the horizontal, and the carpenter pulls directly along the horizontal. Typesetting math: 29% Part A With what force \texttip{F_{\rm pull}}{F_pull} must the carpenter pull on the crowbar to remove the nail? Express the force in terms of \texttip{F_{\rm nail}}{F_nail}, \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta}. You did not open hints for this part. ANSWER: Now, imagine that \texttip{F_{\rm pull}}{F_pull} is not large enough to dislodge the nail. In other words, the nail stays in place, and, if the surface below the crowbar weren’t present, the crowbar would rotate around the point of contact with the nail. This makes it natural to take the pivot point to be the point where the crowbar is in contact with the nail. (But you are always free to choose the pivot point to be any fixed point, even one some distance from the object.) Part B What is the magnitude of the normal force that the surface exerts on the crowbar, \texttip{F_{\rm bar}}{F_bar}? Express your answer for the normal force in terms of \texttip{F_{\rm pull}}{F_pull}, \texttip{\theta }{theta}, \texttip{L_{\rm n}}{L_n}, and \texttip{L_{\rm h}}{L_h}. Take the upward direction to be positive. You did not open hints for this part. ANSWER: Three bars are shown in the figure. Both bars A and B have \texttip{F_{\rm pull}}{F_pull} acting on them in the horizontal direction. Bar C has \texttip{F_{\rm pull}}{F_pull} = \texttip{F_{\rm bar}}{F_bar} = Typesetting math: 29% \texttip{F_{\rm pull}}{F_pull} strictly perpendicular to the bar. \texttip{L_{\rm h}}{L_h}, \texttip{L_{\rm n}}{L_n}, and \texttip{\theta }{theta} are the same quantities in each case. Part C Let the magnitude of the torque about the bend in the crowbars be denoted \texttip{\tau _{\mit A}}{tau_A}, \texttip{\tau _{\mit B}}{tau_B} and \texttip{\tau _{\mit C}}{tau_C} for each of the three cases shown. Which of the following is the correct relationship between the magnitude of of the torques? You did not open hints for this part. ANSWER: Tipping Crane \tau_A > \tau_B > \tau_C \tau_A > \tau_B = \tau_C \tau_A = \tau_B = \tau_C \tau_A < \tau_B = \tau_C \tau_A < \tau_B < \tau_C \tau_A = \tau_B > \tau_C \tau_A = \tau_B < \tau_C Typesetting math: 29% Learning Goal: To step through the application of \Sigma \vec{\tau} = 0 to prevent a crane from tipping over. A crane of weight \texttip{W}{W} has a length (wheelbase) \texttip{c}{c}, and its center of mass is midway between the wheels (i.e., the mass of the lifting arm is negligible). The arm extending from the front of the crane has a length \texttip{b}{b} and makes an angle \texttip{\theta }{theta} with the horizontal. The crane contacts the ground only at its front and rear tires. Part A While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle \texttip{\theta _{\rm max}}{theta_max} between 0 \degree and 90 \degree that the crane arm can make with the horizontal without tipping the crane over. Is this correct? ANSWER: Part B Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load by observing the minimum angle \texttip{\theta _{\rm min}}{theta_min} that the crane arm makes with the horizontal. Is this correct? ANSWER: yes no Typesetting math: 29% Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H yes no Typesetting math: 29% Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of the crane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defying Newton's laws. What's going on here? ANSWER: Part I Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at 70^\circ above the horizontal, the crane begins to tip slowly forward. Consider the following possible actions: Release the brake on the lifting cable so that the load accelerates 1. downward. 2. Release the lifting arm so that \texttip{\theta }{theta} decreases rapidly and the load accelerates downward. 3. Increase \texttip{\theta }{theta} while simultaneously letting out the lifting cable so that the load accelerates downward. 4. Put the crane wheels in gear and accelerate the crane forward. None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground? ANSWER: Spinning Situations Suppose you are standing on the center of a merry-go-round that is at rest. You are holding a spinning bicycle wheel over your head so that its rotation axis is pointing upward. The wheel is rotating counterclockwise when observed from above. Newton's laws don't apply to torques. The rear wheels exert a downward force on the front wheels. The crane is not accelerating so forces don't matter. The earth exerts forces on the crane and the load. all but 1 all but 2 all but 3 all but 4 all of them Typesetting math: 29% For this problem, neglect any air resistance or friction between the merry-go-round and its foundation. Part A Suppose you now grab the edge of the wheel with your hand, stopping it from spinning. What happens to the merry-go-round? You did not open hints for this part. ANSWER: Twirling a Baton A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120{\rm kg} and length 80.0{\rm cm} . Part A Initially, the baton is spinning about a line through its center at angular velocity 3.00{\rm rad/s} . What is its angular momentum? Express your answer in kilogram meters squared per second. It remains at rest. It begins to rotate counterclockwise (as observed from above). It begins to rotate clockwise (as observed from above). Typesetting math: 29% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm kg \cdot m^2/s Typesetting math: 29%

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