2017.02.02 Homework 10_ Rotational Dynamics.pdf - Homework...

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Unformatted text preview: Homework 10: Rotational Dynamics Due: 11:59pm on Tuesday, April 4, 2017 To understand how points are awarded, read the Grading Policy for this assignment. Torque Magnitude Ranking Task The wrench in the figure has six forces of equal magnitude acting on it. 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. Hint 1. Definition of torque Torque is a measure of the "twist" that an applied force exerts on an object. Mathematically, torque is defined as τ = rF sin θ, where r is the magnitude of the displacement vector from the rotation axis to the point of application of the force of magnitude F , and θ is the angle between this displacement and the applied force, as shown in the figure. The direction of a torque can be either counterclockwise (as above) or clockwise. This is determined by the direction the object will rotate under the action of the force. Hint 2. Maximum torque Based on the mathematical definition of torque, torque is maximized when the force is large in magnitude, located a ⃗ large distance from the axis of interest, and oriented perpendicular to the displacement r , which is often referred to as the lever arm of the force. ANSWER: Reset Help largest smallest B D F A C E The correct ranking cannot be determined. Correct Calculating Torque I Part A Calculate the net torque about point O for the two forces applied as in the figure . The rod and both forces are in the plane of the page. As usual, take out­of­the­page torques to be positive and into­the­page torques to be negative. ANSWER: τ = ­28.0 N ⋅ m Correct Calculating Moment of Inertia I Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods . Part A Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure). ANSWER: I = 6.40×10−2 kg ⋅ m2 Correct Part B Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure). ANSWER: I = 3.20×10−2 kg ⋅ m2 Correct Part C Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O. ANSWER: I = 3.20×10−2 kg ⋅ m2 Correct Newton's 2nd Law for Rotation II A bucket of water of mass 15.4 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.340 m with mass 11.2 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.0 m to the water. You can ignore the weight of the rope. In addition, assume that the rope rolls off of the cylinder without slipping. Before beginning the problem, draw an extended force diagram for the cylinder and a force diagram for the bucket. Part A Is the bucket rotating, translating, or both, or neither? Use this to decide, from the following choices, how the analysis of the bucket should be done. ANSWER: We need to apply Newton's 2nd law for rotation to the bucket. We do not need to apply Newton's 2nd law nor Newton's 2nd law for rotation to the bucket. We need to apply Newton's 2nd law to the bucket. We need to apply both Newton's 2nd law and Newton's 2nd law for rotation to the bucket. Correct Part B Is the cylinder rotating, translating, or both, or neither? Use this to decide, from the following choices, how the analysis of the cylinder should be done. ANSWER: We need to apply Newton's 2nd law to the cylinder. We need to apply Newton's 2nd law for rotation to the cylinder. We do not need to apply Newton's 2nd law nor Newton's 2nd law for rotation to the cylinder. We need to apply both Newton's 2nd law and Newton's 2nd law for rotation to the cylinder. Correct Part C Use your answer to parts (A) and (B), as well as your force diagrams, to write an equation for the bucket and an equation for the cylinder. Your equations should contain the unknowns abucket , αcylinder and T , the tension in the rope. Using the relation between αcylinder and arope , as well as your two equations, solve for the acceleration of the bucket. ANSWER: abucket = ­7.19 m/s2 Correct Part D Assuming the bucket is released from rest, how long does it take to hit the ground? ANSWER: 1.67 s Correct Newton's 2nd Law for Rotation III The figure shows a simple model of a seesaw. These consist of a plank/rod of mass mr and length 2x allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass m1 is attached to the left end of the rod, and a small sphere of mass is attached to the right end. The and a small sphere of mass m2 is attached to the right end. The spheres are small enough that they can be considered point particles and assume the rod is uniform. Part A What is the moment of inertia I of this assembly about the axis through which it is pivoted? Express the moment of inertia in terms of mr , m1 , m2 and x. Keep in mind that the length of the rod is 2x. Hint 1. How to approach the problem The moment of inertia of the assembly about the pivot is equal to the sum of the moments of inertia of each of the components of the assembly about the pivot point. That is, the total moment of inertia is equal to the moment of inertia of the rod plus the moment of inertia of the particle of mass m1 plus the moment of inertia of the particle of mass m2 , all measured with respect to the pivot point. Hint 2. Find the moment of inertia due to the sphere of mass m1 What is the moment of inertia of the particle of mass m1 measured about the pivot point? Express your answer in terms of given quantities. Hint 1. Formula for moment of inertia Consider an object consisting of particles with masses mi . Let ri be the distance of the ith particle from the axis of rotation. Then the moment of inertia I of the object about the axis of rotation is given by 2 I = ∑ mi r i i . ANSWER: I1 = m1 x 2 Hint 3. Find the moment of inertia due to the sphere of mass m2 What is the moment of inertia of the particle of mass m2 measured about the pivot point? Express your answer in terms of given quantities. ANSWER: I2 = m2 x 2 Hint 4. Find the moment of inertia of the rod What is the moment of inertia of the rod about the pivot point? Express Ir in terms of mr and x. Hint 1. General formula for the moment of inertia of a rod Consider a rod of total length L and mass mr , pivoted about its center. (In this problem, L equals 2x.) What is the moment of inertia of the rod about its pivot point? ANSWER: mr L 2 (1/3)m r L (1/4)m r L 2 2 (1/12)m r L 2 ANSWER: Ir = 1 3 mr x 2 ANSWER: I = 2 (x )(m1 + m2 + mr 3 ) Correct Part B Find the net torque on the rod. As usual, take out­of­the­page torques to be positive and into­the­page torques to be negative. Express the net torque in terms of some or all of the variables mr , m1 , m2 , x and g. Hint 1. How to approach the problem The forces acting on the system (spheres and rod) are the weights of the spheres and the rod, and the reaction force from the pivot. Find the torque due to each of these forces about the pivot point and add them with the correct signs. Hint 2. Find the torque due to mass 1. Find the torque about the pivot due to the particle of mass m1 Express your answer in terms of given quantities. Keep in mind that the positive direction is out­of­the­ page. ANSWER: τ1 = m 1 gx Hint 3. Find the torque due to mass 2. Find the torque about the pivot due to the particle of mass m2 . Express your answer in terms of given quantities. Keep in mind that the positive direction is out­of­the­ page. ANSWER: τ2 = −m 2 gx Hint 4. Find the torque due to the gravitational force and the normal force. Besides the two masses, there are two more forces to consider: the normal force acting at the pivot and the gravitational force acting on the rod. The normal force acts at the pivot point, so its distance from the pivot point is zero, and thus this force contributes zero torque. The gravitational force acts at the rod's center of mass, which is also at the pivot point. Therefore, the torque due to the gravitational force about the pivot point is also zero for the rod. ANSWER: τnet = m 1 gx − m 2 gx Correct Part C Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol I for the moment of inertia of the assembly, whether or not you have answered the first part correctly.) What is the angular acceleration α of the rod immediately after it is released? As usual, take the out­of­the­page direction to be positive. Express α in terms of some or all of the variables mr , m 1 , m 2 , x, I and g. Hint 1. How to approach the problem The forces acting on the system (spheres and rod) are the weights of the spheres and the rod, and the reaction force from the pivot. Find the torque due to each of these forces about the pivot point and add them with the correct signs. Finally, use Newton's second law for rotational motion: τ = I α. Hint 2. Find the torque due to the sphere of mass m1 Find the torque about the pivot due to the sphere of mass m1 . Express your answer in terms of given quantities. Keep in mind that the positive direction is counterclockwise. Hint 1. Formula for torque ⃗ The torque about the pivot point due to a force F is , where πvot is the vector from the pivot point to the point where the force is applied. The other symbols have their usual meanings. If you are using any of the latter two expressions, you must remember that if the force tends to cause a clockwise rotation, you need to include a negative sign in your expression since the torque due to such a force is taken to be negative (by convention). τ = rpivot F sin ϕ = rpivot F ⊥ = (moment arm)F r ⃗ ANSWER: τ1 = m 1 gx Hint 3. Find the torque due to the sphere of mass m2 Find the torque about the pivot due to the particle of mass m2 . Express your answer in terms of given quantities. Keep in mind that the positive direction is counterclockwise. ANSWER: τ2 = −m 2 gx Hint 4. Torque due to forces acting on the rod Besides the two masses, there are two more forces to consider: the normal force acting at the pivot and the gravitational force acting on the rod. The normal force acts at the pivot point, so its distance from the pivot point is zero, and thus this force contributes zero torque. The gravitational force acts at the rod's center of mass, which is also at the pivot point. Therefore, the torque due to the gravitational force about the pivot point is also zero for the rod. Hint 5. Relating the angular acceleration to the net torque Let the net torque acting on the system about the pivot point be denoted by τπvot . Find an expression for τπvot . Express your answer in terms of the system's moment of inertia I and its resulting angular acceleration α. (Use I in your answer, not the expression for I you found in Part A.) ANSWER: τπvot ANSWER: = Iα m 1 g−m 2 g α = x(m 1 +m 2 + mr ) 3 Correct Substituting for I , the value obtained in Part A yields (m 1 −m 2 )g α= ( mr 3 . +m 1 +m 2 )x A large angular acceleration is often desirable. This can be accomplished by making the connecting rod light and short (since both mr and x appear in the denominator of the expression for α). For a seesaw, on the other hand, mr and x are usually chosen to be as large as possible, while making sure that the "rod" does not get too heavy and unwieldy. This ensures that the angular acceleration is quite low for better safety. Rotational Kinetic Energy A uniform sphere with mass 23.5 kg and radius 0.390 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. Part A If the kinetic energy of the sphere is 169 J , what is the tangential velocity of a point on the rim of the sphere? Express your answer with the appropriate units. ANSWER: v = 6.00 m s Correct Gravitational Potential Energy of Extended Objects A uniform 1.55 m ladder of mass 5.70 kg is leaning against a vertical wall while making an angle of 50.5 ∘ with the floor. A worker pushes the ladder up against the wall until it is vertical. ( Drawing a diagram of the ladder before and after the push will help! ). Part A Find the change in the gravitational potential energy of the ladder during the push. Hint 1. Center of Mass Remember that for an extended object, the gravitational potential energy is the same as if all the mass were located at the center of mass of the object, so that Ug = mgycm . This means that we need to find the change in vertical position only of the center of mass . Find the initial vertical position of the center of mass (before the ladder is pushed), and then the final vertical position of the center of mass (after the ladder is vertical against the wall). ANSWER: y cm,i ,y cm,f = 0.598,0.775 m ANSWER: ΔUg = 9.89 J Correct Conservation of Energy The pulley in the figure has radius 0.160 m and a moment of inertia 0.560 kg ⋅ m2 . The rope does not slip on the pulley rim. Ignore air resistance and friction. The diagram shows the initial state of the system at rest. At some time, the masses are released from rest and the 4.00 kg mass falls while the 2.00 kg mass rises. The final time of interest is when the 4.00 kg mass is just about to strike the ground. Part A Regarding just the 2.00 kg mass, select which energy types change from the initial state to the final state. ANSWER: Ug Krot Ktrans Uel Uint Correct Part B Regarding just the 4.00 kg mass, select which energy types change from the initial state to the final state? ANSWER: Krot Uint Ktrans Ug Uel Correct Part C Regarding just the pulley, select which energy types change from the initial state to the final state? ANSWER: Krot Uint Ktrans Ug Uel Correct Part D Which of the following statements is true regarding the application of the law of conservation of energy to the system? ANSWER: The total combined energy of all three objects (the two masses and pulley) is conserved. The energy of the 4.00 kg mass is by itself conserved. The total combined energy of the 2.00 kg mass and the 4.00 kg mass is conserved. The energy of the 2.00 kg mass is by itself conserved. The total energy of the pulley is by itself conserved. Correct The energy is only conserved if you are considering a closed system. In this case, the system of all three objects (the two masses and pulley) is a closed system! Therefore to apply conservation of energy you must consider all three objects in your system. Part E Which statement best describes the speed of each of the masses in comparison with the speed of various points on the pulley? ANSWER: The speed of each mass is completely unrelated to the tangential speed of points on the pulley. The speed of each mass is equal to the tangential speed of points halfway from the center of the pulley to its rim. The speed of each mass is equal to the tangential speed of points on the rim of the pulley. Correct Part F Use your answers to parts A through E and conservation of energy to calculate the speed of the 4.00­kg block just before it strikes the floor. ANSWER: v = 2.65 m/s Correct Calculating Angular Momentum A woman with a mass of 53.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.540 rev/s about an axis through its center. The disk has a mass of 111 kg and a radius of 4.00 m . Part A Calculate the magnitude of the total angular momentum of the woman­plus­disk system. (Assume that you can treat the woman as a point.) ANSWER: 5890 kg ⋅ m2 /s Correct Conservation of Angular Momentum I Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10 14 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 9.0×105 km (comparable to our sun); its final radius is 18 km . Assume its mass is 3.98 × 10 30 kg (twice the mass of our Sun). Part A Is there a net external torque acting on the star as it shrinks in size? What does this tell you about the total angular momentum of the star? ANSWER: There is a net external torque, which causes the angular momentum of the star to change. There is no net external torque, which causes the angular momentum of the star to change. There is a net external torque, which causes the angular momentum of the star to be conserved. There is a no net external torque, which causes the angular momentum of the star to be conserved. Correct Part B How and why did the moment of inertia of the neutron star change as it collapsed? ANSWER: The moment of inertia decreased because the mass got further from the axis of rotation. The moment of inertia increased because the mass got further from the axis of rotation. The moment of inertia increased because the mass got closer to the axis of rotation. The moment of inertia decreased because the mass got closer to the axis of rotation. Correct Part C If the star initially (before collapsing) rotated once every 32 days, use conservation of angular momentum to solve for the angular speed (in rad/s!) of the neutron star after collapse. (Hint: You will first need to convert the number of initial rotations per day into radians per second). ANSWER: ωf = 5680 rad/s Correct Conservation of Angular Momentum II A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 44.0 kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.400 kg , traveling perpendicular to the door at 11.0 m/s just before impact. Part A Consider the system to be the door plus the mud. Is there a net external torque on the system during the collision of the door and mud? What does this tell you about the total angular momentum of the system? ANSWER: There is no net external torque on the system, which causes the angular momentum of the system to be conserved. There is a net external torque on the system, which causes the angular momentum of the system to be conserved. There is no net external torque on the system, which causes the angular momentum of the system to change. There is a net external torque on the system, which causes the angular momentum of the system to change. Correct Part B Assume the mud sticks to the door. Use conservation of angular momentum to solve for the angular speed of the door after the mud sticks to it. ANSWER: ω = 0.149 rad/s Correct Score Summary: Your score on this assignment is 98.9%. You received 116.67 out of a possible total of 118 points. ...
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