Physics 21: Homework 8 Hints

Transcription

1 Physics 21: Homework 8 Hints 1. This problem echoes a practically identical problem done in class. The important notion is to determine the distance, r j/a from the rotation axis, A, to the position of each point mass, j, with mass m j. (a) Since all rotation axes in this part are running perpendicular to the plane of the triangle, the distance from the axis to each point mass is simply the distance from each mass to the point of intersection of this axis with the plane of the triangle. i. The answer is: I A = 13ml 2. ii. The answer is: I B = 6ml 2. iii. The answer is: I C = 5ml 2. iv. The answer is: I P = 9 2 ml2. v. The position of the CM needs to first be found. A convenient reference from which it may be found is from the position of B. The answer can be obtained either by then figuring out the distance between each mass and the CM and then using the definition of the MI, or by using the Parallel-Axis Theorem to use one of the results from above to relate to the desired MI. The answer is: I CM = 28 9 ml2. (b) In this case, all rotation axes are in the plane of the triangle. In this case, the emphasis on finding the perpendicular distance between an axis and each point mass is paramount. If a point mass happens to be on one of these axes, then its perpendicular position from this axis is clearly zero. i. The answer is: I AB = 5ml 2. ii. The answer is: I BC = ml 2. iii. The answer is: I AC = 3 2 ml2. iv. The answer is: I PB = 3ml Recall that the rod is a continuous object, so that the continuous version of the MI formula must be utilized. (a) Note that all points on the rod are a (perpendicular) distance, r, away from the rotation axis. The answer is: mr 2. (b) The MI about an axis running perpendicular to the rod and through its CM is I CM = 1 12 ml2. Since the parallel axis, A, is a distance, r, away from this CM axis, then by the Parallel-Axis Theorem: I A = I CM + md 2 A/CM = 1 12 ml2 + mr 2 = 1 12 m ( l r 2). (c) Assume that the rod is placed on an x axis, with its left edge at x = 0 and its right edge at x = l. For the linear density function, λ (x), we must have λ (x = 0) = λ 0, λ (x = l) = 3λ 0, and the profile be linear in x. Thus, using, for example, the point-intercept form of the equation of a straight line, it must therefore be true that ( λ (x) = λ ) l x. i. To determine this, enforce the rule that m = dm, using the fact that dm = λds, which (in this case) becomes dm = λ (x) dx. ii. For each of these cases, the integral becomes I P = x 2 dm = x 2 λ (x) dx. However, it is typically best to choose the axis of rotation as the origin. In this case, the profile for λ (x) shown above will be slightly altered to accommodate the choice of origin. Thus, with the caveat that λ (x) has been appropriately amended to reflect these changes, the limits of integration will, therefore, also have to be adjusted for each of the cases. 3. As discussed and used in class, note that the MI of a system of objects about some axis is the sum of the MI s of the constituents. As such, the constituents may nicely be separated as the rod, the sphere on the left, and the sphere on the right. 1

2 (a) In this approximation, the spheres may be treated as point masses and, thus, may each be considered to have all of the mass concentrated at the CM of each. The exact result would be [ I CM = 1 ( M + 6m 1 + 2R ) ] 2 l l However, depending on how much of an approximation is desired, the zeroth-order approximation would give whereas a first-order approximation would result in I CM 1 12 I CM 1 12 (M + 6m) l2, [ M + 6m ( R )] l 2. l The second-order approximation would just be the exact result quoted above. (b) Since the MI of a solid, uniform sphere is I (sphere) CM = 2 5 mr2, then using the Parallel-Axis Theorem will allow one to find it about an axis that is a distance d A/CM = R + l 2 away, which is what is required in this case to determine the MI of the system about the CM of the system. The exact result is: I CM = 1 12 [ M + 6m ( R l R 2 l 2 )] l A similar problem was done in class when we determined the MI of a solid, uniform disk. In other words, the integration is still done by invoking the MI of a hoop about an axis perpendicular to the hoop and running through its center. For such a hoop, of radius r and mass dm, the infinitesimal contribution to the MI of this annulus, di CM, is di CM = r 2 dm = r 2 σda = r 2 σ2πrdr, where da is the infinitesimal area of this infinitesimally thin hoop, and σ is the areal density. However, the contrasting feature in the case of this annulus is that the areal density must be determined by taking the total mass of the annulus, M, and dividing by its total area, π ( b 2 a 2). With r [a, b], upon integrating, the result becomes I (annulus) CM = 1 2 M ( b 2 + a 2). 5. In this case, suppose we draw an xy coordinate system, with the origin situated at the center of the disk, +x running to the right, and +y running up. In Quadrant I, y = y (x) = R 2 x 2 is the equation governing the quarter circle. Consider breaking the circle up into infinitesimal strips that are parallel to the y axis. In other words, consider infinitesimal strips, of mass dm, at the position x that have infinitesimal width dx and a length 2y = 2 R 2 + x 2. Since the areal density of the disk is σ = M/ ( πr 2), then the infinitesimal mass of each strip becomes dm = σda = M πr 2 2ydx = M πr 2 2 R 2 x 2 dx. Using Problem 2a, the MI of each infinitesimal strip may be written as After integration, the result is: I = MR 2 /4. di = x 2 dm = 2M πr 2 x2 R 2 x 2 dx. 6. This problem may be tackled either by doing a force analysis on the block, a torque analysis on the pulley, and the use of constant-acceleration kinematics for the block assuming a no-slip condition between the rope and the pulley; or via the use of energy conservation on the pulley and on the block. (a) Since the acceleration of the block will be constant in this process, this just boils down to a constantacceleration kinematics problem. 2

3 (b) Use the conservation of energy on the block and the pulley, noting that the work done by tension will cancel out when combining the separate energy equations. This is because of the no-slip condition, as well as the fact that the tension exerted on the pulley is equal, but opposite, to the tension exerted on the block. (c) The no-slip condition between linear and angular speed may be used here. Note that the pulley travels with an angular speed that is related (via the no-slip condition) to the speed with which the rope unravels from the pulley. Since the rope and the block travel as one, this unravelling speed of the rope will be identical to the speed of the falling block. (d) Technically, the net torque on the pulley is provided by the static friction between the rope and the rim of the pulley. Since the no-slip condition is valid, this effectively translates into simply using the tension provided by the bit of rope that is instantaneously unraveling from the pulley on its right-most position. Thus, if the string were to slip perfectly, no net torque will be provided to the pulley by this tension. In fact, this really translates into the tension being zero! Since the axle running through the pulley is deemed frictionless, this would imply that the pulley would simply continue at the angular speed it had before the perfect slippage began. Ultimately, when the rope starts slipping, the block is actually in free fall, while the pulley is spinning at whatever speed it had when the block reached the halfway point. 7. The important idea here is that since friction between the axle and the rod is nonexistent, then the mechanical energy of the system is conserved. However, care needs to be taken to only account for changes in gravitational potential energy via the change in height of the CM. (a) The answer is: ω (φ) = 3g sin (φ) /l. (b) The answer is: v CM (φ) = 3gl sin (φ) /4. (c) For the case of the simple pendulum, the height changes simply come from the height change of the pendulum s bob itself. The point of this problem goes to show that the problem cannot be condensed into one where the entire mass of the rod is concentrated to its CM. This is not because of taking into account the gravitational potential energy changes in the problem with the rod, as the terms will be identical whether or not the rod was thought of as a point mass concentrated at its CM; however, the issue sprouts when trying to account for the kinetic energy of these two separate systems. The rod s kinetic energy is definitely not the kinetic energy of its CM, as the fact that the speed of the points on the rod depend on how far away each point is from the pivot cannot be accounted by simply looking at the CM speed. 8. For this problem, the use of the geometric definition of the cross product is most useful. Namely, that for vectors A and B that have lengths A and B, respectively, and have an angle between their tails of θ, the magnitude of the cross product is A B = AB sin (θ). Of course, the direction is given by the right-hand rule for cross products. (a) Quoting only two significant figures, the answer is: 9.6,. (b) The answer is: 0. (c) The answer is: 12,. 9. Recall that ˆx ŷ = ẑ, and the use of cyclic permutations of this, as well as the use of the anti-commutative nature of the cross product, will give you all other relevant cross products between ˆx, ŷ, and ẑ. Also, recall that ˆx ˆx = ŷ ŷ = ẑ ẑ = 0. (a) The answer is: ŷ. (b) The answer is: ŷ. (c) The answer is: 0. (d) The answer is: ˆx. (e) The answer is: 0. 3

4 10. The area of the mentioned parallelogram is simply the magnitude of the cross product between these vectors. To determine the answer, both the geometric and algebraic implications of the cross product must be utilized. The answer is: A gram = 11; θ = Note that the magnitude of all of the moment arms is exactly half the length of a diagonal of the square, namely l/ 2. The tails of the forces F 1 and F 2 make angles of 3π/4 relative to their respective moment arms, while the tail of F 4 makes an angle of γ + π/4 relative to its moment arm. Of course, F 3 and its moment arm are at right angles to one another. The answer is: 1.79 N m,. 12. In essence, the scales read the contact forces acting on the body at the position of the head and the feet. These forces are both pointing vertically upward, while the person s weight is pointing downward. (a) Since this is a static-equilibrium situation, the contact forces must balance the weight of the person, so that the person s weight is simply w = 700 N. This weight acts at the position of the person s CM, which we assume is a distance, x, from the person s feet. Analyzing torques from the feet as the reference point will output the result. (b) In this case, the horizontal tensions in the rope must balance each other out, since the only other force acting on the body is the person s weight and is, therefore, vertical. Thus, the vertical components of the tensions must balance the weight of the person, which is now known from Part (a). Just analyzing forces is not enough, since the problem consists of three unknowns (the tensions in the ropes on either side and the angle that the left-hand rope makes relative to the ceiling), yet the analysis of the forces only outputs two independent equations. Thus, the torque must also be analyzed. It is easiest to pick the reference point as the attachment of the left-hand rope to the board, as this will eliminate the torque contribution from the tension of the left-hand rope altogether. As such, the torque equation will allow solving for the tension in the right-hand rope immediately, which could then be used to determine the remaining unknowns. (Another, more clever way to solve the problem is to note that the vertical tensions must necessarily play the role of what the normal forces did from Part (a). As such, those vertical tensions may be set equal to the readings on each scale, outputting two equations with three unknowns. Finally, by using the force balance horizontally, a final equation can be obtained. Thus, it is not even necessary to use the torque equation in this context, so long as the similarities can be realized from Part (a).) The answers are: T B = 400 N, T A = 450 N, and α = The hinges on the gate serve not only to prevent the gate from accelerating downward due to the weight of the gate, they also prevent the gate from angularly accelerating due to its gravitational torque. (a) Since the weight of the block is vertical, the fact that the hinges provide horizontal reactive forces must necessitate that these forces balance each other out. (b) Since the gravitational torque is aiming to make the gate rotate clockwise, the upper hinge is being pulled out while the bottom hinge is being pushed in. A good way to see this is to get rid of one of the hinges to see the effect on the other. If A is eliminated, then the gate will rotate clockwise about B, so that the top of the gate would want to pull away from A; however, if B is eliminated, then the gate will rotate clockwise about A, so that the bottom of the gate would want to push into B. (c) A very useful way to determine these horizontal forces is to compute torques from references A and B, separately. This will eliminate the torque from all vertical reactive forces, as well as the horizontal reactive force from the hinge at which the reference is chosen. Thus, this only leaves the torque due to the other hinge s horizontal reactive force, and the torque due to the gate s weight. (d) As mentioned above, the vertical reactive forces must, as a combination, balance the weight of the gate. (e) As mentioned in Part (b), if hinge A is taken away, the gate will rotate clockwise about hinge B. To prevent this from happening, a counter-torque must be provided, which stems from the rope. Indeed, note that if the rope, for example, contracts, then the gate will want to rotate in an anti-clockwise fashion relative to B. It helps to analyze the torques from hinge B, which will immediately output the value of the tension without having to analyze forces. 14. Note that since the ladder is uniform, its CM is at the physical center of the ladder, which is where the weight of the ladder is technically acting. 4

5 (a) The ladder s weight acts downward at its CM, while the ladder feels normal forces at the points of contact with the floor and the wall. The normal force at the wall is pointing left, while the normal force at the floor is pointing up, which follows from the fact that normal forces always act normal (or perpendicular) to the (local) surface of contact. A nice way to get a feel for all of the relevant friction-force directions is to imagine what the tendency of movement would be if both the wall and the floor were made frictionless. In this circumstance, the bottom of the ladder would slip to the left, while the top of the ladder would simultaneously slip downward. Thus, in general, a static-friction force would act rightward at the point of contact with the floor, while a static-friction force would act upward at the point of contact with the wall. (b) Note that if the wall was frictionless, the normal force from the floor could supply the vertical force to balance the weight of the block, while the static-friction force from the floor may conceivably balance the normal force from the wall. However, if the floor was made frictionless, then there is no horizontal force that can balance the normal force from the wall. Thus, the ladder is bound to slip. (c) Note that the ladder can be made stable when the wall is frictionless, as discussed in Part (b). i. The answer is: f s = Mg/ [2 tan (φ)]. ii. Unlike in Part (ci) above, the static friction is maximized in this case. The result follows by determining the value of the normal force on the ladder by the floor, which is the contact force that governs the friction force from the ground. iii. In this case, the person introduces an extra weight to the problem, which acts vertically down at the position of the person on the ladder. As such, this introduces an additional force to be balanced vertically, as well as an additional torque upon referencing the point of contact with the floor that must be balanced by a counter-torque from the normal force by the wall. A. The answer is: l/2. B. The answer is: φ min = This problem is mostly made difficult due to the geometry. However, carefully inspecting all of the relevant angles and moment-arm lengths should condense the problem into a more manageable one. It is also useful to tap into your experience to help make the effects of this problem more tangible. For example, if you have ever used a dolly to move a heavy object (e.g., a refrigerator dolly to move a refrigerator or a washing machine), you know that you have to tip the dolly to be able to roll it along a ramp or a floor. The degree to which the dolly is rotated makes a big difference on the amount of force you need to provide in order to make it move linearly. (a) If the mass of the stone is taken as M, then the answer is: F = ( ) Mg. (b) Particularly from the experience to which we alluded, you may know that there is a specific angle at which the stone would balance. This happens to be the exact point where the weight of the stone, as an extended line, intersects the pivoting corner. Up to this angle, the torque due to the weight provided for an anti-clockwise rotation of the block to land it back onto its width. Thus, the force from the rope had to provide the counter-torque in order to keep the object from rotating. However, once the weight no longer provides a torque, there is no more need for the rope to provide one. Thus, as the angle increases, the force provided by the man on the rope decreases. (c) This angle can now be geometrically determined based on the analysis described in Part (b). The answer is: θ b = (d) This is a circumstance that a person rotating a dolly to be able to transport a refrigerator would likely want to avoid. Going beyond this angle would make the gravitational torque want to, instead, make the block rotate so that it lands on its length (i.e., clockwise relative to the figure). In other words, the person should move out of the way if this was to happen! 16. A similar problem will be done in class, with the sole distinction of not including friction between the block and the ramp. Although it is not clear exactly how the system will move, it can really only move in a clockwise sense (i.e., with the block moving up the incline, the pulley rotating clockwise, and the ball moving down) or in an anti-clockwise sense (i.e., with the block moving down the incline, the pulley rotating anti-clockwise, and the ball moving up). So, it is wise to just pick one of these senses of movement as positive, so that all of the 5

6 relevant accelerations are taken to be positive. Upon use of such compatible coordinates, if it is found that the outcome for the accelerations is negative, then the incorrect direction was chosen. Even though the incorrect direction may have been chosen, the physics would nonetheless be correct. As seen in class, note also that the tensions on either side of the pulley cannot be equal, for if they were, then the pulley would not angularly accelerate out of rest. 17. In this problem, since the disks are fused together, the pulley will move as one. (a) To determine the mass of the pulley, use the fact that the disks are made of the same material, but have different radii. Indeed, note that because the disk L has triple the radius of S and because the area of a circle scales as the square of the radius then disk L has nine times the mass of S. (b) Since the MI of a system is the sum of the MI s of the mutually-exclusive constituents, then the MI of this pulley system is the sum of the MI s of each disk about their common axis of symmetry. Note that the MI of a disk of mass M and radius R about its axis of symmetry is MR 2 /2. (c) Although one might be tempted to immediately invoke the dynamical equations to solve this problem, it is worthwhile to note that it is known exactly how many revolutions are performed by the pulley in the given time. Furthermore, note that the pulley will have a constant angular acceleration since all torques will be derived from constant forces that are acting at positions with unchanging moment-arm values and orientations. Thus, the angular acceleration of the pulley may be determined using angular kinematics at a constant angular acceleration. Finally, since it is presumed that the strings do not slip relative to the disks around which each respectively wraps, then this angular acceleration may be related to the linear acceleration of the blocks via a no-slip condition. Nonetheless, since the angular acceleration of all points on the pulley is the same, the linear acceleration of points at different radii on the pulley are not the same. Thus, the accelerations of the blocks are not the same. (d) Imagine holding the system at rest. Under such a circumstance, the tension on each block must be balanced by each block s weight. However, since the weights of the blocks are the same, the tensions must also be the same. Finally, since the tension on the larger disk acts at a point that will have a larger moment arm, then this allows one to see that the sense of rotation must be anti-clockwise upon release, even though it will no longer be true that the tension in each rope is equal to the weight of the block attaching each. This is because there will be an acceleration for each block, so that Newton s 2nd Law can no longer allow for the forces to balance on each block under these circumstances. (e) To solve this problem, one only needs to use the linear version of Newton s 2nd Law for the hanging blocks. (f) The main point of this part is that if one side of the law when calculated independently from the other side does not equal the other side, then there must be something missing. Indeed, when the torque on the pulley due to each tension is summed up vectorially, the outcome will be greater than the product of the pulley s MI about its axle and its angular acceleration. As such, there must be an additional torque that acts opposite to the torque due to the rope attaching to the larger pulley. The tension torques subtracted by the net torque will provide the additional clock-wise torque supplied by the friction from the axle. 18. This problem is effectively a combination of a variety of examples done on the kinematics and dynamics of rolling motion during class. Revisit your notes to reproduce those results by walking through this problem carefully. The purpose of the problem was to provide such a walk-through of the lecture materials in the first place. Optional (Problems on Angular Momentum and Its Conservation) 1. The fact that the disks eventually spin at some common angular velocity implies that the non-rotating disk must have felt a net torque (and, thus, an angular acceleration) to allow it to start spinning in the first place. Of course, this net torque must be supplied by the rotating disk. Indeed, if there was no friction present between the disks, Disk 1 would maintain its non-rotating state even while maintaining contact with the rotating Disk 2. Thus, it is a frictional interaction between the disks that is responsible for the net torque on the non-rotating disk in order to get it to begin spinning. In turn, this frictional force aims to slow down the rate at which Disk 2 spins. The collision between these disks may be thought as the rotational analog of the linear perfectly inelastic (or sticking) collision. For the same reasons which, of course, may be proved the mechanical energy of the 6

7 system of disks is not conserved during the collision process. However, since the frictional torques between the disks are internal, this allows for conserving the angular momentum of the system (e.g., about the axle). 2. This problem involves the use of the definition of the angular momentum, L A = r A p, of a particle with linear momentum, p, about some reference point A. From t = 0, φ = 0 π/2, so that the particle is rotating about the reference due to it technically sweeping an angle about this point, just as it would if it were moving in a circle or an ellipse about this point. Part (a) is completely geometrical and does not even involve the use of angular momentum in its determination. For Parts (b) and (c), the full angular momentum (magnitude and direction) is not changing, which means that the instantaneous and average values are identical! 3. This problem invokes the definition of the angular momentum, L Q = r Q p, of a particle with linear momentum, p, about some reference point Q. The important step in analyzing each part is to perform the cross-product properly. In all the cases in this problem, the angular momentum (if nonzero) will point into the page,, since all points of reference are on the half-plane to the left (or below) the line defined by the velocity of the particle. The interesting aspect of this problem is that even though the angular momentums have different values depending on the references chosen, the change in those angular momentums will be zero so long as the particle maintains its velocity (speed and direction)! (a) The answer is: mvb sin (φ),. (b) The answer is: mva cos (φ),. (c) The answer is: mv [a cos (φ) + b sin (φ)],. (d) If one were to write the equation of the line determined by the particle s velocity at the point indicated, then the x-intercept of this line would be the point at which the angular momentum would be zero. (e) If one were to write the equation of the line determined by the particle s velocity at the point indicated, then the y-intercept of this line would be the point at which the angular momentum would be zero. 4. For Part (a), the bead is attached to the string and moving in a circle, while in Part (b), the string has broken, making the bead move at constant velocity along the line defined by the tangent of the circle at the top. (a) Even though the angular momentum of an extended body is necessary to invoke for just such an object, recall that the derivation of the result arose from applying the definition of the angular momentum for a point particle. So, if it is clear what the angular velocity of the system is, then the extended-body version can just as easily be applied for a point particle. This is a useful thing to know, especially when dealing with problems that involve both extended bodies and point-like objects. i. The answer is: mvr,. ii. The identical answer from Part (ai) follows noting that this bead s MI about the center of the circle is mr 2. (b) After the string breaks, the problem practically boils down to the same one as Problem 2. Note that for Part (bii), only the component of r that is perpendicular to the line defined by the bead s velocity will matter for its angular momentum about the center of the circle. 5. Note that this is a problem for which the MI of the system (i.e., of the combined system of the beads and the rod) is changing as the beads slide out. (a) Recall that in order for an object to be able to change direction, a centripetal force must be supplied. In this case, the centripetal force for each bead must point radially inward (i.e., toward the rotation axis); however, there is no force that acts radially inward for each bead. As a result, each bead will tend to move purely tangential to its trajectory at some point in time, which amounts to each bead moving radially outward. 7

8 (b) The contact force from the rod that acts on each bead will aim to speed each bead up, since this force acts completely parallel to each bead s (tangential) direction of movement. As a result, the bead s will speed up as they move farther away from the rotation axis. In turn, by Newton s 3rd Law, the contact force exerted on the rod due to each bead acts in opposition to the instantaneous velocity of any point on the rod. This will cause a torque on the rod that points opposite to the angular velocity of the rod. Since the torques due to the beads (both of which act in the same direction) will amount to the net torque on the rod, then the rod will have an angular acceleration that opposes its angular velocity. Thus, the rod will further slow down as the beads slide farther away from the axis of rotation. (c) If we consider the combined system of the beads and the rod, then the torques to which we alluded in Part (b) will be internal to the system. Since there are no additional torques acting on the system vertically (relative to the figure), then the total angular momentum of the system must be conserved. Thus, even though the angular speeds of the beads increase while the angular speed of the rod decreases, the moment of inertia compensates in the process to keep the total angular momentum the same. Since the beads were practically at the center of the rod, then they initially had no angular momentum relative to the rotation axis. Thus, the initial angular momentum was only due to the rod. As the beads slide out, at each instant of time the angular speeds of the rod and the beads is the same, since they rotate rigidly. The answer is: ω = ( M M + 6m ) ω 0. (d) Referring to the discussion in Part (b), note that once the beads slide off, they no longer interact with the rod via a contact force. Thus, there is no longer a net torque on the rod. (e) The initial kinetic energy is the rotational kinetic energy of the rod (since the beads are practically on the rotation axis and, thus, technically not rotating). The final kinetic energy is the final rotational kinetic energy of the rod added to the rotational kinetic energies of the beads having reached their respective edges. Of course, the angular speeds of the rod and beads are the same at each instant of time. 6. As mentioned in the problem, it is important to emphasize that the spring exerts not only a force on the small block to propel it tangentially, but also exerts the exact same force (since the spring is massless) in the opposite direction to the wall on which it is anchored. This, in turn, exerts a force on the turntable. (a) Because the energy initially resides in a spring for which we may easily account via the potential energy of the spring then since no nonconservative forces do work in this problem, the mechanical energy of the block-turntable system will be conserved in this explosion. (b) Since the spring places an internal force on the block-turntable system, then the torques resulting from these forces on each respective constituent of the system are internal. Since there are no external torques on the block-turntable system, then the angular momentum of this system will be conserved. (c) The answers are: k v = d m ( 2 m r 2 M R + 1 ) 2 ω = 2 d. k R M ( ) 2 + M R 2 m r 2 (d) The essence of this part stems from the block and the turntable perfectly slipping relative to one another. The condition described in this part would only apply if the block and the turntable were rotating as a unit. Since they are moving separately after the explosion, and, moreover, not interacting with one another thereafter, then there will not be such a condition of constraint linking the motions together in this manner. 7. This problem will be analyzed in some depth in class. The text in the problem is fairly self-explanatory, and should provide ample hints and ideas to move forward. Had this analysis been implemented in the ballisticpendulum lab for the predictions, the outcomes for the experiment would have been much more in-line with the theory. 8