A 500.0 g bird is flying horizontally at 2.25 m/s, not paying much at- tention, when it suddenly flies into a stationary vertical bar, hitting it two-thirds of the way up from the bottom. The bar is uniform and very thin, is 0.750 m long, has a mass of 1.50 kg, and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterwards (but soon recovers and flies away). a) Using conservation of total angular momentum, determine the angular velocity of the bar just after it is hit by the bird. b) What numerical fraction of the bird's initial kinetic energy is converted to other forms of energy during the collision?
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- Consider a stick of mass, M = 14.0 kg and length, L = 1.60 m hanging vertically initially. A bullet of mass, m = 0.040 kg and speed, v = 70 m/s strikes and embeds itself at the bottom end of the stick. Magnitude of acceleration due to gravity is g = 10 m/s². Note: The moment of inertia of a rod about it's center of mass 1 = ML². You may need to use the parallel axis theorem. Closeup Part B: @max M What is the maximum angle, Omax achieved by the stick (with the bullet embedded in it)? Incorrect 2.17⁰ Correct: 6.97⁰ Incorrect 83° Incorrect 9.87⁰ Incorrect None of the aboveAs part of a carnival game, a my = 0.653 kg ball is thrown at a stack of 23.8 cm tall, mo = = 0.363 kg objects and hits with a perfectly horizontal velocity of Ubi = 10.8 m/s. Suppose that the ball strikes the topmost object. Immediately after the collision, the ball has a horizontal velocity of bf = 3.10 m/s in the same direction, the topmost object has an angular velocity of @= 1.63 rad/s about its center of mass, and all the remaining objects are undisturbed. Assume that the ball is not rotating and that the effect of the torque due to gravity during the collision is negligible. If the object's center of mass is located r = 16.7 cm below the point where the ball hits, what is the moment of inertia I, of the object about its center of mass? What is the center of mass velocity Uo.cm of the tall object immediately after it is struck? (1) 0 Io = b.J Vo.cm = kg-m² m/sA 45 kg figure skater is spinning on the toes of her skates at 1.5 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 70 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head. In this latter orientation, she can be modeled as a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.What is her new rotation frequency, in revolutions per second?Express your answer in revolutions per second.
- The moment of inertia I of a cheap door of mass M = 4.00 kg (about an axis going through the hinges at the door frame) is I= (1/3) M · R², where R= 0.960 m is the width of the door. The door is initially open and at rest. The door suddenly is struck by a huge and heavy dart of mass m = 0.300 kg traveling perpendicular to the plane of the door at a speed vi = 20.0 m/s. The dart perforates the wooden door getting permanently stuck at the point of impact, which happened to be right next to the free vertical edge of the door (close to the handle). Because the dimensions of the dart are so small (even though they are exaggerated in the picture for clarity) compared to its distance R to the rotational axis (the distance from the hinge to the free vertical edge next to which the dart strikes the door, which is the width of the door R) we can treat the dart as a point mass. I remind you that we learned in class that the moment of inertia of a point mass is I, = m-R², where R is the…A 0.005 00-kg bullet traveling horizontally with a speed of 1.00 3 103 m/s strikes an 18.0-kg door, embedding itself 10.0 cm from the side opposite the hinges as shown in Figure P11.30. The 1.00-m wide door is free to swing on its frictionless hinges. (d) At what angular speed does the door swing open immediately after the collision just after the bullet embeds itself in the door? (e) Calculate the initial momentum of the bullet–door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.A weather vane initially at rest has a moment of inertia of 0.104 kg · m2 about its axis of rotation. A 68.0 g piece of clay is thrown at the vane and sticks to it at a point 14.0 cm from the axis. The initial velocity of the clay is 27.5 m/s, directed perpendicular to the vane. Find the angular velocity of the weather vane just after it is struck.
- A solid sphere of mass, M = 5.0 kg, and radius, R = 0.100 m is placed on two blocks so that it’s centre of mass lies at the origin as shown on the right. A bullet of mass, m = 0.100 kg, and with an x coordinate of b = 0.04 m strikes the sphere from below with a vertical velocity, v = 65 m/s and embeds in the sphere coming to rest at the location (-b,0) in the diagram below. Since there is no momentum in the horizontal direction, after the bullet hits the sphere , it will rise up to some maximum height and then fall back to its initial position on the two blocks. While the sphere is in the air it will rotate. What is the magnitude of the angle (in degrees) through which the sphere will rotate before it makes contact with the two blocks again?A rod of mass M = 3.25 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 65 g, moving with speed v = 5.25 m/s, strikes the rod at angle θ = 51° from the normal at a distance D = 2/3 L, where L = 1.3 m, from the point of rotation and sticks to the rod after the collision. 1. What is the angular speed ωf of the system immediately after the collision, in radians per second?Two 3.40 kg balls are attached to the ends of a thin rod of length 44.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (the figure), a 73.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 2.97 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle (deg) will the system rotate before it momentarily stops? Putty wad Rotation axis (a) Number i Units (b) Number i Units Units i (c) Number
- Two 3.50 kg balls are attached to the ends of a thin rod of length 36.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (the figure), a 42.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.25 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle (deg) will the system rotate before it momentarily stops? Putty wad Rotation axisA 45 kg figure skater is spinning on the toes of her skates at 1.3 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 71 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head. In this latter orientation, she can be modeled as a 45 kg, 20-cm-diameter, 200-cm-tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer in revolutions per second. ► View Available Hint(s) W₂ Submit Provide Feedback ΑΣΦ ? rev/sA rod of mass M = 3.45 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 55 g, moving with speed v = 4.3 m/s, strikes the rod at angle θ = 63° from the normal at a distance D = 2/3 L, where L = 1.05 m, from the point of rotation and sticks to the rod after the collision. Part (a) What is the initial angular momentum of the ball, in kilogram meters squared per second, right before the collision relative to the pivot point of the rod? Part (b) What is the total moment of inertia If with respect to the hinge, of the rod-ball-system after the collision, in terms of the variables from the problem statement? Part (c) What is the angular speed ωf of the system immediately after the collision, in radians per second?