A ballet dancer spins about vertical axis at 90 r.p.m. with arms outstretched. With the arms folded, the momentum of inertia about the same axis change by 25%. Calculate the new frequency of revolution,
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![A ballet dancer spins about vertical axis
at 90 r.p.m. with arms outstretched. With the
arms folded, the momentum of inertia about
the same axis change by 25%. Calculate the new
frequency of revolution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9eceee6e-273e-4b45-be95-fb11d98e847a%2F657697e2-5fa4-4903-bd8d-33c8eff9ab92%2Fnz9gwp_processed.jpeg&w=3840&q=75)
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- A diver (m = 60 kg) jumps from a diving board. At takeoff, his angular momentum about the transverse axis is 30 kg⋅m2/s. His radius of gyration about the transverse axis is 0.5 m at this instant. During the dive, he tucks and reduces his radius of gyrations about the transverse axis to 0.2 m. At takeoff, what is the diver’s moment of inertia about his transverse axis?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…The velocity of the 8-kg cylinder is 0.3 m∕s at a certaininstant. The mass of the grooved drum is 12 kg, its centroidal radiusof gyration is k = 210 mm, and the radius of its groove isri = 200 mm. The frictional moment at O is a constant3 N∙m. Find the frictional force if final speed is given to be 2.5.
- 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?A uniform solid disk of mass m = 3.03 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 6.07 rad/s. (a) Calculate the magnitude of the angular momentum of the disk when the axis of rotation passes through its center of mass. kg · m2/s (b) What is the magnitude of the angular momentum when the axis of rotation passes through a point midway between the center and the rim? kg · m2/s Need Help? Read ItIn the figure, a 7.90 g bullet is fired into a 0.430 kg block attached to the end of a 0.140 m nonuniform rod of mass 0.578 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about A is 0.0761 kg·m2. Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) If the angular speed of the system about A just after impact is 3.64 rad/s, what is the bullet's speed just before impact?
- The figure shows a small piece of clay colliding into a disk. The disk is initially at rest, but can rotate about a pivot fixed at its center. The collision is completely inelastic.1) The disk has mass M = 4.73 kilograms and radius R = 0.755 meters. The clay has mass m = 279 grams and is moving horizontally at vi = 3.66 m/s just before colliding with and sticking to the disk. The clay strikes the edge of the disk at a location of b = 0.505 meters offset from the center of the disk. Note that the size of the clay is negligible compared to the radius of the disk. 2) Calculate the angular velocity (rad/s) of the disk just after the collision.The figure shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 1.9 s. If R = 0.7 m and m = 4.4 kg, calculate the angular momentum about that axis. -Rotation axis | 2R Number i Units >A block of mass m = constant speed v = 5.00 m/s, slides along a horizontal, frictionless surface and collides with and sticks to the end of an initially vertical, stationary thin rod, of mass M= 5.00 kg, and length L = 0.500 m. See the figure below. The 0.65 kg, moving at other end of the rod is attached to a horizontal frictionless pivot which allows the mass-rod system to rotate to an angle 0 with the vertical direction. What is the maximum angle Omax the rod makes with the vertical? (The moment of inertia of a uniform thin rod about an axis through one end is given by ML2/3) L
- A uniform solid disk of mass m = 2.99 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 5.99 rad/s. (a) Calculate the magnitude of the angular momentum of the disk when the axis of rotation passes through its center of mass. |kg · m2/s (b) What is the magnitude of the angular momentum when the axis of rotation passes through a point midway between the center and the rim? |kg · m2/sA 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 = 5.63 m/s, strikes the rod at angle θ = 55° from the normal at a distance D = 2/3 L, where L = 1.25 m, from the point of rotation and sticks to the rod after the collision. 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? 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? c) What is the angular speed ωf of the system immediately after the collision, in radians per second?An ice skater is spinning at 5.4 rev/s and has a moment of inertia of 0.44 kg ⋅ m2. Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.4 rev/s. Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to 0.75 rev/s. Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.25 rev/s. What is the magnitude of the average torque that was exerted, in N ⋅ m, if this takes 17 s?