12.3 The cross-section geometry of a machine part is shown in Figure 12.3, with all dimensions given in mm. Given that the centroid is located at (7.45, 65.6) mm from the origin, determine the moment of inertia of the cross section about a horizontal axis drawn through the centroid r-20 C 60 10 20 40 Figure 12.3
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- D1 Sign D2 A 2.40kg pole is attached to a wall with a hinge as shown. A 5.50kg sign is suspended at the other end of a pole, where D2 = 3.50m. A cable attached to the wall forms an angle a = 14.0° with the wall, and is attached to the pole a distance D = 1.60m from the hinge. Assume the origin is at the hinge. What is the torque on the pole due to the weight of the sign? 183 NmA 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 25 degrees to the rod and causes a cw torque. What is the net torque about the pivot? Choose – 6.3 Nm –1.7 Nm 0.6 Nm – 0.6 NmWhen a diver gets into a tuck position by pulling in her arms and legs, she increases her angular speed. Before she goes into the tuck position, her angular velocity is 5.5 rad/s and she has a moment of inertia of 1.7 kg · m². Once she gets into the tuck position, her angular speed is 15.1 rad/s. Determine her moment of inertia when she is in the tuck position. Assume the net torque on her is zero. kg · m2
- 2. Let F = 4 i -6 j + 2 k and Let r = 3 i -3 j - 5 k. Find the torque in both Cartesian and polar coordinates. (Find Torque by cross product, find the magnitude of the torque, find the SOLID angle theta and phi since the vector will be in 3 dimensions.) ____________________________________________________________________________________________________________________________________________________________________________ If you could clearly circle the answer I need, that would be great! Thanks!The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center ( Ibar=ml² where l is the length of the bar). When the skater's hands and arms are brought in and wrapped around their body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. The hands and arms have a combined mass 10 kg. When outstretched, they span 2.1 m. When wrapped, they form a cylinder of radius 21 cm. The moment of inertia about the rotation axis of the remainder of the body is constant and equal to 0.6 kg-m2. If the original angular speed is 0.5 rev/s, what is the final angular speed? 00f= rev/s Determine how much the figure skater's rotational KE increased. [NOTE: We express energy in J. To get J, you must convert your angular velocities to the appropriate mks. units which are rad/s.] ΔΚΕ- =I’ve asked this question already but the explanation didn’t make any sense Could you explain it in an easier way? ☹️
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