2) A uniform slender rod has a length, L, cross sectional area, A and a planar mass density distribution of p (kg/m^2). Determine the second moment of mass about an axis passing through the mass center perpendicular to the longitudinal axis. ANS. mL^3/12 N Z N dm X
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- The uniform lamina of density 2.2 kg/m2. for a=2.2 m,b=4.8 m,c=8 m,d=8.9 m 1-The x-co-ordinate of total area centroid is: (a.8.27 - b.13.17 - c.7.05 - d.8.88 - e.6.13) 2- The y-co-ordinate of total area centroid is: (a.1.96 - b.2.68 - c.3.04 - d.1.61 - e.3.57) 3- The total mass is : ( a. 163.00 - b. 326.00 - c. 244.50 - d. 81.50 - e. 40.75 )When using T = ½ I + ½ m vG 2 for kinetic energy of a planer rigid body, what point must the mass moment of inertia I be found about?A spacecraft must be maneuvered through an angle of 6O-deg in 30 s. The spacecraft moment of inertia is 8000 kg-m^2; the thrusters are located at a radius of 2.2 m from the center of mass. What is the minimum thrust level for the thrusters? Answer : 1.90 N. Don't give me wrong answers. correct answer is 1.90....show me how
- Determine and locate the resultant of the system shown in fig (3). If you know that the values of Ax=240 N. Ay=2000 N. F=160 N, N=2400N, W=400 N and M=2N.m2 Kinetic energy of a rigid body can be found by integrating the square of the velocity of differential mass elements over the entire mass of the body. True Falseit is known that the shaft in the shape with a rotating disk on it rotates parallel to the horizontal plane. rotation speeds are wr=955 wp=296 rpm. Find the mass of the rotating disk to remain in a parallel position.gravitational acceleration g=9.81 m 2. mass moment of inertia of disk I=0.090479 kgm 2
- Q4:-A rotating shaft carries four masses 176.6 N, 137.3 N, 157 N and 117 N at radii 5 cm, 6 cm, 7 cm and 6 cm respectively. The second, third and fourth masses revolve in planes 8 cm, 16 cm and 28 cm respectively measured from the plane of the first mass (at the left end). They are angularly located at 60o, 135oand 270orespectively measured clockwise from the first mass (at the positive x-axis). The shaft is dynamically balanced by two masses both located at 5 cm radii and revolving in planes midway between those of 1st and 2nd masses and midway between those of 3rd and 4th masses. Determine the magnitude of the masses and their respective angular positions.The figure below is a system consisting of two masses MA = 1.82 kg and MB = 3.6 kg, whose distance from the pore axis is R₁ = R₂ = 15 cm, while the angular positions are 0₁ = 30° and 0 = 150° . Determine the weight and position of the balancing mass placed on planes C and D if the distance from the axis RC = RD = 15 cm.A shaft with 3 meters span between two bearings carries two masses of 10 kg and 20 kg acting at the extremities of the arms 0.45 m and 0.6 m long respectively. The planes in which these masses rotate are 1.2 m and 2.4 m respectively from the left end bearing supporting the shaft. The angle between the arms is 60°. The speed of rotation of the shaft is 200 r.p.m. If the masses are balanced by two counter-masses rotating with the shaft acting at radii of 0.3 m and placed at 0.3 m from each bearing centers, estimate the magnitude of the two balance masses and their orientation with respect to the X-axis, i.e. mass of 10 kg Note: It must be solved using the graphic method, not the equations method.
- A shaft with 3 meters span between two bearings carries two masses of 10 kg and 20 kg acting at the extremities of the arms 0.45 m and 0.6 m long respectively. The planes in which these masses rotate are 1.2 m and 2.4 m respectively from the left end bearing supporting the shaft. The angle between the arms is 60°. The speed of rotation of the shaft is 200 r.p.m. If the masses are balanced by two counter-masses rotating with the shaft acting at radii of 0.3 m and placed at 0.3 m from each bearing centers, estimate the magnitude of the two balance masses and their orientation with respect to the X-axis, i.e. mass of 10 kg.5) In ice figure skating, a couple execute a “top” (see picture). The centre of mass of the woman (58 kg) is situated 1.3 m from the axis of rotation which is vertical and passes through the centre of mass of the man (85 kg). They are spinning at a constant angular velocity equal to 3.1415 rad/s and the man and woman have moments of inertia, about their own centres of mass, equal to 1.6 and 2.5 kg.m2 respectively. Then the woman grabs the neck of the man. At this point, her moment of inertia decreases to 1.4 kg.m2 and her body centre of gravity is 0.9 m from the axis of rotation. Determine the new angular velocity. Hints: This is a conservation of angular momentum problem, and needs the parallel axis theorem to determine moments of inertia about the axis of rotation. The skaters are moving as one body with one angular velocity, but they each have their own moments of inertia given relative to their own CoMs. For the man, that’s fine…the axis they’re rotating about passes through his…Beam has a mass of 75.0 kg. F1 = 85.0 N F2 = 150.0 N F3 = 125.0 N a) Locate the center of gravity of the beam. Determine b) the net torque on thebeam about point O; c) the required magnitude of F3 to keep the beamstationary