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- Find IOG, The mass moment of inertia for the center of mass about the z′ axis is Iz′z′G=494.6 kg⋅m2Consider the objects labeled A, B, C, and D shown in the figure. A 米」 B C D Each object is composed of identical thin sticks of uniformly distributed mass 3.43 kg and length 0.303 m. What are the moments of inertia IA, IB, Ic, and Ip of the objects about a rotation axis perpendicular to the screen and passing through the black dot displayed on each object? IA = kg-m2 IB = %3D kg-m² Ic = kg-m? ID = kg-m?082 α1 A x Part C As shown, a server at a restaurant is carrying a tray that has a glass of water resting on it. (Figure 3) The glass and tray exert a force with magnitude F₁ = 11.0 N on the server's hand. If l₁ = 0.420 m, l₂ = 0.310 m, l3 = 0.0400 m, a₁ = 145°, and a2 = 130°, what is the moment of force about the shoulder joint at point A? Assume that moments acting in the counterclockwise direction about point A are positive whereas moments acting in the clockwise direction are negative. Express your answer numerically in newton-meters to three significant figures. ► View Available Hint(s) ΜΑ = —| ΑΣΦ ↓↑ vec ? N.m
- An ice skater has a moment of inertia of 4.5 kgm2 when her arms are outstretched. With her arms outstretched she is spinning 3.8 radians per second. If she pulls her arms in and decreases her moment of inertia to 2.4 kgm2, how fast is she spinning? A: 7.1 radians/second B: 2.8 radians/second C: 17.1 radians/second D: 6.2 radians/second E: 2.0 radians/secondItem 8 8 of 26 II Review | Constants Part A The parallel axis theorem relates Im, the moment of inertia of an object about an axis passing through its center of mass, to Ip, the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is I, = Iem + Md², where d is the perpendicular distance from the center of mass to the axis that passes through point p, and M is the mass of the object. Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Iem = mL. Find Iend, the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express Iend in terms of m and L. Use fractions rather than decimal numbers in your answer. • View Available Hint(s) {(m) (L²) Iend = Submit Previous Answers v Correct Part B Now consider a cube of mass m with edges of length a.…1 S X₁ X₂ E I TLL s F A person is pulling a rope and this results in a force ♬ that acts on their hand at point H, at an angle of 63 deg above the horizontal. Determine: The magnitude and direction of the moment about the shoulder (point S), in terms of F a) b) The magnitude and direction of the moment about the elbow (point E), in terms of F H V E E
- Q2.A physical pendulum consists of a ring of radius 0.5 m and mass 2 kg. The ring is pivoted at a point on its perimeter. The ring is pulled out such that its centre of mass makes a small initial angle ?0 = 0.2 rad from the vertical and released from rest. (A) Calculate the moment of inertia of the ring about an axis passing through the pivot point O (B) Work out the equation of motion for the vibration of the ring4- A wooden block is at rest on a frictionless horizontal plane, connected to a rod of negligible mass and length l. The rod is attached to the horizontal plane at its other end and can rotate around its attachment point. A lead tree of mass m = 0.5 M and velocity v, moving parallel to the horizontal plane and perpendicular to the solid bar, strikes the block and becomes embedded in the block. Taking into account the angular momentum of the lead+block system, find how much of the kinetic energy lost in the collision corresponds to a fraction of the kinetic energy before the collision.2- Tim has 40 kg mass and is going to play see-saw with his brother, Jack, who has only 15 kg mass. The sea-saw is 3 meters long and assume the moving part is a solid rod of mass 60 kg. pivoted in the middle. Tim has Jack sit on one side and the sea saw has an angle of 20 degrees with horizontal now. Then he suddenly sits on the other side and you know what happens! (Moment of inertia of point-like particle rotating about a radius r is mr² and that of a uniform rod of length L about its COM is ML²) 12 A) Find the total torque about the pivot due to Tim and Jack at the moment Tim jumps on. B) Calculate the total moment of inertia of the system about the pivot. C) Find Jack's angular acceleration once Time sits on the other side.