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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- 4) Use the parallel axis theorem to determine the second moment of mass about the z-axis through the origin for a thin rectangular plate with base b, and height h. Ул ANS. Z 3 [b² + h²] (kg-m^2) h z' bTwo uniform, solid spheres (one has a mass M1= 0.3 kg and a radius R1= 1.4 m and the other has a mass M2 = 2M, kg and a radius R2= 2R;) are connected by a thin, massless rod of length L= 4R1. Find the moment of inertia (in kg.m2) about the axis through the center of sphere 1. (Igphere = MR?). Axis of rotation R, R2 a. 59.74 b. None of the choices O c. 117.13 d. 97.37 O e. 45.04A rod of mass M and length L is nonuniform such that its mass per unit length varies with x according to the expression x= a x*, where a is a constant. Find the x coordinate of the center of mass as a fraction of L. on -L▬ O a. 5L/6 O b. 3L/4 O c. 6L/7 O d. 4L/5
- 5. Consider the flat object bounded by the curves y = 4x and y = Vx +b where (b=2). The area density of the M object is o = yx where the relationship between the total mass M and y is = 0.0309. Find the y-coordinate of the center of mass of this object (do not find the x coordinate of the center of mass). 1 y em Ym = vdm MThe centre of mass of a non-uniform rod of length 4 m whose mass per unit length is K = 3x kg/m, where x is in m. The distance of centre of mass from its one end which is at origin will beThe figure below shows the positions of the Centers of mass of the three disks on a rotating shaft. m. =6 kg, M2=4 kg, m3=5 kg, r-=100 mm, r2=200 ver3 = 300. B to dynamically balance the shaft it is desirable to place balancing masses of 3 kg each in their planes. Balancing find the positions of the center of mass (004) of mass A in the unit of cm and degrees. The D value in the figure is 200 mm. mlg my m1 37 53 200 mm 150mm 200 mm ma Please choose one n
- Choose the following systems that are equivalent to the one shown. Choose all that apply. (Remember conditions for equivalency require the sum of forces to be equal and the sum of moments about on arbitrary point to be equal) 50 N DA OB. OC OD. 1m 1 m 1 m 50 N 1 m 50 N 1 m 100 N 1 m 50 N 1m 50 N 1 m 50 N-m 1 m 100 N-m 1 mDetermine the components of the resultant along i-j axis.Z 50 N 3 m y 300 Nm 70 N 2 m 3 m 40 N 2 m 30 N Questions: (a) Reduce the system of forces and moments acting on the structure to a single resultant force (Fro) and a couple (Mro) acting at the origin of coordinates. (b) Determine the "parallel component” (M₁₁) of the couple (Mro), i.e., the compo- nent lying parallel to (F). State (M₁₁) in vector form. (c) Determine the corresponding "perpendicular component” (M₁). State (M₁) in vector form. (d) Reduce the force and couple at the origin to a "wrench resultant" and determine a precise and complete description of the line of action of the "wrench resultant."
- eRana H st Class ineering Mechanics 20-2021 Highway and Transport. Engineering Department College of Engineering Mustansiriyah University Lec.Rana Hashim Home Work H.W 4 Determine the equivalent resultant force of the system and specify its location measured from point A. 24N S00 N 500 N SON 1mQuestion 4 The moment of the force F=960-lb about the point O in Cartesian vector form is. Not yet answered Marked out of 15.00 P Flag question 1 ft 4 ft 2ft 4 ft. Select one: O a. Mo = 1600 – 3200k оь. Мо —-2005 + 400k %3D O c. Mo = -200j 400k O d. Mo = 200j + 400kFind the mass and center of mass of the lamina bounded by the graphs of the equations for the given density. x² + y² = 36 x 20 y zo p=k(x² + y²) m = (x, y) =