A uniform steel rod of length 0.8 m and mass 1.3 kg has two point masses on each end. The point mass on the left end is of 1.7 kg and the one on the right end is 3.5 kg.  Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through the center of mass of this system.

Take an equilateral triangular sheet of side, and remove the "middle" triangle (1/4 of the area). Then remove the "middle" triangle from each of the remaining three triangles, and so on, forever. Let the final object have mass. The moment of inertia of final object about axis passing through 'O' and perpendicular to plane of object is ml²/x. Then the value of x is

A 117 kg horizontal platform is a uniform disk of radius 1.67 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.11 m from the center, and a 29.5 kg dog sits on the platform near the person 1.39 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis. moment of inertia: kg · m?

Two kids, each with a mass of 25.0 kg, are standing at the edge of a merry-go-round. Treat both kids as point masses. The merry-go-round is a uniform disk with a mass of 97.5 kg and a radius of 1.25 meters, which is free to rotate about its center of mass without friction. The moment of inertia of a solid disk rotating about its center of mass is given by I = MR2 /2 , and the moment of inertia of a point mass is given by I = mr2 . The merry go round is initially rotating at 12.5 revolutions per minute (rpm). a) If both kids walk to the center of the merry-go-round, what will the final angular speed be, in rpm? Hint: use conservation of angular momentum.  b) What is the change in the total kinetic energy of the disk and the two kids during this process? Give your answer in units of Joules (J)

A uniform thin rod of mass ?=3.47 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass ?=0.277 kg, are attached to the ends of the rod. What must the length ? of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ?=0.987 kg·m2 ?

Can you please help me solve this..   Three masses A = 3kg , B = 4kg and C = 5kg are connected by rods of negligible mass to form an equilateral triangle of side 8m as shown. Determine the individual moment of inertia and total moment of inertia of the system about axis XY.   Solve for: (a) Total moment of inertia of the system (b) Moment of inertia of object C (in kg-m^2) (c) Moment of inertia of object B (in kg-m^2) (d) Moment of inertia of object A (in kg-m^2)   •Pls answer with complete and detailed solution (3 DECIMAL PLACES all final answers with correct units) •Sketch an illustration/diagram with labels (this is a must; for a better understanding)   Thank you I will give your answer a LIKE for helping me...

A 139-kg platform-oriented horizontally consists of a uniform disk of radius 1.85 m and can rotate about the vertical axis through its center. A 63.3-kg person stands on the platform at a distance of 1.17 m from the center and a 27.5-kg dog sits on the platform near the person, 1.39 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

As a physics demonstration, a special bowling ball is made so that it can be rotated about its center of mass to get a feel for how "big" a moment of inertia of 1 kg⋅m2 is. The average bowling ball has a weight of 15.4 lbs and a circumference of 26.3 inbut it does not have a moment of inertia equal to 1 kg⋅m2.Since the sporting goods manufacturer has no understanding of how "big" 1 kg⋅m21  is, calculate the diameter ? demo of the demo bowling ball, in inches, that it will need to manufacture. Assume that bowling balls are solid, with a constant density.   D of demo required in inches.

Three masses, each m = 12 kg, are located at the corners of an equilateral triangle. Calculate the moment of inertia of the system about an axis running through one of the point masses perpendicular to the plane of the triangle. The length of one side of the triangle is 0.94 m

The moment of inertia of the lower leg and foot about an axis through the knee joint is 0.23 kgm^2. What is the moment of inertia of the leg and foot about the knee joint if a 0.82 kg shoe is worn on the foot? Assume that the shoe's mass is all concentrated in one point 40 cm from the knee. (please report to two decimal places)