What is the moment of inertia about an axis perpendicular to the paper and through the center of mass? Answer in units of kg m2 .

A uniform thin rod of mass M = 3.37 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.219 kg, are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.975 kg-m? ? L = m M. m m L

A rope with a negligible mass has two blocks that are suspended over a pulley, m, and M, as shown in the image.  The pulley can be considered as a uniform solid cylindrical disk. What is the equation for the net torque on the disc? Assume acceleration of blocks are a=(1/2)(9.8m/s2). What is the moment of inertia for this shape? (using the net torque equation from the above question, substitute the moment of inertia) The angular acceleration is correlated to the linear acceleration as α=r. Make the substitution into the torque equation in order to get the equation in terms of acceleration, α only.

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 ?

A uniform sphere made of modeling clay has radius R and moment of inertia I1 for rotation about a diameter. It is flattened to a disk with the same radius R. In terms of I1, what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?

please solve symbolically before using numbers

A disk of radius R and thickness t has a mass density that increases from the center outward, given by p = po (r/R), where r is the distance from the axis of the disk. What is the moment of inertia about the disk axis in terms of M and R? O MR 2MR O sMP 3MR 5 O 2MR 3 5MR 3

A small pine tree has a mass of 19 kg. Its center of mass is located at 0.87 m from the ground. Its trunk is sawed though at ground level, causing the tree to fall, with the severed trunk acting as the pivot point. At the instant the falling tree makes a 16° angle with the vertical, the angular acceleration of the tree is 2.6 rad/s. What is the moment of inertia of the tree?

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of diameter Da = 0.273 m. The mass of the hoop part is mh = : 0.120 kg and the mass of the disk part is md = 0.050 kg. Imagine making a boomerang that has the same total moment of inertia around its center as the sport disk. The boomerang is to be constructed in the shape of an "X," which can be approximated as two thin, uniform rods joined at their midpoints. If the total mass of the boomerang is to be m₁ = 0.235 kg, what must be the length Lы of the boomerang? | I Lb = = Cross-sectional view m

A 0.8 m Mass Moment of Inertia. For each system, find the mass moment of inertia about an axis going through Point A (perpendicular to the paper). c. Point A is in the upper left corner of the object shown. It has a uniform area density of 120 kg/m². 0.4 m 0.8 m A - 0.4 m 120 kg/m² 0.4 m 0.4 m d. The system is 15kg uniform isosceles triangle with a height of 0.33m and a base of 0.26m. Point A is at the lower left corner of the triangle. m = 15 kg 0.26 m 0.33 m