Rotational inertia and torque of four masses at square vertices

[Clutch306]

Four particles, each of mass 0.200kg, are placed at the vertices of a square with sides of length 0.500m. The particles are connected by rods of negligible masses. This rigid body can rotate in a vertical plane about a horizontal axis that passes through one of the particles (A for example, as shown in figure attached.)
(a) What is the rotational inertia I of the body about axis A?
(b) What is the net torque Tnet about axis A (due to the gravitational forces), when the rod AB is horizontal?
(c) What is the angular acceleration a of the body at the instant rod AB is horizontal?

 

[enigma]

Others have said it in your other threads, but you must show what you've tried, Clutch.

https://www.physicsforums.com/showthread.php?t=28

 

[M98Ranger]

(a) The rotational inertia I of the body about axis A can be calculated using the formula I = Σmr², where m is the mass of each particle and r is the distance from the axis of rotation. In this case, the particles have the same mass (0.200kg) and the distance from the axis of rotation is 0.250m (half of the side length of the square). Therefore, the rotational inertia I = 4(0.200kg)(0.250m)² = 0.100kgm².

(b) The net torque Tnet about axis A can be calculated using the formula Tnet = ΣrFsinθ, where r is the distance from the axis of rotation, F is the force acting on the particle, and θ is the angle between the force and the line connecting the particle to the axis of rotation. In this case, the only force acting on the particles is the gravitational force, which is directed towards the center of the earth. Since the particles are arranged in a square, the angle between the force and the line connecting the particle to the axis of rotation is 45 degrees. The distance from the axis of rotation is 0.250m (half of the side length of the square). Therefore, Tnet = 4(0.250m)(0.200kg)(9.8m/s²)sin45° = 3.92Nm.

(c) The angular acceleration a of the body can be calculated using the formula a = Tnet/I, where Tnet is the net torque and I is the rotational inertia. Substituting the values calculated in parts (a) and (b), we get a = (3.92Nm)/(0.100kgm²) = 39.2 rad/s². This means that the body will experience an angular acceleration of 39.2 rad/s² in the direction of the net torque when the rod AB is horizontal.