Rotational Energy of 3 Point Masses along a Rigid Rod

[shenwei1988]

Three point masses lie along a rigid, massless rod of length L = 5.61 m :

- Two particles, both of mass m = 2.1 kg, lie on opposite ends of the rod.
- Mass M = 7.29 kg is in the center of the rod.

Assume the rod lies along the x-axis, and rotates about the y-axis. about a point 1.28 m from one end at constant angular speed ω = 5.38 rad/s.

Find the kinetic energy of this system



could someone explain the ( about a point 1.28m from one end) . dose the distance 1.28 contribute any Kinetic energy to the system?
thank you so much

 

[Doc Al]

Use the distance from the axis of rotation and the angular speed to find the linear speed of each point mass. Use that to find their kinetic energy.

 

[Moetasim]

for your help.

I can explain the rotational energy of this system. The rotational energy of a system is the energy associated with its rotational motion. In this case, the system consists of three point masses lying on a rigid rod and rotating about the y-axis.

To calculate the kinetic energy of this system, we need to consider the rotational energy of each point mass. The rotational energy of a point mass can be calculated using the formula E = 1/2 Iω^2, where I is the moment of inertia and ω is the angular velocity.

In this system, we have two point masses of 2.1 kg each at the ends of the rod and one mass of 7.29 kg at the center of the rod. The moment of inertia for a point mass rotating about an axis perpendicular to its motion is given by I = mr^2, where m is the mass and r is the distance from the axis of rotation.

In this case, the distance from the axis of rotation to the center of mass M (7.29 kg) is L/2 = 5.61/2 = 2.805 m. So, the moment of inertia for M is I = (7.29)(2.805)^2 = 58.52 kg.m^2.

For the two point masses at the ends of the rod, the moment of inertia is calculated using the parallel axis theorem, which states that the moment of inertia of a point mass about an axis parallel to the original axis is equal to the moment of inertia about the original axis plus the product of the mass and the square of the distance between the two axes.

In this case, the distance between the two axes is 2.805 m, so the moment of inertia for each point mass is I = (2.1)(2.805)^2 + (2.1)(1.28)^2 = 13.12 kg.m^2.

Now, we can calculate the kinetic energy of the system by adding the rotational energy of each point mass. The total rotational energy is given by E = 1/2 Iω^2, where I is the total moment of inertia and ω is the angular velocity.

Substituting the values, we get E = 1/2 (58.52 + 13.12 + 13.12)(5.38)^2 = 866.9 J.

To answer your question about the