Consider a model of an amusement ride the carriage weight 10

Consider a model of an amusement ride the carriage weight 100-lbs and moves at a constant speed of 15 ft/sec on a horizontal track. The rigid arms of 12-ft length and negligible weight rotate at a constant angular speed of 1.57 red/sec.

Solution

Here we need to note that the total kinetic energy of the system would be the sum of translation kinetic energy and the rotational kinetic energy.

We know that for a body of mass m travelling with a velocity of v, the kinetic energy is given as 0.5mv²

Also, for the a body with inertia I and rotational speed w, the rotational kinetic energy is given as 0.5Iw²

Now, the net mass of the system here would be 1600 + 1000 = 2600 lbs and the linear speed would be 15 ft/sec

That the net translational kinetic energy = 0.5 x 2600 x 225 = 292500 Units

Also, we will have a rotational kinetic energy given as 0.5 Iw²

That is rotational kinetic energy = 0.5 x 2(800 x 144) x 2.4649 = 283956.48 Units

Therefore the net kinetic energy of the system would be sum of translational and rotational kinetic energy

Hence the total kinetic energy = 576456.48 lbf-ft

NOTE: Here, we have calculated the translational as well as the rotational kinetic energy for the cages. It is important to understand that the arm+cages system is translating as well as rotating, hence we use the linear speed to find the translational kinetic energy and rotational speed for rotational kinetic energy. That is why the arm+cage system has a contribution on the translational kinetic energy.

Missing the above point is one of the common mistakes. You can imagine a rod rotating at its point for case 1 and the throwing that rotating rod with some velocity for case2. Clearly case 2 will have more kinetic energy and for such situations you will have to calculate the translational as well rotational KE so as to find the total.