Three solid cylinder rods roll down a hill with the same linear speed. Rod R1 has a 1inch diamter, Rod R2 has a 2inch diameter, and Rod R3 has a 3inch diameter. They all have the same masses. Rank in order the rotational energies of R!, R2, and R3.
a. E1 > E2 > E3
b. E3 > E2 > E1
c. E1 = E2 = E3
I say C based on the eaqution k=1/2(I)(omega)^2
You're on the right track! The equation you mentioned, ke = 1/2 * I * ω^2, is indeed the equation for rotational kinetic energy.
To determine the rotational energies of the three rods, we need to consider their moments of inertia. The moment of inertia, represented by I, depends on the distribution of mass around the axis of rotation. In the case of a solid cylinder, the moment of inertia is given by I = (1/2) * m * r^2, where m is the mass and r is the radius of the cylinder.
Now, let's analyze each of the rods:
Rod R1 has a diameter of 1 inch, which means its radius (r) is 1/2 inch. So the moment of inertia for R1 would be I1 = (1/2) * m * (1/2)^2 = (1/8) * m.
Rod R2 has a diameter of 2 inches, so its radius is 1 inch. The moment of inertia for R2 would be I2 = (1/2) * m * 1^2 = (1/2) * m.
Rod R3 has a diameter of 3 inches, so its radius is 1.5 inches. The moment of inertia for R3 would be I3 = (1/2) * m * (1.5)^2 = (9/8) * m.
Since all three rods have the same mass, we can directly compare their rotational energies by considering the moments of inertia.
Comparing the moments of inertia, we can see that I1 < I2 < I3. Since rotational kinetic energy is directly proportional to the moment of inertia, we can conclude that the rotational energy rankings would be:
E1 < E2 < E3.
Therefore, the correct answer is: b. E3 > E2 > E1.
So, the rods with greater rotational energy are R3, followed by R2, and then R1.