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
To determine the rotational energies of the three rods, we can use the equation you mentioned:
E = (1/2) I ω^2
Where E represents the rotational energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid cylinder can be calculated using the formula:
I = (1/2) m r^2
Where m is the mass and r is the radius of the cylinder.
Since all three rods have the same mass, we only need to compare the radii to determine the order of their rotational energies.
Rod R1 has a diameter of 1 inch, which means its radius is 0.5 inches or 0.5/12 = 0.0417 feet.
Rod R2 has a diameter of 2 inches, which means its radius is 1 inch or 1/12 = 0.0833 feet.
Rod R3 has a diameter of 3 inches, which means its radius is 1.5 inches or 1.5/12 = 0.125 feet.
Now, let's calculate the rotational energy for each rod:
For Rod R1:
E1 = (1/2) (0.5 m r^2) ω^2
For Rod R2:
E2 = (1/2) (0.5 m r^2) ω^2
For Rod R3:
E3 = (1/2) (0.5 m r^2) ω^2
Notice that the mass and the angular velocity are the same for all the rods, so we can remove them from the equation.
Comparing the radii, we find that the order from smallest to largest is R1 < R2 < R3.
Since the moment of inertia (I) is directly proportional to the square of the radius, the order of the rotational energies will be the same as the order of the moment of inertia:
E1 < E2 < E3
Therefore, the correct answer is:
a. E1 < E2 < E3