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 for rotational kinetic energy, which is given as:
K = (1/2) * I * ω^2
Where:
K is the rotational kinetic energy
I is the moment of inertia
ω is the angular velocity
In this case, we have three solid cylinder rods rolling down the hill with the same linear speed. 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 of the rod
r is the radius of the rod
Now, let's analyze each rod individually and compare their rotational energies:
Rod R1: Diameter = 1 inch
Radius = 1/2 inch = 0.5 inch
Moment of inertia (I1) = (1/2) * m * (0.5)^2
Rod R2: Diameter = 2 inches
Radius = 1 inch
Moment of inertia (I2) = (1/2) * m * (1)^2
Rod R3: Diameter = 3 inches
Radius = 1.5 inches
Moment of inertia (I3) = (1/2) * m * (1.5)^2
Since all three rods have the same masses, we can assume that the mass term cancels out. Now, we can compare the moments of inertia of each rod:
I1 < I2 < I3
For rotational energy, we need to consider angular velocity. However, since all rods are rolling down the hill with the same linear speed, they will have the same angular velocity. Therefore, the term ω^2 will be the same for all rods.
Comparing the rotational kinetic energy for each rod using the equation K = (1/2) * I * ω^2, we can conclude that:
E1 < E2 < E3
Therefore, the correct ranking in order of the rotational energies of R1, R2, and R3 is:
b. E3 > E2 > E1