A hollow tube (I = MR2) rolls without slipping along the surface. The ratio of its translational kinetic energy to its rotational kinetic energy is
a. 1
b. 2
c. 3
d. 1/2
e. 1/3
I say A becasue both add up to get the total kinetic energy.
Translational KE is (1/2) M V^2
Rotational KE is (1/2)I w^2
= (1/2) M R^2*(V/R)^2 = (1/2) M V^2
Your answer is correct but your reasoning is faulty. The ratio could be anything and they would still add up to give total KE
The Answer is "I don't know. Ask I".
To determine the ratio of translational kinetic energy to rotational kinetic energy for a hollow tube rolling without slipping, we need to consider the formulas for these types of kinetic energy.
Translational kinetic energy (K_trans) is given by the formula:
K_trans = (1/2) * m * v^2
Rotational kinetic energy (K_rot) is given by the formula:
K_rot = (1/2) * I * ω^2
Where:
- m is the mass of the hollow tube
- v is the velocity of the hollow tube
- I is the rotational inertia of the hollow tube (given as I = MR^2, where M is the mass and R is the radius of the hollow tube)
- ω (omega) is the angular velocity of the hollow tube
Since the hollow tube rolls without slipping, there is a relationship between the linear velocity and angular velocity:
v = ω * R
Substituting this relationship into the formulas for translational and rotational kinetic energies, we get:
K_trans = (1/2) * m * (ω * R)^2
K_rot = (1/2) * I * ω^2
We can simplify these equations:
K_trans = (1/2) * m * ω^2 * R^2
K_rot = (1/2) * (M * R^2) * ω^2
Now, to find the ratio K_trans / K_rot, we can plug in the values into these equations:
K_trans / K_rot = [(1/2) * m * ω^2 * R^2] / [(1/2) * (M * R^2) * ω^2]
Many terms cancel out:
K_trans / K_rot = (m * ω^2 * R^2) / (M * R^2 * ω^2)
The ω^2 terms also cancel out:
K_trans / K_rot = m / M
Therefore, the ratio of translational kinetic energy to rotational kinetic energy for a hollow tube rolling without slipping is equal to the mass of the tube divided by the mass of the tube. This simplifies to 1:1, or in other words, the ratio is 1.
Hence, the correct answer is a. 1. Both translational and rotational kinetic energy contribute equally to the total kinetic energy of the rolling hollow tube.
(Note: In the provided options, option a is the correct answer.)