A 2.0 kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.77 m high and 5.0 m long.
(a) When the cylinder reaches the bottom of the ramp, what is its total kinetic energy?
(b) What is its rotational kinetic energy?
(c) What is its translational kinetic energy?
I will be happy to critique your thinking.
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To find the answers to these questions, we need to understand the concepts of potential energy, kinetic energy, and rotational motion.
(a) First, let's calculate the potential energy of the cylinder when it is at the top of the ramp. The formula for potential energy is given by:
PE = mgh
Where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Given:
m = 2.0 kg
g = 9.8 m/s^2
h = 0.77 m
Substituting these values into the formula, we get:
PE = 2.0 kg * 9.8 m/s^2 * 0.77 m = 15.092 J
Next, let's calculate the total kinetic energy of the cylinder when it reaches the bottom of the ramp. The total kinetic energy is the sum of the rotational kinetic energy and the translational kinetic energy.
The formula for rotational kinetic energy is given by:
KE_rot = (1/2) * I * ω^2
Where KE_rot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The formula for translational kinetic energy is given by:
KE_trans = (1/2) * m * v^2
Where KE_trans is the translational kinetic energy, m is the mass, and v is the velocity.
To calculate the rotational kinetic energy, we need to find the moment of inertia. The moment of inertia for a solid cylinder is given by:
I = (1/2) * m * r^2
Where I is the moment of inertia, m is the mass, and r is the radius.
Given:
r = 0.08 m
Substituting these values into the formula, we get:
I = (1/2) * 2.0 kg * (0.08 m)^2 = 0.016 kg * m^2
Now, to find the translational kinetic energy, we need to find the velocity of the cylinder when it reaches the bottom of the ramp. We can use the principle of conservation of energy, which states that the total mechanical energy at any point is equal to the total mechanical energy at any other point. The potential energy at the top of the ramp is converted to kinetic energy at the bottom of the ramp. Therefore, we can equate the potential energy to the sum of the rotational kinetic energy and the translational kinetic energy:
PE = KE_rot + KE_trans
Substituting the values we have:
15.092 J = (1/2) * 0.016 kg * m^2 * ω^2 + (1/2) * 2.0 kg * v^2
Since the cylinder is rolling without slipping, the velocity of the center of mass is related to the angular velocity by:
v = ω * r
Substituting this relationship into the equation, we get:
15.092 J = (1/2) * 0.016 kg * m^2 * ω^2 + (1/2) * 2.0 kg * (ω * r)^2
Simplifying this equation, we get:
15.092 J = (1/2) * 0.016 kg * m^2 * ω^2 + (1/2) * 2.0 kg * ω^2 * (0.08 m)^2
Now, solve for ω:
15.092 J = (1/2) * 0.016 kg * m^2 * ω^2 + (1/2) * 2.0 kg * ω^2 * (0.08 m)^2
15.092 J = 0.5 * 0.016 kg * m^2 * ω^2 + 0.04 kg * m^2 * ω^2
15.092 J = 0.008 kg * m^2 * ω^2 + 0.04 kg * m^2 * ω^2
15.092 J = 0.048 kg * m^2 * ω^2
Rearranging the equation and solving for ω^2:
ω^2 = 15.092 J / 0.048 kg * m^2
ω^2 = 314.42 s^-2
ω ≈ 17.72 s^-1
Now that we have the angular velocity, we can find the translational velocity:
v = ω * r
v = 17.72 s^-1 * 0.08 m
v ≈ 1.42 m/s
Finally, we can calculate the total kinetic energy:
KE_total = KE_rot + KE_trans
KE_total = (1/2) * 0.016 kg * m^2 * (17.72 s^-1)^2 + (1/2) * 2.0 kg * (1.42 m/s)^2
KE_total = 2.827 J + 4.032 J
KE_total ≈ 6.859 J
Therefore, the total kinetic energy of the cylinder when it reaches the bottom of the ramp is approximately 6.859 joules.
(b) To find the rotational kinetic energy, we substitute the values we have into the formula for KE_rot:
KE_rot = (1/2) * I * ω^2
KE_rot = (1/2) * 0.016 kg * m^2 * (17.72 s^-1)^2
KE_rot = 0.141 J
Therefore, the rotational kinetic energy of the cylinder when it reaches the bottom of the ramp is 0.141 joules.
(c) To find the translational kinetic energy, we substitute the values we have into the formula for KE_trans:
KE_trans = (1/2) * m * v^2
KE_trans = (1/2) * 2.0 kg * (1.42 m/s)^2
KE_trans = 4.032 J
Therefore, the translational kinetic energy of the cylinder when it reaches the bottom of the ramp is 4.032 joules.