A uniform cylinder (I = mR²) of diameter 0.20 m and mass 12 kg rolls without slipping down a 37º inclined plane. The gain in translational kinetic energy of the cylinder when it has rolled 5 m down the incline of the plane is approximately A) 24 J B) 0.12 kJ C) 0.24 kJ D) 0.35 kJ E) 0.59 kJ

Question

Quizplus · Accepted Answer

The gravitational potential energy (GPE) is converted into both translational kinetic energy (TKE) and rotational kinetic energy (RKE) as the cylinder rolls down the incline. The moment of inertia (I) for a uniform cylinder is 0.5mr^2, where m is the mass and r is the radius (half the diameter).
The initial GPE of the cylinder is mgh, where h is the height of the incline. The final TKE of the cylinder is 0.5mv^2, where v is the final velocity after rolling 5 m down the incline. The final RKE of the cylinder is 0.5Iω^2, where ω is the final angular velocity.
Using conservation of energy, we can equate the initial GPE to the final TKE and RKE:
mgh = 0.5mv^2 + 0.5Iω^2
Substituting the given values, we get:
(12 kg) (9.81 m/s^2) (sin 37º) (5 m) = 0.5 (12 kg) (v^2) + 0.5 (0.5) (12 kg) (0.1 m)^2 (v/0.1 m)^2
Simplifying and solving for v, we get:
v ≈ 5.80 m/s
The initial angular velocity is zero since the cylinder starts from rest, so the final angular velocity is:
ω = v/r = 5.80 m/s / 0.1 m = 58 rad/s
The gain in TKE is the difference between the final and initial TKE:
ΔTKE = 0.5mv^2 - 0.5m(0)^2 = 0.5mv^2
Substituting the given values, we get:
ΔTKE ≈ 0.24 kJ

Therefore, the best choice is C.

A Uniform Cylinder (I = MR2) of Diameter 0

A Uniform Cylinder (I = MR²) of Diameter 0