A sphere of mass 1.3 kg and radius 0.5 m is attached to the end of a massless ro

Question

A sphere of mass 1.3 kg and radius 0.5 m is attached to the end of a massless rod of length 3.0 m. The rod rotates about an axis that is at the opposite end of the sphere (see below). The system rotates horizontally about the axis at a constant 272 rev/min. After rotating at this angular speed in a vacuum, air resistance is introduced and provides a force 0.18 N on the sphere opposite to the direction of motion. What is the power (in W) provided by air resistance to the system 147.0 s after air resistance is introduced? (Enter the magnitude.) Axis of rotation

Explanation / Answer

Let us assume the sphere as solid, so moment of inertia about its center is
I_s = (2/5)MR² = (2/5) * 1.3kg * (0.5m)² = 0.13 kg·m²

And the moment of inertia of the assembly, about an end opposite the sphere and using the parallel axis theorem on the sphere, is
I = mL²/3 + I_s + M(R+L)² = 0 + 0.13kg·m² + 1.3kg*(3.5m)² = 16.06 kg·m²

again -
= F * r = I*
0.18N * 3.5m = 16.06kg·m² *
= 0.039 rad/s²

now as given, = 272rev/min * 2 rads/rev * 1min/60s = 28.48 rad/s

= - *t = 28.48 rad/s - 0.039*147.0s = 22.75 rad/s

initial KE = ½I² = ½ * 16.06 kg·m² * (28.48rad/s)² = 6513 J
And, final KE = ½I² = ½ * 16.06 kg·m² * (22.75rad/s)² = 4156 J

Therefore, friction work = initial KE - final KE = 6513 - 4156 = 2357 J

Therefore, power = work / time = 2357 J / 147s = 16.03 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.