A thin, cylindrical rod = 27.0 cm long with a mass m = 1.20 kg has a ball of dia

Question

A thin, cylindrical rod = 27.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.

After the combination rotates through 90 degrees, what is its rotational kinetic energy?
What is the angular speed of the rod and ball?
What is the linear speed of the center of mass of the ball?

Explanation / Answer

L = 27.0 cm, m = 1.20 kg, radius of the ball r = 5.00 cm, M = 2.00 kg, initial potential energy E = mg(L/2) + Mg(L + r) After the combination rotates through 90 degrees, its rotational kinetic energy = E = mg(L/2) + Mg(L + r) = 7.86 J the angular speed of the rod and ball is w moment of inertia I = mL^2/3 + [2Mr^2/5 + M*(L + r)^2] (use parallel axis theorem) I = 0.236 kgm^2 E = Iw^2/2 so w = sqrt(2E/I) = 8.16 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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