As shown in the figure, ball 1 with m1=4.00 kg is connected through a mass less

Question

As shown in the figure, ball 1 with m1=4.00 kg is connected through a mass less rod with ball 2 with m2=3.00kg. The system is rotating at angular speed omega=8.00 rad/s about a vertical axle going through the center of the rod.

(a) How far away is the center of the mass of the system from the axle?
(b) Find the kinetic energy of the system. (l=mr^2 can be directly used to calculate the rotational inertia about the axle)
(c) What is the net force on each ball? (hint: try to apply F= m(v^2/r) and the relation between omega and v)

Everyone agrees on C. No one agrees on A and B. Is it really complicated?

Explanation / Answer

Initially let the 4 kg be on the left and 3 kg be on the right of the center of the rod . Let the C.M of the system be at a distance d from the center of the rod and near to the 4 kg mass Taking moment about the right end of the rod 4L = 7(L/2 + d) d = 0.071m from the axle nearer to 4 kg mass. --------------------------- b) I = mr^2 = 7*0.071² = 0.036 kgm² --------------------- C v²/r = r ?² = L8²/2 =32 L On the 4kg mass the force is 4*32 L = 128 L N On the 3kg mass the force is 3*32 L =96 L N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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