Ok, i\'ve found many questions that is similar to this one, but no answers on th

Question

Ok, i've found many questions that is similar to this one, but no answers on the second question. So here we go again:

A small 10 g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50 g and is 100 cm in length. The bug jumps off in horizontal direction, perpendicular to the bar, with a speed of 20 cm/s relative to the table.

a) What is the angular speed of the bar just after the frisky insect leaps?
- i've solved this part and got w = 0.12 rad/s, which is the correct answer according to the book.

b) What is the total kinetic energy of the system just after the bug leaps?
THIS is the part that i just can't get right!
I've used K = 1/2*m_bug*v_bug^2+1/2*I*w^2, but i get 2,012*10^-4, and the answer should be 3.2*10^-4.

I use m_bug=0.01 kg, v_bug = 0.2 m/s, m_bar = 0.05 kg

Can anyone help me see where i am going wrong in this?

Explanation / Answer

Initial momentum of the system was zero.
The gain in momentum of the bug is mu 
The loss in momentum of the rod gives an angular momentum to the rod = muL.
The angular momentum of the rod about one end is I = *ML^2/3

Equating the two

= 3mu / (ML) = 3*10*20/(50*100) = 0.12 rad /s
= 0.12 rad /s 

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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