A 20.0 gram block zoom to the East with a velocity of 18.0 m/s. The block collid

Question

A 20.0 gram block zoom to the East with a velocity of 18.0 m/s. The block collides with a uniform density rod at rest pivoted about its center. The rod is 0.800 m long and has a mass of 0.300 kg. The block sticks to the edge of the rod, and both rotate after the collision. a.) Calculate the total angular momentum about the pivot. b.) Calculate the total moment of inertia of the rod and block when stuck together. c.) Calculate the angular velocity of the rod and block after the collision, in rad/s

Explanation / Answer

a)

total angular momentum L = m*v*L/2

m = mass of the block

v = speed of the block

L = length of the rod

L = 0.02*18*0.8/2

L = 0.144 kg m^2/s

=-====================

(b)

total moment of inertia I = (1/12)*M*L^2 + m*(L/2)^2

I = (1/12)*M*L^2 + (1/4)*m*L^2

I = ((1/12)*0.3*0.8^2) + ((1/4)*0.02*0.8^2)

I = 0.0192 kg m^2

-------------------

c)

final angular momentum Lf = I*wf

from conservation of angular momentum

totla momentum remainsconstant

0.0192*wf = 0.144

wf = 7.5 rad/s <<<<<<<-----answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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