Question

A 415 gram, 1.6 meter long wooden rod is pivoted at its center. A 4 gram bullet traveling a 188 m/s strikes it perpendicular to its length and half way between the center and the end. If the bullet exits with half its entrance velocity, what is the angular speed of the rod after the bullet leaves to the nearest hundredth of a radian/s? answer is 1.70

In this problem use 6x1024 kg and 6.4x106 m for the mass and radius of the earth, respectively, and 7.3x10-5 radians /s for its angular velocity. What is the mass of an asteroid (as a fraction of the mass of the earth to 5 decimal places) if it stops the earths rotation when it strikes the earth tangentially at the equator traveling opposite to the earth's rotation at a speed of 44.9 km/s? answer is 0.00416

Answer #1

**1)
let**

**M = 415 grams = 0.415 kg
L = 1.6 m
m = 4 grams = 0.004 kg
vi = 188 m/s
vf = vi/2 188/2 = 94 m/s**

**let w is the angular speed of the rod just after bullet
passes through it.**

**Apply conservation of angular momentum**

**final angular momentum = initial angular
momentum**

**I_rod*w + m*vf*L/4 = m*vi*L/4**

**I_rod*w = m*(vi - vf)*L/4**

**w = m*(vi - vf)*L/(4*I_rod)**

**= m*(vi - vf)*L/(4*M*L^2/12)**

**= 0.004*(188 - 94)*1.6/(4*0.415*1.6^2/12)**

**= 1.70 rad/s**

**2)**

**let m is the mass of the asteroid.**

**Apply conservation of angular momentum**

**angular momentum of the earth = angular momentum of the
asteroid in the opposite direction
I_Earth*w = m*v*Re**

**(2/5)*Me*Re^2*w = m*v*Re**

**m = (2/5)*Me*Re*w/v**

**= (2/5)*6.4*10^6*7.3*10^-5/(44.9*10^3)*Me**

**= 0.00416*Me**

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