PHYS 1443-501, Spring 2025, 2nd -Term Exam
Questions and Answers (University of Texas)
Name:
[ 1 – 20 points] A uniform rod of length 150 cm and mass 900g is attached at one end
to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown
in the figure. Answer the following series of questions, assuming that the rod is released
from rest in the horizontal position, and the magnitude of the gravitational acceleration g
is 9.80m / s2 .
a) What is the line density of the rod?
1. 6.00kg/cm3 2. 0.600kg/m
3. 0.600kg/cm 4. 6.00kg/m3
Solution: Since line density is the density per unit length, the line density for this
rod is m/L=0.900/1.50=0.600kg/m.
b) What is the moment of inertia of the rod in this motion?
1. 0.267kg 2. 1.35kg m
m 2
3. 1.35kg m2 4. 0.675kg m2
Solution: Moment of inertia of this rod when rotates about the axis at one end is
I=ML2/3=0.900x(1.50)2=0.675kg.m2.
c) What is the potential energy of the rod before the release?
1. 13.2k
m/ s 2. 13.2J
g
3. 6.62J 4. mgh
, Solution: Potential energy is mgh. The height of the CM when it is in its equilibrium
position is L/2. Therefore the potential energy
U=mgh=mgL/2=0.9x9.8x1.5/2=6.62J.
d) What is the angular speed of the rod when the rod reaches the bottom?
1. 4.43 / s 2. 3.38 / s
3. 6.62m / s 4. 16.3 / s
Questions and Answers (University of Texas)
Name:
[ 1 – 20 points] A uniform rod of length 150 cm and mass 900g is attached at one end
to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown
in the figure. Answer the following series of questions, assuming that the rod is released
from rest in the horizontal position, and the magnitude of the gravitational acceleration g
is 9.80m / s2 .
a) What is the line density of the rod?
1. 6.00kg/cm3 2. 0.600kg/m
3. 0.600kg/cm 4. 6.00kg/m3
Solution: Since line density is the density per unit length, the line density for this
rod is m/L=0.900/1.50=0.600kg/m.
b) What is the moment of inertia of the rod in this motion?
1. 0.267kg 2. 1.35kg m
m 2
3. 1.35kg m2 4. 0.675kg m2
Solution: Moment of inertia of this rod when rotates about the axis at one end is
I=ML2/3=0.900x(1.50)2=0.675kg.m2.
c) What is the potential energy of the rod before the release?
1. 13.2k
m/ s 2. 13.2J
g
3. 6.62J 4. mgh
, Solution: Potential energy is mgh. The height of the CM when it is in its equilibrium
position is L/2. Therefore the potential energy
U=mgh=mgL/2=0.9x9.8x1.5/2=6.62J.
d) What is the angular speed of the rod when the rod reaches the bottom?
1. 4.43 / s 2. 3.38 / s
3. 6.62m / s 4. 16.3 / s
