Concept Questions with Answers
8.01
W14D1
3-Dimensional Rotational
Motion and Gyroscopes
8.01
Week 14D1
Today s Reading Assignment:
MIT 8.01 Course Notes
Chapter 22 Three Dimensional Rotations and
Gyroscopes
Sections 22.1.-22.3
Concept
Question: Rotating Vector
!
A vector A(t) of fixed length A is rotating about the z-axis at
an angular! speed ω. At t = 0 it is pointing in the positive ydirection. A(t) is given by
1.
2.
3.
4.
5.
6.
7.
8.
!
A(t) = Asin(! t) î + Acos(! t) ĵ
!
A(t) = ! Asin(" t) î + Acos(" t) ĵ
!
A(t) = Asin(! t) î " Acos(! t) ĵ
!
A(t) = ! Asin(" t) î ! Acos(" t) ĵ
!
A(t) = Acos(! t) î + Asin(! t) ĵ
!
A(t) = ! Acos(" t) î + Asin(" t) ĵ
!
A(t) = Acos(! t) î " Asin(! t) ĵ
!
A(t) = ! Acos(" t) î ! Asin(" t) ĵ
Concept Q. Ans. : Rotating Vector
Answer 2: At t = 0, it is pointing in the positive y-direction
(answers 1 and 2 satisfy this condition) and rotating
counterclockwise where ωt is the angle with respect to the
positive y-direction (figure at time t such that 0 <t <π/2). By
vector decomposition
!
A(t) = ! Asin(" t) î + Acos(" t) ĵ
Concept
Question: Rotating Vector
!
A vector A(t) of fixed length A is rotating about the z-axis at
an angular speed
ω. At t = 0 it is pointing in the positive y!
direction. dA(t) / dt is given by
1.
2.
3.
4.
5.
6.
7.
8.
!
dA(t) / dt = ! Asin(! t) î " ! Acos(! t) ĵ
!
dA(t) / dt = !" Asin(" t) î + " Acos(" t) ĵ
!
dA(t) / dt = ! Asin(! t) î " ! Acos(! t) ĵ
!
dA(t) / dt = !" Asin(" t) î ! " Acos(" t) ĵ
!
dA(t) / dt = ! Acos(! t) î + ! Asin(! t) ĵ
!
dA(t) / dt = !" Acos(" t) î + " Asin(" t) ĵ
!
dA(t) / dt = ! Acos(! t) î " ! Asin(! t) ĵ
!
dA(t) / dt = !" Acos(" t) î ! " Asin(" t) ĵ
Concept Q. Ans. :
Time Derivative of Rotating Vector
Answer 8: At time t
!
A(t) = ! Asin(" t) î + Acos(" t) ĵ
Therefore
!
dA / dt = !" Ascos(" t) î ! " Asin(" t) ĵ
!
and is perpendicular to A(t)
Concept Question: Gyroscope
For the simple gyroscope problem we just solved,
if the mass of the disk is doubled how will the new
z-component of the precession angular velocity Ωz be
related to the original Ωz,0?
1)  Ωz = 4 Ωz,0
2) Ωz = 2 Ωz,0
3) Ωz = Ωz,0
4) Ωz = (1/2) Ωz,0
5) Ωz= (1/4) Ωz,0
Concept Question Answer: Gyroscope
Answer 3. Both the torque and the angular momentum
are proportional to the mass of the spinning wheel, so
they cancel form both sides of the torque equation and
thus the precessional angular speed is independent of
mass.
.