Physics 8
Spring 2012
NAME:
TA:
Quiz 3 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
Recall that the position vector of a particle in two dimensions is
~r = xî + y ĵ.
1. Noting that the x-component of a vector r may be given in terms of the magnitude, r,
and angle, θ, as x = r cos θ, while the y-component may be written as y = r sin θ, show
that the position vector of a particle on a circle of constant radius R with its center at
the origin is
~r = R cos θî + sin θĵ ,
where θ is the angle that the position vector makes with the x-axis.
2. If the particle moves with constant speed v starting on the x-axis at t = 0, find an
expression for θ in terms of the velocity, v, radius, R, and time t. Write the position
vector with this value of θ.
3. In terms of the velocity, v, and radius r, what is the period, T , to complete a full
circle?
4. Differentiate the position vector twice with respect to time to find the acceleration,
determine its magnitude, and show that the direction is towards the center of the
circle.
Hint: You might find the following derivatives useful:
d
dt
d
dt
(A cos (at)) = −aA sin (at)
(A sin (at)) = aA cos (at) ,
where A and a are constants.
————————————————————————————————————
Solution
1. Because the particle is moving around at a constant radius, then the magnitude of
the position vector is constant, and r = R. Thus, the x-component is x = R cos θ,
while the y-component is y = R sin θ. Plugging these results into the position vector
equation we find
~r = xî + y ĵ = R cos θî + R sin θĵ = R cos θî + sin θĵ ,
as claimed.
1
2. The particle moves along at a constant speed, which is the speed that it moves around
the circle. As it moves around the circle, it sweeps out an angle which depends on the
speed. In a time t, the particle moves along the circle a distance s = vt. If the radius
is R, then the distance sweeps out an angle θ = s/R, and so
θ=
v
s
= t ≡ ωt,
R
R
where the angular velocity, ω ≡ v/R. So, the angle grows linearly with time. Thus,
the position vector is
v i
h v t î + sin
t ĵ .
~r = R cos
R
R
3. The time it takes to go completely around the circle is the period, T , and is simply
the total distance around, divided by the speed of the particle. The total distance is
just the circumference of the circle, 2πR, and so
T =
2πR
.
v
4. Taking one derivative of the position vector gives the velocity vector, ~r˙ = ~v , and using
the hint above gives
i
~r˙ = R − Rv sin Rv t î + Rv sin Rv ĵ
h
i
v
v
= v − sin R t î + cos R t ĵ .
Taking one more derivative gives the acceleration, ~r¨ = ~v˙ = ~a, and so
h
i
v
v
v
v
¨
~r = v − R cos R t î − R sin R t ĵ
h
i
2
= − vR cos Rv t î + sin Rv t ĵ
Now, the position vector points along cos θî+sin θĵ, while the acceleration points along
the negative of this, meaning that the acceleration vector points towards the center, as
expected. Furthermore, the magnitude of the acceleration vector is
a = |~
a|
q
2
2
2
− vR cos Rv t
+ − vR sin
=
q
v4
2 v t + v 4 sin2 v t
=
2 cos
R
R
R
R
q
v4
=
R2
=
2
v
t
R
v2
.
R
So, the magnitude of the centripetal acceleration, for a particle moving around a circle
of constant radius R, at speed v, is
aC =
2
v2
.
R