Physics 130
West Chester University

Syllabus || Schedule || Problems || Labs

PS4 Solutions
Angular Kinematics

Note: if you see funny symbols, your browser may not be standards compliant, or you may not have the proper fonts. To get these fonts, click here.

Chap 8: 4,5,6,9*,18,20,28,33,40,43*,48

4) Given: T = 0.033s is the period of the neutron star. Find the angular velocity.

Solution
This one is plug and chug:

\omega = {{2 \pi} \over T}
\omega = {{2 \pi } \over 0.033s}
\omega = 190 rad/s


5) Given (for a CD player):
\Delta t = 78 min
\omega_o = 480 rpm
\omega_f = 210 rpm
Find the angular velocity in rad/s2

Solution
To get an angular velocity in rpm is relatively easy.

\alpha = {{\omega_f - \omega_o} \over {\Delta t}}

\alpha = {{210 rpm - 480 rpm} \over {78 min}}

\alpha = -3.46 rpm/min

However, we must now convert this to rad/sec. Note the conversion:
1 rpm = {{1 rot} \over {1 min}} \times {{2 \pi} \over {1 rot}} \times {{1 min} \over {60 s}} = 0.1047 rad/s

Thus 1 rpm/min is:
1 rpm/min = {{{0.1047 rad} \over s} \over {60 s}} = 0.00174 rad/s^2

We can now multiply our answer by this conversion factor:
{{0.00174 rad/s^2 } \over {1 rpm/min}} \times -3.46 rpm/min = 0.006 rad/s^2


6) Given a Ferris wheel with the following:
\omega_f = 0.24 rad/s
\alpha = 0.030 rad/s^2
What is the time it takes to accelerate from rest?

Solution This can be solved using the definition of angular acceleration:

\alpha = {{\omega_f - \omega_o} \over {\Delta t}}
Simple manipulation to solve for change in time gives us:
\Delta t = {{\omega_f - \omega_o} \over \alpha}
\Delta t = {{0.24 rad/s - 0.00 rad/s} \over {0.030 rad/s^2}}
\Delta t = 8.0 s


9*) Given a space station with two rings, A and B such that
r_A = 3.2 \times 10^2 m
r_B = 1.1 \times 10^3 m
If ring A rotates such that a distance of s_A = 2.2 \times 10^2 m is covered, what's the corresponding length of arc for ring B?

Solution
This is just an arc-length problem. To solve it, we work backwards to find the angle which subtends the arc in question:

s_A = r_A\theta
\theta = {s_A \over r_A} = {{2.2 \times 10^2 m} \over {3.2 \times 10^2m}}
\theta = 0.6875 rad
But this is the same angle that subtends the arc for ring B. Thus we can reapply the arc-length formula to find the length of the corresponding arc at distance rB
s_B = r_B\theta = (1.1 \times 10^3 m )(0.6875 rad)
s_B = 7.6 \times 10^2 m


18) Given a ball that spins more slowly with time such that:
\omega_o = 18.5 rad/s
\omega_f = 14.1 rad/s
\Delta \theta = 85.1 rad
Find the time it takes to slow down.

Solution
This is a two part problem. First we need to find the angular acceleration:

\alpha = {{\omega_f^2-\omega_o^2} \over {2 \Delta \theta}}
\alpha = {{(14.1 rad/s)^2 - (18.5 rad/s)^2} \over {2(85.1 rad)}}
\alpha = -0.8427 rad/s^2

Now we can apply the definition of angular acceleration to find the change in time:
\alpha = {{\omega_f - \omega_o} \over {\Delta t}}
\Delta t = {{\omega_f - \omega_o} \over \alpha}
\Delta t = {{14.1 rad/s - 18.5 rad/s} \over {-0.8427 rad/s^2}}
\Delta t = 5.2 s


20) Given (for a centrifuge):
\omega_o = 420 rad/s
\omega_f = 1420 rad/s
\Delta t = 5.00 s

What is the magnitude of the angular acceleration?
Solution
Use the definition of angular acceleration:
\alpha = {{\omega_f - \omega_o} \over {\Delta t}}
\alpha = {{1420 rad/s - 420 rad/s} \over {5.00 s}}
\alpha = 200 rad/s^2

What is the angle through which the rotor turns?
Solution
\Delta \theta = {{\omega_f^2 - \omega_o^2} \over {2 \alpha}}
\Delta \theta = {{(1420 rad/s)^2 - (420 rad/s)^2} \over {2(200 rad/s^2)}}
\Delta \theta = 4600 rad


28) Given
r_{gal} = 2.2 \times 10^20 m
\omega = 1.2 \times 10^{-15} rad/s
What is the tangential velocity of the Sun?
Solution
v = r\omega
v = (2.2 \times 10^20 m)(1.2 \times 10^{-15} rad/s
v = 2.64 \times 10^5 m/s

What is the period of rotation (in years)?
Solution
First let's get this in seconds:
T = {{2 \pi} \over \omega} = {{2 \pi} \over {1.2 \times 10^{-15} rad/s}}
T = 5.234... \times 10^{15} s
To get this into years, we need the number of seconds in a year: N
N = 60 s/min \times 60 min/hr \times 24 hr/day \times 365 day/year
N = 31,536,000 s/year
So the period in years is T/N or T = 1.7 \times 10^8 years


33) Given
R_{earth} = 6.38 \times 10^6 m
T = 23.9 hr = 86040 s
What is the tangential velocity of a person on the equator?
Solution
v ={ {2\pi R_{earth}} \over T}
v = {{2 \pi 6.38 \times 10^6 m} \over {86040 s}}
v = 466 m/s

What is the latitude (angle from the center from the earth) at which the tangential speed is 1/3 of the above answer?
Solution
To solve this, one must use trigonometry with the figure in your book. One must find the distance from the Earth's axis to the surface of the Earth at this latitude.
r' = R_{earth}cos \theta
But since the Earth rotates at the same angular velocity, regardless of latitude, we can write:
v' = r'\omega = {1 \over 3} v
where we have written the last part to include the constraint that we're looking for a velocity that's one third of that at the equator. By substitution:
R_{earth} \omega cos \theta = {1 \over 3} v
or
v cos \theta = {1 \over 3} v
cos \theta = {1 \over 3}
\theta = cos^{-1}( {1 \over 3})
\theta = 70.5^o


40) Given that the Earth has a period of 3.16 x 107s, and an orbital radius of 1.50 x 1011m, what is a) the angular velocity of the planet about the Sun, b) the orbital tangential velocity, and c) the angular acceleration of the planet towards the Sun?

Solution
a) The angular velocity may be obtained directly from the period:

\omega = {{2\pi} \over T}
\omega = {{2\pi} \over {3.16 \times 10^7 s}}
\omega = 1.99 \times 10^{-7} rad/s

b) The tangential velocity is now directly obtained:
v = r \omega = (1.50 \times 10^{11} m)(1.99 \times 10^{-7} rad/s)
v = 2.98 \times 10^4 m/s

c) Finally, having the tangential velocity, we can obtain the centripetal acceleration.
a_c = {{v^2} \over r} = {{2.98 \times 10^4 m/s}^2 \over {1.50 \times 10^{11}m}}
a_c = 0.00593 m/s^2


43) Given a rectangle (L1 x L2) spun with constant angular acceleration from a corner about an axis such that one corner B was at a distance L_1 and the far corner from the axis A was at a distance given by the hypoteneuse of the rectangle, and where it is known that the rectangle rotates such that the centripetal acceleration of corner A is twice that of corner B: What is the ratio of L1 to L2?

Solution
The accelerations of the corners are related by:

a_A = 2a_B
Recall that centripetal acceleration may be represented as:
a_A = r_A \omega^2
a_B = r_B \omega^2
where the radii noted above are given by:
r_B = L_1
and
r_A = \sqrt{L_1^2 + L_2^2}
because rA is the hypoteneuse of the rectangle. Bringing these equations together we have:
r_A \omega^2 = 2 r_B \omega^2
\sqrt{L_1^2 + L_2^2} = 2 L_1
after dividing out the angular velocity and substituting for the radii.
L_1^2 +L_2^2 = 4 L_1^2
L_2^2 = 3L_1^2
{1 \over 3} = {\left ( L_1 \over L_2 \right )} ^2
{L_1 \over L_2} = \sqrt{ \left ( {1 \over 3} \right) }
{L_1 \over L_2} = 0.577


48) Given a bicycle wheel that spins with an angular velocity of 9.1 rad/s and which has a radius of 0.45 m for a time of 35 minutes, how far does the bicycle go in this time?

Solution
The tangential velocity may be calculated immediately.

v = r \omega = 0.45m \times 9.1 rad/s
v = 4.09 m/s
This tangential velocity is the same speed the wheel would impart to the bicycle. If we multiply this by the time in seconds (35 min x 60 sec = 2100 s), we can find the distance travelled.
\Delta x = v \Delta t = 4.09 m/s \times 2100 s
\Delta x = 8600 m
First calculate the tangential (linear) velocity