| Finding g for an
unknown planet If an unknown planet
has 1/2 the radius and twice the mass, what is the value of g on the
planet.
g
unknown =
(2
/ (1/2)2 )(g Earth)
= 8 (g
Earth)
Explanation: The above comes directly from
the universal gravity equation. Gravity force is directly proportional
to mass and inversely proportional to the square of the distance
between the centers of mass of the objects |
|
Angular
or rotational motion kinematics
A fan
is switched on. It has an angular acceleration of 10 rad/s2.
After 10 seconds, what is the fan's angular velocity and displacement?
w = at
+ wo
= (10 rad/s2)(10 s) + 0
= 100 rad/s
q = 1/2 at2 + wot
= 1/2 (10 rad/s2)(10 s)2
+ 0
= 500 radians
|
| Elliptical Orbits
For an elliptical orbit of a mass =m around a
planet, the velocity at the closest point = vc and the
closest distance = Rc with the furthest distance = Rf.
What is the kinetic energy at the furthest distance?
First find the velocity vf
vf = (Rc/ Rf) vc
K = 1/2 m v 2
= 1/2 m [(Rc/ Rf)
vc] 2 |
|
Pendulums A pendulum with a mass of m
and length of L swings with a frequency of 3 Hz. What will the
frequency be for a pendulum with a mass of 3m and a length of 1/4 L?
f = 1/(2p) (g / L)^0.5
= 1/(2p)
[g / (1/4 L)]^0.5
= 1/(2p)
[4 g / L]^0.5
=
(2) 1/(2p)
[g / L]^0.5
but 1/(2p)
[g / L]^0.5 = original f, therefore
f =
2 (3 Hz)
= 6 Hz |
|
Frequency of a Mass
and Spring System
You have two linear springs with spring
constants of k and 3k. You also have 2 masses m and 2m. What is the
ratio of the highest to lowest resonant frequencies possible using
these components.
For parallel springs: Ktotal
= (3k + k)
= 4k
for series springs: 1 / Ktotal
= 1 / (3k) + 1 / k
Ktotal = 3 / 4 k
highest freq =
1/(2p) [4k/ m]^0.5
lowest freq =
1/(2p) [(3 / 4) k / (4m)]^0.5
| (ratio high to low) = |
1/(2p)
[4k/
m]^0.5
|
|
1/(2p)
[(3 / 4) k / (4m)]^0.5 |
| (ratio high to low) = |
[4]^0.5
|
|
[3 / 16]^0.5 |
| (ratio high to low) = |
8
|
| (3)^0.5 |
|
|
Rotational Momentum
A magical massless ice skater holds a dumbbell
with a mass of m in each hand as she spins around with a rotational
velocity of wo.
The distance between the center of mass of the 2 dumbbells is 2R. She
pulls her arms in so that the distance is R. Find her new rotational
velocity.
from conservation of momentum:
Lbefore = Lafter
Ibefore
wo=
Iafter w
2 [m (R)2] wo
= 2 [m (1/2 R)2] w
wo
= 1/4 w
w = 4
wo
|