**E. Circular Motion
Homework Part I**

**7.1-7.3**

** **

Things you might need to know for some of these problems: \

1 mile = 1609 meters

**M _{earth} =
5.98 x 10^{24} kg R_{earth}
= 6.378 x 10^{6} m (equator)
**

**M _{sun} = 2.0
x 10^{30 }kg R_{sun}
= 7 x 10^{8} m (equator)**

**M _{moon} =
7.4 x 10^{22} kg R_{moon}
= 1.75 x 10^{6} m (equator)**

**R _{earth-sun }=
1.5 x 10^{11} m (average distance) Known as 1 A.U. (astronomical unit)**

**R _{earth-moon}
= 3.8 x 10^{8} m (average distance)**

** **

**7.1 Angular Measures**

**1, 2, 5, 7, 8, 10, 15**

10: How many degrees
will the Earth turn in those 3 months?
It takes the Earth 12 months to make a complete (360^{o})
circle. Look up the distance from Earth
to Sun in your textbook (this distance is called 1 astronomical unit (1 AU))

**7.2 Angular Speed
and Velocity**

**19, 20, 21, 22, 24, 30, 31, 34**

21: It is so important that you know the difference between these two ideas.

22: Again, you need to know this.

30: V_{tangent}/r
= w. And then, Dt = Dq/w
where Dq = 2p rad
since it maintains the angular velocity (w) over one complete
revolution (2prad)

34: Aw come on…. You wanted to know this anyway. Weren’t you curious?

**7.3 Uniform Circular Motion and Centripetal Acceleration**

** **

**36, 37, 40, 46, 48, 49, 50, 51, 53, ****55 (you
have got to at least try this one!), 56 **

40: Think about it! What is the direction of acceleration for any rotating object? The bobs will always move in the direction of acceleration. Just like a helium balloon on a string in your car will lurch forward when you step on the gas (accelerating in the forward direction). But be careful, because acceleration in this case is NOT in the forward direction!

46: a_{c} =
rw^{2}

48: a) v = (d/t)=(2pr)/t b) F_{c} = ma_{c} = m(v^{2}/r) c)
No, there has to be some kind of upward component tension force to
balance the downward force of gravity.

49: q =
tan^{-1}(mg/F_{c})

50: To keep the
water from coming out of the bucket, you have to have a velocity which will
match the gravitational force to the centripetal force. Mg = Fc = m(v^{2}/r). Therefore, v = _{}

51. f_{s} = m_{s}N
= m_{s}mg
= F_{c} = m(v^{2}/r)
such that v = _{} (that “s” should be
subscripted. And that “u” should be a “m”. But the darn computer wouldn’t let me do
that!)

53: a) A normal force by the loop must act on the block to provide extra centripetal force. Minimum speed corresponds to a minimum normal force of zero so the only force on the block is gravity. Gravity is the only source of centripetal force in this case.

F_{c} = mg = m(v^{2}/r) so that v = _{}

b)
From the energy conservation equation:
U + KE = U_{o} + KE_{o} such that:

mgh + 1/2 m(0)^{2}
= +mg(2r) + 1/2 mv^{2}

Therefore, h = 2r + 1/2 (rg/g) = (5/2) r

55: The key to this
problem is the positioning of the x and y axis. Typically, we have been referring to the **ramp** as being the
**x axis** and the **normal force** to be the **y axis** (since the
normal force always acts ^ to the ramp).
But now, we are interested in the __force acting towards the center of
the curve__ (which is N sin q). We will say,
in this case, that the x axis is along the line of N sin q. And
therefore, the y axis will be along N cos q. (Look at that figure
7.28 on p. 246 – can you see how N sin q is ^ to N
cos q?)

As the car
travels faster and faster, it produces a greater force **against** the curve
and therefore **increases** the N sin q force acting on it
towards center (remember, the magnitude of the normal force is a direct
response to the pushing on the ground by the object – mantra: “you can’t hit the ground any harder than it
hits you back”) .

Think about
it: If its V_{tangential}
increases, it is “trying” to go **into** the track, but since the track is
rigid, it cannot expand the radius of its turn. (What if this were the space shuttle, not on a track, but in
orbit around the Earth. If the space
shuttle were to speed up in the V_{tangential} direction its orbit
would grow in radius right?)

Okay, now how do you solve this confounded problem?

Along the y
axis, Ncosq
= mg (look at the drawing!). Along the
x axis, N sin q
is the force which acts centripetally (F_{c}). N sin q is the only force keeping
the car in its circular acceleration.

We can re-arrange N cos q = mg to say: N = mg/cosq

We can say also
(along the x axis) that F_{c} = N sin q

Since N = mg/cosq, we
can substitute this in the F_{c} equation above to say:

F_{c}
= (mg/cosq) sin q and by trig identity (sin q /cosq =
tan q)

Therefore…….. F_{c}
= mg tan q

We also know
that F_{c} = mv^{2}/r

So, let’s set
the two equations we have for F_{c} equal to each other

mg tan q = mv^{2}/r (how about that,
the m’s cancel!!!!!!)

So, tan q
= v^{2}/gr

56: The static friction between you and the wall provides the upward force to balance your weight. The normal force on you by the wall provides the centripetal force.

N = F_{c}
= mrw^{2}. So…..
f_{s} = m_{s}N = m_{s}mrw^{2}
= mg

Therefore, w = _{}

**Even Answers:**

2: c

8: 1.4 x 10^{2}
m

10: 2.4 x 10^{8}
km

20: d

22: Remember the earth turns
counterclockwise in north and clock in south

24: 3.1 rad/s

30: 0.42 sec

34: 0.120 sec

36: b

40: In the direction of the acceleration: inward

46: 2.31 x 10^{-3}
m/s^{2}

48: a) 7.85 m/s b) 10.3 N c) no

50: 3.1 m/s 56: 3.6 rad/sec