Using the “Clicker”
If you have a clicker now, and did not do this last time,
please enter your ID in your clicker.
First, turn on your clicker by sliding the power switch, on
the left, up. Next, store your student number in the clicker.
You only have to do this once.
Press the * button to enter the setup menu.
Press the up arrow button to get to ID
Press the big green arrow key
Press the T button, then the up arrow to get a U
Enter the rest of your BU ID.
Press the big green arrow key.
Moving with the Earth
Remember: for uniform circular motion, the acceleration has
magnitude v2/r = w2r, where w = v/r = 2p/T
So let’s calculate the acceleration:
1) caused by the Earth’s rotation about its axis.
2) caused by the Earth’s orbit around the Sun.
Note: 1 yr = 60*60*24*365 s = 3.15 x 107 s = p x 107 s
(good approx)
rorb = 1.50 x 1011 m,
rrot = 0.7(6.38x106 m) = 5 x 106 m
worb = 2p/T = 2 x 10-7 rad/s, wrot = 365 worb = 7.30 x 10-5 rad/s
agrav = g = 9.8 m/s2
arot = w2r = (50 x 10-10 )(5 x 106) = 0.025 m/s2
aorb = w2r = (4 x 10-14)(1.5 x 10-11) = 0.006 m/s2
Gravitron (or The Rotor)
In a particular carnival ride, riders are pressed against the
vertical wall of a rotating ride, and then the floor is removed.
Which force acting on each rider is directed toward the
center of the circle?
1.
2.
3.
4.
5.
A normal force.
A force of gravity.
A force of static friction.
A force of kinetic friction.
None of the above.
Gravitron (see the worksheet)
Gravitron simulation
Sketch a free-body diagram for the rider.
Apply Newton’s Second Law, once for each direction.
Gravitron (work together)
Sketch a free-body diagram for the rider.
He’s
blurry
because
he is
going so
fast!
FS
Axis of
rotation
FN
a
mg
y
x
Apply Newton’s Second Law, once for each direction.
y direction: FS - mg = may = 0 (he hopes)
x direction: FN = max = m (v2/r)
Vertical circular motion
Examples
• Water buckets
• Cars on hilly roads
• Roller coasters
Ball on a string
When a ball with a weight of 5.0 N is whirled in a vertical
circle, the string, which can withstand a tension of up to
13 N, can break.
Let’s see how to answer questions such as:
Why?
Where is the ball when the string is most likely to break?
What is the minimum speed of the ball needed to break
the string?
Ball on a string – free-body diagrams
Sketch one or more free-body diagrams, and apply
Newton’s Second Law to find an expression for the
tension in the string.
At the top
At the bottom
Ball on a string – free-body diagrams
Sketch one or more free-body diagrams, and apply
Newton’s Second Law to find an expression for the
tension in the string.
Do the bottom first
Assume
same
speed
ma =
mv2/r =
8N
mg =
5N
So, T = 3 N
(Actually, v will be
smaller at the top)
At the top
Breaks at
T = 13 N
Weight
= mg
=5N
At the bottom
T + mg = mv2/r
T – mg = mv2/r
T = mv2/r – mg
T = mv2/r + mg
So, ma
= mv2/r
=8N
when it
breaks
A water bucket
As long as you go fast enough, you can whirl a water
bucket in a vertical circle without getting wet.
What is the minimum speed of the bucket necessary to
keep the water in the bucket?
The bucket has a mass m, and follows a circular path of
radius r.
If you go too slow, the string will go slack, and the water
and the bucket will stay together along a parabolic free
fall path.
Free-body diagram for the water
bucket
Sketch a free-body diagram for the bucket (or the water),
and apply Newton’s Second Law.
Ma =
Mv2/r
“Toward
center” is
down
Mw g
or
Mb+wg
FN on water, or T on
bucket + water
Mg + FN = Mv2/r
But critical speed is when FN or T = 0
So Mg = Mv2min /r or vmin = (rg)1/2
Roller coaster
On a roller coaster, when the coaster is traveling fast at
the bottom of a circular loop, you feel much heavier than
usual. Why?
Draw a free-body diagram and apply Newton’s Second
Law.
Roller coaster
On a roller coaster, when the coaster is traveling fast at
the bottom of a circular loop, you feel much heavier than
usual. Why?
Draw a free-body diagram and apply Newton’s Second
Law.
FN
FN – mg = mv2/r
ma =
m(v2/r)
so
FN = mg + mv2/r
The faster you go, the larger
the normal force has to be.
The normal force is equal to
your apparent weight.
mg
Driving on a hilly road
As you drive at relatively high speed v over the top of a
hill curved in an arc of radius r, you feel almost weightless
and your car comes close to losing contact with the road.
Why?
Draw a free-body diagram and apply Newton’s Second
Law.
r
Driving on a hilly road
As you drive at relatively high speed v over the top of a
hill curved in an arc of radius r, you feel almost weightless
and your car comes close to losing contact with the road.
Why?
Draw a free-body diagram and apply Newton’s Second
Law.
Mv2/r
Mg
FN -> 0
FN – Mg = M(-v2/r)
loses contact when
FN = 0 at v = (rg)1/2
Warning to
drivers: Your
braking is
worst at the
crest of a hill.