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## PowerPoint Slideshow about ' Using the “Clicker”' - peta

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### Using the “Clicker”

### Moving with the Earth

### Gravitron (or The Rotor)

### Gravitron (see the worksheet)

### Gravitron (work together)

### Ball on a string

### Ball on a string – free-body diagrams

### Ball on a string – free-body diagrams

### A water bucket

### Free-body diagram for the water bucket

### Roller coaster

### Roller coaster

### Driving on a hilly road

### Driving on a hilly road

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.

Remember: for uniform circular motion, the acceleration has magnitude v2/r = w2r, where w =v/r = 2p/T

So let’s calculate the acceleration:

caused by the Earth’s rotation about its axis.

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 106m

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

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?

A normal force.

A force of gravity.

A force of static friction.

A force of kinetic friction.

None of the above.

Gravitron simulation

Sketch a free-body diagram for the rider.

Apply Newton’s Second Law, once for each direction.

FS

He’s blurry because he is going so fast!

Axis of rotation

FN

a

y

mg

x

Sketch a free-body diagram for the rider.

Apply Newton’s Second Law, once for each direction.

y direction: FS- mg = may = 0 (he hopes)

x direction: FN = max = m (v2/r)

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?

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

Do the bottom first

Assume same speed ma = mv2/r = 8 N

Breaks at T = 13 N

mg = 5 N

So, T = 3 N

So, ma = mv2/r = 8 N when it breaks

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

(Actually, v will be smaller at the top)

Weight = mg = 5 N

T + mg = mv2/r

T = mv2/r – mg

T – mg = mv2/r

T = mv2/r + mg

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.

FN on water, or T on bucket + water

Ma = Mv2/r

Mw g or Mb+wg

Sketch a free-body diagram for the bucket (or the water), and apply Newton’s Second Law.

Mg + FN = Mv2/r

But critical speed is when FNor T= 0

So Mg = Mv2min/r or vmin= (rg)1/2

“Toward center” is down

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.

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

ma = m(v2/r)

FN – mg = mv2/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

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

Warning to drivers: Your braking is worst at the crest of a hill.

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

FN -> 0

Mg

FN – Mg = M(-v2/r) loses contact when FN = 0 at v = (rg)1/2

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