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Centripetal Force Examples

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Centripetal Force Examples

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Centripetal Force Examples

Physics 6A

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Ride your bike around a curve and you will notice that if you go too fast, your tires will slip and you will fall.

Why does this happen?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Ride your bike around a curve and you will notice that if you go too fast, your tires will slip and you will fall.

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Ride your bike around a curve and you will notice that if you go too fast, your tires will slip and you will fall.

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a level curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a level curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

We will need to find a formula relating v and R. A diagram may help.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a level curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

We will need to find a formula relating v and R. A diagram may help.

View from above

Notice that the friction force points toward the center of the curve. It is the centripetal force.

friction

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

1) What is your maximum speed if the radius of the curve is 2R?

We will need to find a formula relating v and R. A diagram may help.

View from above

Notice that the friction force points toward the center of the curve. It is the centripetal force.

friction

We know a formula for friction as well:

Maximum static friction will give maximum speed.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

1) What is your maximum speed if the radius of the curve is 2R?

We will need to find a formula relating v and R. A diagram may help.

View from above

Notice that the friction force points toward the center of the curve. It is the centripetal force.

friction

We know a formula for friction as well:

Maximum static friction will give maximum speed.

Solve for vmax

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

1) What is your maximum speed if the radius of the curve is 2R?

We will need to find a formula relating v and R. A diagram may help.

View from above

Notice that the friction force points toward the center of the curve. It is the centripetal force.

friction

We know a formula for friction as well:

Maximum static friction will give maximum speed.

Solve for vmax

If R is doubled, vmax increases by √2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

2) What is your maximum speed if the radius is R, but the road is wet, so that your coefficient of static friction is only 1/3 of the value when the road is dry?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

2) What is your maximum speed if the radius is R, but the road is wet, so that your coefficient of static friction is only 1/3 of the value when the road is dry?

We can use our formula from part 1)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Why does this happen?

Static friction is not strong enough to keep your tires from slipping on the pavement.

OK, let’s say you are riding your bike around a curve and your maximum speed is v when the radius of the curve is R. Here are a couple of multiple choice questions:

1) What is your maximum speed if the radius of the curve is 2R?

2) What is your maximum speed if the radius is R, but the road is wet, so that your coefficient of static friction is only 1/3 of the value when the road is dry?

We can use our formula from part 1)

If µs decreases to µs/3 then vmax will decrease by √3.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

This carnival ride is a giant metal cylinder which will spin around and pin the occupants to the wall. The fun part is when the floor drops out from below and the patrons see a spike-filled pit of angry crocodiles awaiting them should they fall. As safety inspector, your problem will be to determine when it will be unsafe to ride. The given information is this: Radius of cylinder = 20m. Speed of rotation = 10 rpm.

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

This carnival ride is a giant metal cylinder which will spin around and pin the occupants to the wall. The fun part is when the floor drops out from below and the patrons see a spike-filled pit of angry crocodiles awaiting them should they fall. As safety inspector, your problem will be to determine when it will be unsafe to ride. The given information is this: Radius of cylinder = 20m. Speed of rotation = 10 rpm.

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

This carnival ride is a giant metal cylinder which will spin around and pin the occupants to the wall. The fun part is when the floor drops out from below and the patrons see a spike-filled pit of angry crocodiles awaiting them should they fall. As safety inspector, your problem will be to determine when it will be unsafe to ride. The given information is this: Radius of cylinder = 20m. Speed of rotation = 10 rpm.

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

The normal force is the force of the wall pushing inward. This is a centripetal force (it points toward the center of the circle).

friction

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

The normal force is the force of the wall pushing inward. This is a centripetal force (it points toward the center of the circle).

We can write down our formula for centripetal force:

friction

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

The vertical forces must balance out if the person wants to avoid the crocodile pit, so we can write down a formula:

friction

What type of friction do we want – static or kinetic?

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

The vertical forces must balance out if the person wants to avoid the crocodile pit, so we can write down a formula:

friction

By putting the maximum force of static friction in our formula, we are assuming the man is just on the verge of sliding.

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

The vertical forces must balance out if the person wants to avoid the crocodile pit, so we can write down a formula:

friction

We can replace N with the expression we found earlier.

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

Now that we have this formula, how do we use it?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

Notice that the mass canceled out, so based on the given information we should solve for µ.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

The radius and speed are given, but the speed is in rpm, so we will need to convert it to m/s.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

The radius and speed are given, but the speed is in rpm, so we will need to convert it to m/s.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

Substitute the values for g, R and v

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

We can start by drawing a free-body diagram of the forces on the person.

friction

Normal

mg

So if the coefficient is 0.44 the person will be on the verge of sliding down into the pit.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Wheel of Doom!

This carnival ride is a giant metal cylinder which will spin around and pin the occupants to the wall. The fun part is when the floor drop out from below and the patrons see a spike-filled pit of angry crocodiles awaiting them should they fall. As safety inspector, your problem will be to determine when it will be unsafe to ride. The given information is this: Radius of cylinder = 20m. Speed of rotation = 10 rpm.

a) Will leather-clad Biker Bob (mass = 100kg ;coeff. of static friction = 0.6) be safe?

b) How about Disco Stu, a 75kg man wearing a silk shirt and polyester pants (µs=0.15)?

Biker Bob is safe (his 0.6 coefficient is larger than 0.44 , so static friction is enough to hold him in place)

friction

Disco Stu is doomed! (his 0.15 coefficient is too small, so static friction fails to hold him in place)

Normal

mg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

GRAVITY

Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity.

There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to every other mass. Look around the room – everybody here is attracted to you!

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

GRAVITY

Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity.

There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to every other mass. Look around the room – everybody here is attracted to you!

Newton’s law of gravitation gives us a formula to calculate the attractive force between 2 objects:

m1 and m2 are the masses, and r is the center-to-center distance between them

G is the gravitational constant – it’s tiny: G≈6.674*10-11 Nm2/kg2

Use this formula to find the magnitude of the gravity force.

Use a diagram or common sense to find the direction. The force will always be toward the other mass.

m1

r

F2 on 1

m2

F1 on 2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

We should start by defining our coordinate system.

Let’s put the origin at planet H and say positive is to the right.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

We can also draw the forces on planet H in our diagram.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Our formula will find the forces (we supply the direction from looking at the diagram):

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Our formula will find the forces (we supply the direction from looking at the diagram):

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Our formula will find the forces (we supply the direction from looking at the diagram):

This is negative because the force points to the left

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Our formula will find the forces (we supply the direction from looking at the diagram):

This is negative because the force points to the left

This is positive because the force points to the right

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example:

Three planets are aligned as shown. The masses and distances are given in the diagram.

Find the net gravitational force on planet H (the middle one).

1012 m

3 x 1012 m

FApes on H

FDP on H

Planet of the Apes:

mass=6 x 1024 kg

Planet Hollywood:

mass=6 x 1020 kg

Daily Planet:

mass=3 x 1025 kg

Our formula will find the forces (we supply the direction from looking at the diagram):

This is negative because the force points to the left

This is positive because the force points to the right

Net force is to the left

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Add the forces to get the net force on H:

GRAVITY

One more useful detail about gravity:

The acceleration due to gravity on the surface of a planet is right there in the formula.

Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet.

m

Rplanet

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

GRAVITY

One more useful detail about gravity:

The acceleration due to gravity on the surface of a planet is right there in the formula.

Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet.

We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth)

m

Rplanet

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

GRAVITY

One more useful detail about gravity:

The acceleration due to gravity on the surface of a planet is right there in the formula.

Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet.

We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth)

m

Rplanet

This part is g

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB