Normal Force on a Hill, Centripetal Force, Roller Coaster Problem, Vertical Circular Motion, Physics

Summary

This educational video script explores the concept of normal force in physics, using examples of a 5 kg box moving at 15 m/s on a circular path. It explains how normal force varies at different points, with greater force needed at the top of the curve due to the box's upward turn. The script calculates normal force at points A and B, revealing that at point B, if the box's speed is too high, it could lose contact with the road. It also discusses the minimum speed required for a roller coaster to prevent passengers from falling out when upside down at the top of a 15-meter radius circle, concluding with the formula for calculating this speed.

Takeaways

• 📏 The normal force at point A is greater than at point B due to the need for the ground to support the box's weight and provide the centripetal force for the turn.
• 🔢 At point A, the normal force is calculated as the sum of the centripetal force (mv^2/r) and the weight force (mg), resulting in 611.5 Newtons for a 5 kg box moving at 15 m/s with a radius of curvature of 2 meters.
• 🔄 At point B, the normal force is the difference between the weight force and the centripetal force, which can potentially be negative, indicating the box would lose contact with the road if moving too fast.
• 🚀 The maximum speed at which a vehicle can maintain contact with the road at point B is found by setting the normal force to zero and solving for velocity, resulting in 4.43 meters per second for the given scenario.
• 🎢 For a roller coaster traveling upside down at the top of a vertical circle, the normal force must be at least the difference between the weight force and the centripetal force to prevent passengers from falling out.
• 🔄 The minimum speed required for a roller coaster at the top of a 15-meter radius circle to prevent passengers from falling out is calculated to be 12.12 meters per second.
• ⚖️ The normal force is influenced by both the weight of the object and the centripetal force required for circular motion, with different directions and implications at the top of a hill versus on a flat road.
• 🛤️ At point B, if the centripetal force exceeds the weight force, the object will lose contact with the road, which is a critical factor in determining the maximum safe speed for vehicles on curved roads.
• 🎯 The concept of normal force is crucial in understanding how objects move on curved paths, whether it's a box on a road or a roller coaster on a track, and is essential for safety and design considerations.
• 📉 The normal force can be negative, which in practical terms means the object is no longer in contact with the surface, a key consideration in the design of roads and tracks for vehicles and amusement park rides.

Q & A

What is the normal force at point A for the 5 kg box moving at 15 m/s?

The normal force at point A is 611.5 Newtons, calculated as the sum of the centripetal force (5 kg * (15 m/s)^2 / 2 m) and the weight force (5 kg * 9.8 m/s^2).

What is the formula for calculating the normal force at point A?

The formula for calculating the normal force at point A is F_n = m * v^2 / r + m * g, where m is the mass, v is the velocity, r is the radius of curvature, and g is the acceleration due to gravity.

Why is the normal force greater at point A than at point B?

The normal force is greater at point A because, in addition to supporting the weight of the box, it must also provide the centripetal force to turn the box upward.

What is the significance of the negative centripetal acceleration at point B?

At point B, the centripetal acceleration is negative because it points in the opposite direction of the positive y-axis, indicating that the box is moving away from the center of the circle.

How is the normal force at point B different from that at point A?

At point B, the normal force is the difference between the weight force and the centripetal force, whereas at point A, it is the sum of the two.

What does a negative normal force at point B indicate?

A negative normal force at point B indicates that the centripetal force exceeds the weight force, suggesting that the box could lose contact with the road and fly off.

What is the maximum speed at which the box can maintain contact with the road at point B?

The maximum speed is calculated when the normal force is zero, which is when the centripetal force equals the weight force (mg). The formula is v = sqrt(rg), where r is the radius of curvature and g is the acceleration due to gravity.

How can you find the minimum speed required for a roller coaster to prevent passengers from falling out at the top of a vertical loop?

The minimum speed is found when the normal force is zero, which occurs when the centripetal force equals the weight force. The formula is v = sqrt(rg), where r is the radius of the loop and g is the acceleration due to gravity.

What is the difference between the normal force calculations for a box on a hill and a roller coaster at the top of a loop?

For a box on a hill, the normal force is the difference between the weight force and the centripetal force, while for a roller coaster at the top of a loop, it is the difference between the centripetal force and the weight force.

Why must a roller coaster maintain a minimum speed when traveling upside down at the top of a loop?

A roller coaster must maintain a minimum speed to ensure that the centripetal force is at least equal to the weight force, preventing the roller coaster from falling out of the loop.

Outlines

00:00

📏 Calculating Normal Force at Points A and B

The paragraph discusses calculating the normal force on a 5 kg box moving at 15 m/s at points A and B on a circular path. At point A, the normal force is greater because it must support the box's weight and provide the upward force to change direction. The formula for normal force at A is given by the sum of the centripetal force (mv^2/r) and the weight force (mg). Using a radius of 2 meters, the normal force at A is calculated to be 611.5 N. At point B, the situation is different as the centripetal force acts downwards, potentially leading to a negative normal force if the speed is high enough, indicating the box would lose contact with the road.

05:01

🚗 Maximum Speed to Maintain Road Contact

This section explores the concept of maximum speed for a vehicle to maintain contact with the road at point B. If the centripetal force exceeds the weight force, the vehicle will lose contact with the road. The maximum speed is determined by setting the normal force to zero and solving for velocity, resulting in a formula v = sqrt(rg). For a radius of 2 meters and g = 9.8 m/s^2, the maximum speed is calculated to be 4.43 m/s. This speed is critical; exceeding it will cause the vehicle to lose contact with the road.

10:02

🎢 Minimum Speed for a Roller Coaster at the Top of an Inverted Loop

The paragraph focuses on the minimum speed required for a roller coaster to prevent passengers from falling out when it's upside down at the top of a loop with a radius of 15 meters. The normal force in this scenario is the difference between the weight force and the centripetal force. The minimum speed is found by setting the normal force to zero and solving for velocity, which is also given by v = sqrt(rg). The calculated minimum speed is 12.12 m/s, indicating that the roller coaster must travel at this speed or faster to ensure passengers remain safely inside the loop.

15:03

🏁 Conclusion of the Physics Problem

The final paragraph summarizes the physics problem and its solution, emphasizing the importance of understanding the relationship between normal force, centripetal force, and weight in different scenarios. It highlights the critical speeds for maintaining contact with a road and ensuring safety in an inverted roller coaster loop. The presenter thanks the viewers for watching and hopes the explanation was clear.

Keywords

💡Normal Force

Normal force is the force exerted by a surface that supports an object against gravity and prevents it from falling through the surface. In the video, it is discussed in the context of a box moving on a curved path. At point A, the normal force must support the weight of the box and provide the centripetal force for the turn, making it greater than at point B where the box is simply following the curvature without needing additional force to change direction.

💡Centripetal Acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, always directed towards the center of the circle. It is crucial in the video for calculating the normal force at points A and B of the box's trajectory. At point A, the centripetal acceleration contributes to the normal force, while at point B, it subtracts from the weight force to determine the normal force.

💡Newton's Second Law

Newton's second law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). This law is applied in the video to determine the net force acting on the box at points A and B, which includes the normal force and gravitational force.

💡Gravitational Force

The gravitational force, often denoted as 'mg' where 'm' is mass and 'g' is the acceleration due to gravity (9.8 m/s²), is the force that pulls objects towards the center of the Earth. In the video, it is one of the forces considered when calculating the normal force at points A and B of the box's path.

💡Radius of Curvature

The radius of curvature refers to the radius of the circular path that an object is following. In the video, it is given as 2 meters for both points A and B and is used in the formula for centripetal acceleration to calculate the normal force experienced by the box at these points.

💡Tangential Speed

Tangential speed is the speed of an object moving along the tangent to its path of motion. In the video, the box has a tangential speed of 15 meters per second, which is used in conjunction with the radius of curvature to calculate the centripetal force and subsequently the normal force at points A and B.

💡Maximum Speed

The maximum speed discussed in the video is the highest speed at which an object can travel without losing contact with a surface, such as a car on a road. At point B, the video calculates the maximum speed the box can have without flying off the road, which is when the centripetal force equals the gravitational force.

💡Minimum Speed

The minimum speed is the lowest speed required for an object to maintain contact with a surface or to perform a certain motion, such as a roller coaster car at the top of an upside-down loop. The video calculates the minimum speed needed for a roller coaster to prevent passengers from falling out when at the top of a circular loop.

💡Upward Normal Force

Upward normal force is the component of the normal force that acts in the upward direction, counteracting gravity. At point A, the video explains that the normal force is upward, supporting the weight of the box and providing the necessary centripetal force for the turn.

💡Downward Weight Force

Downward weight force is the force due to gravity acting on an object, pulling it downward. In the video, this force is considered when calculating the normal force at points A and B, where it either adds to or subtracts from the centripetal force, depending on the direction of motion.

Highlights

Calculation of normal force on a moving box at point A and B.

At point A, the normal force is greater due to the need for upward turning.

At point B, the normal force is expected to be less as the box would just fall off.

Formula for calculating normal force at point A: Normal force = centripetal force + weight force.

Radius of curvature is given as 2 meters for both points A and B.

Calculation of normal force at point A yields 611.5 Newtons.

At point B, the normal force is the difference between weight force and centripetal force.

A negative normal force at point B indicates the box would lose contact with the road.

The maximum speed at which the box can maintain contact with the road at point B is calculated.

The maximum speed is determined when the normal force equals zero.

Formula for maximum speed: Speed = √(rg), where r is radius and g is gravitational acceleration.

The calculated maximum speed is 4.43 meters per second.

If the box travels faster than the calculated speed at point B, it will lose contact with the road.

Minimum speed required for a roller coaster to prevent passengers from falling out at the top of a vertical circle.

Difference between normal force calculations at the top of a hill versus inside a vertical circle.

Formula for minimum speed of the roller coaster: Speed = √(rg), with r as radius and g as gravitational acceleration.

The calculated minimum speed for the roller coaster is 12.12 meters per second.

The roller coaster must maintain a speed greater than or equal to the calculated minimum to stay on the track.

Transcripts

calculate the normal force of a five

kilogram box

moving at a speed of 15 meters per

second

at points a and b shown below

so here we have the box

now once it reaches point a

what is the normal force at point a and

what about at point b

is the normal force going to be greater

at a

or at b what would you say

the normal force is going to be larger

at a

because

not only does the ground have to support

the weight of the box

but it has to cause the box to turn

upward

so which is harder to do so the normal

force is gonna be greater at a at b

it's gonna be easier because it's just

gonna fall off

but let's calculate the normal force at

these two points

so at a

the normal force points in the upward

direction

and we have a downward weight force

and the centripetal acceleration

points in the positive y direction

because as you can see this is the

circle

the centripetal acceleration will always

points towards the center of the circle

so the net force

in the y direction for position a is

going to be the upward normal force

minus mg

and the net force in the y direction

based on newton's second law

is mass times acceleration

and what we have here is a centripetal

acceleration

which is v squared over r

so to get the normal force by itself i

need to move this to the other side

so at point a

the normal force

is the sum

of the centripetal force

and the weight force mg

now i need to give you the radius of

curvature

so let's say

it's two meters

for a and b

with that information go ahead and

calculate the normal force

so at position a the normal force is

going to be

mv squared over r so we have a mass of 5

a speed of 15

a radius of 2

plus the weight force which is 5 times

9.8

so it's 15 squared times 5

divided by 2.

so this is 562.5

and 5 times 9.8 is 49.

so the normal force at position a

is 611.5

newtons

now what about at position b

how can we calculate the normal force

the normal force is still going in the

positive y direction

and we still have a downward

weight force

now this is the circle

the centripetal acceleration is going to

point towards the circle

which means

it's pointing in the negative y

direction this time

instead of the positive y direction

so once again the net force in the y

direction

is going to be the upward

normal force it's going in the positive

y direction so it's positive

minus

the downward weight force

it's negative because it's going in the

negative y direction

now let's replace this with m a but the

acceleration is negative so it's going

to be negative

m a

now let's add mg to both sides

so i have mg

minus ma is equal to the normal force

now

let's replace the acceleration

with v squared over r

because once again this is the

centripetal acceleration

so the normal force

at point b

is the

difference between the weight force

and the centripetal force

at point a is the sum

so at point b it's a lot lighter

in fact if the vehicle is moving fast

enough

the centripetal force could exceed the

weight force

and if that happens

the vehicle is going to fly off the road

which is probably going to happen in

this example

the mass is 5

g is 9.8

and m

that's still 5 v is 15

r is

2.

so notice that we get a negative

normal force

so if you get this answer

what it really means is that the normal

force is really equal to zero

if the car is moving fast enough

it's not going to maintain contact with

the road instead it's just going to go

off in a tangent it's going to fly off

the road and it loses contact if it

loses contact with the road

then there is no normal force the normal

force is zero it doesn't exist

and so you can't get a negative answer

if you see that it just means that the

car it's it's off the road

that's what that really means

now it turns out that there's a maximum

speed

in which the car

can maintain contact with the road if

the car goes too fast

it's going to fly off

so what is the maximum speed at which

the car will maintain contact with the

road at point b

how can we get that answer

this will occur

when the normal force is zero

so at point b the normal force is

mg

minus mv squared over r

so when the normal force is equal to

zero

mv squared over r

is equal to mg you just got to take this

and move it to that side

now let's cancel the mass

so the speed is independent of mass

so v squared divided by r is equal to g

and now let's multiply both sides by r

so v squared is equal to rg

so now all we got to do at this point is

take the square root of both sides

so v is the square root of rg so this

will give us the maximum speed

that the block can have or the boxing

have

at point b without losing contact with

the road

so it's going to be the square root of

the radius which is 2

times g which is 9.8

so the maximum speed is

4.43 meters per second

so

if the box

travels at a speed that's higher than

this number at point b

it's going to lose contact with the road

the faster it's going the more it's

going to fly off the road

so just to give you a visual

illustration

so let's say here's the box if it's

moving at four meters per second

it's just gonna slide down with the road

it's gonna follow the curvature of the

road

however

let's say if it's moving at 50 meters

per second

it's going to lose contact with the road

now let's consider this problem

what is the minimum speed that a roller

coaster must have

when traveling upside down at the top of

a circle of radius 15 meters to prevent

the passengers from falling out

so i want to make a distinction

so in the previous problem

we said at point a the normal force

is the sum

of the centripetal force and the weight

force

and at the top of a circle

the normal force

is the difference

between the weight force

and

the centripetal force

now what we have

is a vertical circle but this time we

have a roller coaster that is inside the

circle

upside down

so what is the normal force in this case

and then how can we use that to find the

minimum speed

for the box at the top of the hill

there's a maximum speed at which you can

safely maintain contact with the road

with the roller coaster

if it's not going fast enough

the roller coaster could fall off

so

it has to maintain a minimum speed in

order to safely

make the trip not to lose contact with

the the tracks

if it goes fast enough it's gonna

it's gonna stay on a circle it's not

gonna fall down

so let's calculate the normal force for

this situation

unlike this problem where the normal

force was extended in a positive y

direction

in this problem

the normal force

extends in a negative y direction

towards the center of the circle

and the weight force

is also in the negative y direction

and the centripetal acceleration vector

also points towards the center of the

circle

so let's write an expression

to get the normal force but let's make

some space first

so the net force in the y direction

is going to be the normal force

which is negative because it's going in

the negative y direction

and the wave force is negative as well

now the net force we know is m a

but the acceleration is in the negative

y direction so we're going to put a

negative sign in front of it

now let's replace the centripetal

acceleration

with v squared over r

and now let's isolate the normal force

so i'm going to take this and move it to

this side so it becomes positive

fn

and i'm going to take this term move it

to this side

so it becomes positive

mv squared over r

so for the roller coaster

notice that the normal force

is mv squared over r

minus mg

it's the difference

between the centripetal force

and

the weight force

but in the other problem

at the top of the hill

the normal force is the difference

between the weight force and

the centripetal force

notice the difference between these two

equations

what is the difference that you see

notice that this

must have a maximum speed

and the roller coaster in order to make

contact with

the tracks it must have a minimum speed

notice that the weight force and the

centripetal force are written in the

reverse order

in the first example

with the box on top of the hill

if it's going too fast

this term the centripetal force will

exceed the weight force and so you're

going to get a negative normal force

which means that it loses contact with

the road

now

for this one there's no issue if it goes

too fast

because if it goes very fast you're

going to have a very high centripetal

force

and so the normal force is going to be

positive

if it goes too slow

the centripetal force will be less than

the weight force

and so if this term is greater you can

have a negative normal force which means

that the roller coaster is falling down

so if you want to find the minimum speed

that the roller coaster must have

the threshold is when the normal force

is zero

if you get a negative answer

it fell

if you get a positive answer

it's still in contact with the road but

the threshold

is when the normal force is equal to

zero

so we're going to use this equation and

replace fn with 0. so 0

is equal to mv squared over r

minus mg

now we're going to move this to this

side

so once again mg

is equal to mv squared divided by r and

we can cancel the mass

now just like we did before we're going

to multiply both sides by r

so we could cancel r as well

and so rg

is equal to v squared

and now let's take the square root of

both sides

so to calculate the minimum speed

it's equal to the square root of rg and

the same is true if you want to find the

maximum speed at the top of the hill

it's just the square root of rg

but the max speed has to be less than

or equal to the square root of rg

here if you want to find the minimum

speed it's going to be equal to or

greater than the square root of rg

so you can represent it with an

inequality but the threshold is at this

point

so now let's get the answers

so we have a radius of 15

and a gravitational acceleration of 9.8

so the speed is going to be 12.12 meters

per second

so in order for the passengers not to

fall out the speed has to be less than

or equal to

12.12 meters per second

rounded to nearest hundredth place

and so that's it for this problem

hopefully everything makes sense

and

thanks for watching this video

you