Normal Force on a Hill, Centripetal Force, Roller Coaster Problem, Vertical Circular Motion, Physics
Summary
TLDRThis 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
- π 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.
- π’ 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.
- π 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.
- π 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.
- π’ 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.
- π 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.
- βοΈ 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.
- π€οΈ 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.
- π― 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.
- π 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
π 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.
π 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.
π’ 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.
π 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.
Mindmap
Keywords
π‘Normal Force
π‘Centripetal Acceleration
π‘Newton's Second Law
π‘Gravitational Force
π‘Radius of Curvature
π‘Tangential Speed
π‘Maximum Speed
π‘Minimum Speed
π‘Upward Normal Force
π‘Downward Weight Force
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 = β(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 = β(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
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