14-84 Kinetics of Particle: Conservation of Energy Chapter 14: Hibbeler Dynamics | Engineers Academy
Summary
TLDRThis video from Engineers Academy walks through the solution of a dynamics problem involving a 4 kg smooth collar. The instructor applies the law of conservation of energy to determine the maximum distance the collar travels before it momentarily stops. The explanation covers calculating kinetic and potential energies, and using trial and error in Excel to find the precise solution. The video is an excellent resource for students learning problem-solving techniques in dynamics. Viewers are encouraged to subscribe for more problem-solving tutorials.
Takeaways
- 📚 The problem involves determining the maximum distance a 4 kg smooth collar travels before stopping momentarily.
- 🛠️ The problem is solved using the law of conservation of energy, considering both kinetic and potential energies at different points.
- 🌐 The initial conditions are a collar speed of 3 m/s at position s = 0, and the spring's unstretched length is 1 meter.
- ⚖️ Kinetic energy at point A is calculated using the formula 1/2 * mass * velocity^2, resulting in 18 joules.
- 🌀 Elastic potential energy is determined based on the spring stretch at point A, which is 0.5 meters, resulting in 12.5 joules.
- ⚖️ Gravitational potential energy is calculated using the weight of the collar and the height from a datum line, with the value 39.24 * s_max joules.
- 📊 The energy at point B is analyzed, considering that the kinetic energy is zero due to the collar's momentary stop.
- 🔗 Pythagoras' theorem is used to determine the stretch in the spring at point B, factoring in the distance s_max.
- 📈 The complex equation derived for s_max is solved using a trial-and-error method (or Excel), leading to an approximate value of 1.955 meters.
- 🎯 The final takeaway is that the maximum distance the collar travels before stopping momentarily is approximately 1.955 meters.
Q & A
What is the main problem being solved in the video?
-The main problem is determining the maximum distance a 4 kg smooth collar travels before it stops momentarily, given that it has an initial speed of 3 meters per second and is connected to a spring with an unstretched length of 1 meter.
What principle is used to solve the problem?
-The problem is solved using the law of conservation of energy, which states that the total kinetic and potential energy at one point is equal to the total kinetic and potential energy at another point.
How is the kinetic energy at point A calculated?
-The kinetic energy at point A is calculated using the formula \( \frac{1}{2} mv^2 \), where the mass \( m \) is 4 kg and the velocity \( v \) is 3 meters per second.
What types of potential energy are considered in this problem?
-Two types of potential energy are considered: elastic potential energy (due to the spring) and gravitational potential energy (due to the height of the collar).
How is the stretch in the spring at point A determined?
-The stretch in the spring at point A is determined by subtracting the original length of the spring (1 meter) from the stretched length at point A (1.5 meters), resulting in a stretch of 0.5 meters.
How is the elastic potential energy at point A calculated?
-The elastic potential energy at point A is calculated using the formula \( \frac{1}{2} k x^2 \), where \( k \) is the spring constant (100 N/m) and \( x \) is the stretch in the spring (0.5 meters).
What assumption is made about the gravitational potential energy at point B?
-It is assumed that the gravitational potential energy at point B is zero because the datum line is chosen at point B, making the height from the datum line zero.
How is the stretch in the spring at point B determined?
-The stretch in the spring at point B is determined using the Pythagorean theorem, where the stretch is the hypotenuse of a right triangle with sides equal to the maximum distance traveled and 1.5 meters.
Why is the trial and error method used to solve for the maximum distance?
-The trial and error method is used because the resulting equation for the maximum distance is nonlinear and complex, making it difficult to solve algebraically.
What is the final maximum distance the collar travels before stopping?
-The final maximum distance the collar travels before stopping is approximately 1.955 meters.
Outlines
📚 Introduction and Problem Statement
The instructor welcomes students to the Engineers Academy and emphasizes subscribing to the channel. The problem at hand involves a 4 kg smooth collar moving at 3 m/s when s = 0. The task is to determine the maximum distance the collar travels before it stops momentarily. The spring in question has an unstretched length of 1 meter. The instructor sets up the problem using the law of conservation of energy, incorporating both kinetic and potential energies, including gravitational and elastic potential energies. Detailed calculations for the kinetic and elastic potential energies at point A are provided.
⚖️ Energy Considerations at Point B
At point B, the instructor explains that the kinetic energy is zero since the collar momentarily stops. The focus then shifts to calculating the elastic potential energy and the gravitational potential energy at B. Using the Pythagorean theorem, the instructor determines the stretched length of the spring at B and sets up an equation involving these energies. Further simplifications and calculations are discussed, leading to a complex equation involving s_{max}.
📊 Solving for Maximum Distance Using the Equation
The instructor continues to simplify the equation and makes corrections to a previous mistake involving the gravitational potential energy. The equation becomes a quadratic form in s_{max}, but the instructor notes that it cannot be solved directly. A trial-and-error method (trial and error) using an Excel sheet is suggested to find the value of s_{max} that satisfies the equation. The instructor explains how different values of s_{max} affect the left and right sides of the equation and identifies the approximate solution range.
🔍 Refining the Solution for Maximum Distance
The instructor fine-tunes the value of s_{max} using smaller increments, identifying the precise value where the left and right sides of the equation are almost equal. The final value of s_{max} is determined to be 1.955 meters. A graph is plotted to visualize the intersection point of the left and right-hand sides, confirming the solution. The instructor concludes that the collar will travel 1.955 meters before coming to a momentary stop, and encourages students to subscribe to the channel for more problem solutions.
Mindmap
Keywords
💡Conservation of Energy
💡Kinetic Energy
💡Potential Energy
💡Elastic Potential Energy
💡Gravitational Potential Energy
💡Spring Constant (k)
💡Datum Line
💡Pythagorean Theorem
💡Trial and Error Method
💡Momentarily Stops
Highlights
Introduction to the problem of determining the maximum distance a 4 kg smooth collar travels before it stops momentarily.
Explanation of the initial conditions: the collar has a speed of 3 meters per second at s = 0, and the spring's unstretched length is 1 meter.
Application of the law of conservation of energy, combining kinetic energy and potential energy to solve the problem.
Detailed calculation of kinetic energy at point A, including the mass and velocity, resulting in 18 Joules.
Computation of elastic potential energy at point A, considering the spring's stretch and the spring constant, yielding 12.5 Joules.
Determination of gravitational potential energy at point A using the datum line and the equation W × h, resulting in 39.24 × s_max.
Identification that the velocity at point B is zero, implying the kinetic energy at B is also zero, leaving only potential energies to be considered.
Calculation of the stretch in the spring at point B using the Pythagorean theorem to find the length OB, leading to further energy analysis.
Simplification of the elastic potential energy at B, leading to the final equation involving s_max, which requires solving through trial and error.
Correction of an error in the gravitational potential energy calculation by using the correct mass (4 kg), leading to an updated term of 39.24 × s_max.
Introduction of the hit-and-trial method using an Excel sheet to find the value of s_max, where the left and right sides of the equation are equal.
Demonstration of the iterative process in Excel to narrow down s_max to 1.955 meters as the solution.
Graphical representation of the left and right sides of the equation, showing the intersection point that confirms s_max = 1.955 meters.
Conclusion that the collar will travel a distance of 1.955 meters before momentarily stopping.
Encouragement to subscribe to the Engineers Academy channel and like the video for more problem solutions.
Transcripts
hello students i welcome you all to
engineers academy
do subscribe my channel if you haven't
done it yet
now we are going to solve this problem
which says that the 4 kg smooth collar
has a speed of 3 meter per second
when it is at s equals to 0
determine the maximum distance as it
travels before it stops momentarily
and the spring has an unstretched length
of 1 meter
so at a the velocity is 3 meter per
second so we can apply the
law of conservation of energy which says
that the kinetic energy at a
plus the potential energies at e
that will be equal to the kinetic energy
at point
b
plus the potential energies at point b
so since here we have the spring so we
will have the elastic potential energy
and the gravitational potential energy
so this va can be split into two
potential energies
so the kinetic energy at point a is we
can say that that is 1 divided by 2 the
mass is 4
and the velocity at a is 3 meter per
second so that is
3 square
plus uh the elastic potential energy at
a
plus the gravitational potential energy
at a so for gravitational potential in
the energy at a we have to define our
datum line so our datum line is at b
so this is our datum line for the
gravitational potential energy
so now the elastic potential energy at a
let me write it here the elastic
potential energy at a will be equal to 1
divided by 2
k
and the change in length at e or we can
say that the stretch in the spring at a
so that is x a square
so is it is given that the unstretched
length is 1 meter
so
we can say that the original length of
the spring is 1 meter
and at a the stretch length is
at a the stretch length is 1.5 meters
from here we can say that this is 1.5
meters
so now we can say that the when the
caller is at a the stretch in the spring
is xa
which is which will be equal to this
stretch length at a let me write it as a
stretch length at a
minus the original length or the
unstretched length so that is 1.5 minus
1. so it gives us 0.5
0.5 meter so when the collar is at a the
spring is stretched
to
a length of 0.5 meters so we can say
that this is 1 divided by 2
and k is 100 this is given this is 100
so 100
and this is 0.5
squared
so
this is 50 into let me find it this is
100 divided by 2 is 50 so 50 multiplied
by 0.5 square this gives us
12.5 this is 12.5 joule
and similarly the gravitational
potential energy it is
always determined by
w times h and h is the distance from the
datum line
so weight is is we know that the mass is
0.4 multiplied by 9.81
and this is the h
and the s
this is that maximum distance let's say
that this s is
s max
so we have to find that as max so i will
write this as as
max
so this is 0.4 into 0.4 into
9.81
so this is
uh
3.924
so let me write that this is three point
3.924 as max
so
now we can write that this is this is
the elastic potential energy is at a is
12.5 joules
so we'll write it as 12.5
and remember that if this is the datum
line then here the the gravitational
potential energy is positive since we
are going in this direction and this is
the positive direction of the
gravitational potential energy
so the gravitational potential energy is
positive so this is plus 3.924
s max
and now
it is said that determine the maximum
distance as
it travels before it stops momentarily
so the velocity at b is
zero so the velocity at b zero then the
kinetic energy at b is zero as well so
we are left with the
elastic potential energy at b
and the gravitational potential energy
at b
so since the caller is at b and the
datum line is at v
so the height from the datum line is
zero so the gravitational potential
energy at datum is always zero so the
gravitational potential is zero as well
so this is zero so we are left with the
elastic potential energy
now
when the spring is at b we have to find
the stretch length again the original
length is one meter so the stretch
length at b will be equal to this
let's say if this point is also that
will be equal to this ob
length
so let me write that this is that ob
length
and that will be equal to we can find it
by using the pythagoras theorem since
this is the hypotenuse of this right
angled triangle so we can say that this
will be s max square
plus 1.5 square under the square root
and the stretchness in the spring at b
will be equal to the stretch this the
length of the spring at b
minus the unstretched length which is
the original length right so we can say
that this is as
max square plus 1.5 square
under the square root
minus
the unstretched length which is 1 meters
so this is the stretch in the spring
when the collar is at me so this is that
stretch in the spring so now uh
we can write this the elastic potential
energy at v will be equal to 1 divided
by 2
k k is 100
and that will be x xb
square
so now first let me simplify this this
is 4 divided by this is 2
and 2 into 9 is 18
so this is 18 plus
12.5 plus
3.924 as max
and that is 50
into xb square and this is xb so i will
write it as
s max square plus 1.5
square
minus 1
and this is squared since this is xb
square
so now we can further simplify this is
18
plus
12.5 this gives us
30.5 this is
30.5 plus
3.924
s max
and then we can simplify this further
this is let me open up this square term
right
so that will be
50
and then this square since this is a
minus v squared formula so that will be
s max square plus 1.5 square under the
square root
and then the square of this
plus 1 squared
and then minus
2 into this into this
so that is s
max
plus
1.5
into 1. so we will get this right
so this is
30.5 plus 3.924
s max equals to
so this is 50
and then this will cancel out so we will
be left with s max
square plus
1.5 square plus
1 squared is 1 and then minus
2
s max
plus
1.5 under the square root
so
it can further be simplified as this is
30.5
plus
3.924
as max
and then this will be 50
if i multiply this 50 inside so that is
50 s max
square then 50 into 1.5 square so 50
multiplied by 1.5 squared
this gives us 112.5
and then 15 to 1 is 50
and minus
2
and this will be 50 into 2 so that is
100 right so this is 100
and s
max plus
1.5 now i can add this this is
112.5 plus 50
this gives us 162.5
so this is 162.5
and now if i bring this as these two
terms to the other side of equation
so this will be if i bring this to the
other side of equation so that will be
minus
50 as
max square plus
this
3.924 smx
and then this
constant this is 30.5
and then this will come to the other
side so that will be minus
162.5 and
this we will leave this on the right
hand side so this is minus 100
s max plus 1.5 and this is this is
remember that this is 1.5 squared is 1.5
squared
and this is also square this is max
squared as max squared
so further we can add this 30.5
30.5 minus
162.5 this gives us minus 132.
this is minus 132
and now if i multiply this whole
equation with minus sign so this will
become positive this will become
negative
this will become
positive
and this will become positive
now we we have done a small mistake here
this gravitational potential energy is
weight times h and the
mass is 4 kgs right so i have used this
as 0.4
so 4 times 9.81 is 39.24 this is this is
39.24 so
we need to correct ourselves this is
39.24
as max
this is 39.24
and similarly here we have that 39.24
this is also 39.5
to force ultimately we will get this is
39.24
so this is 39.24
so now we have this equation but
ah there is no direct solution for this
equation to find
the s max value
so to find the s max value we have to
use the heat and trial method we need to
plug in various values of x max let's
say if we start from s max
equals to 1 meter and then increment it
by
0.1 meter
and we find the values of the left hand
side and then we find the values of the
right hand side of this equation and
we find the value of s max for which the
left hand side is equal to the right
hand side so that will be our solution
so we can do it by a calculator but that
will take a lot of time
so i have made an excel sheet for this
and
we can find that solution very easily so
from excel sheet we can see that
uh if fs max is equal to 1 meter
so this is the left hand side where i
have inserted that equation this is 50
times s max square minus 39.24 s max
plus 132
so this is our left hand side and the
right hand side is 100 times
100 times square root of that so
when x max is equal to 1 left hand side
is
142.76 and the right hand side is 180
there is difference of
37
similarly when s max is 1.05 the
difference is this
so if we look into this data so
somewhere here is
very small difference that is point zero
point four one one ah
two nine so now we can plug in some
values
let's say if i write that this is one
point
nine
nine five six if you find the values for
nine five six
so that is
246.54 and now the difference will
become
a bit more smaller that is point
zero four but in fact it has increased
it is in the positive sense
now so we can say that approximately our
solution is between 1.95 and 1.956
so now let's say if i write that let's
say if i change this value to
1.952 let's see
so now the difference has been decreased
right so this is
minus 0.25 let's say if we write that
this is let's say
uh 1.953
so that is 0
minus 0.18 and let's say if we make it
point
1.954
so it has been increased again
okay so it is decreased right so that is
okay now if i write 1.955
so that is minus
0.02
and now if i
make it 1.956
so now if we look into this so the the
difference has been increased right so
we can say that this is our solution for
s max
this is our solution for s max
and now if i plot a graph between the s
max and the left hand side and the right
hand side
so if we plot a graph so they will
intersect where our solution exists so
this is the
plot of left hand side this the blue one
is left hand side and the orange one is
right hand side and they intersect at
1.955
so the s max value for this problem is
as max is
we can say that x max value the solution
is
1.955 meter so the distance
uh the caller will travel to have a
momentarily stop to have a velocity at
point b equals to zero so that distance
will be one point nine five five meters
so this is the solution of this
particular problem i hope this will help
you in your learning
do subscribe engineers academy if it
helps in your learning
also like this video for the solution of
such more problems from hebler dynamics
you
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