14-84 Kinetics of Particle: Conservation of Energy Chapter 14: Hibbeler Dynamics | Engineers Academy

Summary

This 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

00:00

📚 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.

05:02

⚖️ 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}.

10:06

📊 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.

15:08

🔍 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.

Keywords

💡Conservation of Energy

The law of conservation of energy states that the total energy in an isolated system remains constant over time. In the video, this principle is applied to analyze the motion of a collar along a spring. The speaker uses the conservation of energy to equate the kinetic and potential energies at different points, demonstrating how energy transforms from kinetic to potential as the collar moves.

💡Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. In the video, the collar has kinetic energy when it is moving at a speed of 3 meters per second at point A. This kinetic energy is calculated using the formula 1/2 * mass * velocity squared, and is a crucial component in determining how far the collar will travel before coming to a stop.

💡Potential Energy

Potential energy refers to the energy stored in an object due to its position or configuration. In this context, the video discusses two types of potential energy: elastic potential energy, stored in the stretched spring, and gravitational potential energy, which depends on the height of the collar relative to a reference datum line. The conservation of energy equation includes both types of potential energy to find the maximum distance traveled by the collar.

💡Elastic Potential Energy

Elastic potential energy is the energy stored in an elastic object, such as a spring, when it is stretched or compressed. The video calculates the elastic potential energy at two points (A and B) by using the formula 1/2 * spring constant * stretch distance squared. The stretch distance changes as the collar moves, which is essential in determining the collar's motion and the point at which it stops.

💡Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, typically relative to a defined datum line. In the video, this energy is calculated as the product of the object's weight and its height above the datum line. The analysis of the collar’s motion considers changes in gravitational potential energy to find when the collar stops.

💡Spring Constant (k)

The spring constant, denoted as 'k', is a measure of the stiffness of a spring. It defines the relationship between the force applied to a spring and its resulting displacement. In the video, the spring constant is given as 100 N/m, and it is used in calculating the elastic potential energy at various points as the spring stretches or compresses due to the motion of the collar.

💡Datum Line

A datum line is a reference line from which measurements, such as height or distance, are made. In the video, the datum line is set at point B to calculate the gravitational potential energy of the collar. This line serves as the baseline for measuring how the collar’s height changes as it moves along the spring.

💡Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that relates the sides of a right-angled triangle. In the video, this theorem is applied to determine the length of the spring when the collar is at point B, by calculating the hypotenuse of a right triangle formed by the spring's original length and the horizontal distance traveled by the collar.

💡Trial and Error Method

The trial and error method is a problem-solving technique that involves testing various values until the correct solution is found. The video employs this method to determine the maximum distance (s_max) that the collar travels before stopping. Since the exact solution cannot be found analytically, different values of s_max are tested until the left and right sides of the energy equation match closely.

💡Momentarily Stops

When an object momentarily stops, it means that its velocity becomes zero for an instant before it may change direction or continue moving. In the video, the goal is to find the point where the collar, initially moving along the spring, momentarily stops, which corresponds to the maximum distance it travels (s_max). This point is where all the kinetic energy has been converted into potential energy.

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

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