Variable acceleration and terminal velocity

[abdo799]

Homework Statement

A cube is falling from a height of 200 m (initial velocity =0) , it reaches it's terminal velocity . Given that :its mass is 100 kg , cross sectional area 40 cm² and coefficient of drag 2. Putting only weight and drag in consideration , calculate its terminal velocity and time of the fall (g=10)

Homework Equations

W=mg , F_d= C_d*A*1/2*ρ*v² ( ρ
=1)

The Attempt at a Solution

it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , F_r=W-F_D, after working it out , it was something like this , a=(100)/(1000-(0.08 v²) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time

 

[NihalSh]

it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , F_r=W-F_D, after working it out , it was something like this , a=(100)/(1000-(0.08 v²) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time

$$a=\frac{dv}{dt}=\frac{dv}{dt}.\frac{dx}{dx}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dx}{dt}.\frac{dv}{dx}=v.\frac{dv}{dx}$$

Now, you have relation with which you don't need a function of time. And acceleration is a function of velocity.

Edit: check your acceleration again, it should be a little different.

 

[Tanya Sharma]

Homework Statement

A cube is falling from a height of 200 m (initial velocity =0) , it reaches it's terminal velocity . Given that :its mass is 100 kg , cross sectional area 40 cm² and coefficient of drag 2. Putting only weight and drag in consideration , calculate its terminal velocity and time of the fall (g=10)

Homework Equations

W=mg , F_d= C_d*A*1/2*ρ*v² ( ρ
=1)

The Attempt at a Solution

it tried to solve it using calculus , but all the formulas i had was integration or differentiation in respect to time , and i didnt have a function of time . i figured out a relation between velocity and acceleration , F_r=W-F_D, after working it out , it was something like this , a=(100)/(1000-(0.08 v²) , i don't know what to do next , i don't know if i can substitute v with some something to get a function time

Recheck value of F_d.

Your expression of net acceleration also looks incorrect .

mdv/dt = F_d-mg .

Substituting value of F_d ,you will get mdv/(F_d-mg) =dt .

Integrate with proper limits and you will get velocity as a function of time.What do you get ?

 

[NihalSh]

Integrate with proper limits and you will get velocity as a function of time.What do you get ?

He'll need to calculate velocity as a function of distance first, otherwise "velocity as a function of time" information would be useless.

But yes, he does need to calculate "velocity as a function of time".

 

[abdo799]

$$a=\frac{dv}{dt}=\frac{dv}{dt}.\frac{dx}{dx}=\frac{dv}{dx}.\frac{dx}{dt}=\frac{dx}{dt}.\frac{dv}{dx}=v.\frac{dv}{dx}$$

Now, you have relation with which you don't need a function of time. And acceleration is a function of velocity.

Edit: check your acceleration again, it should be a little different.

okay , this is how i got a
F= m/a
100/a = (100*10)-(1/2 * 1 * 2 * (0.4²) v²)
100/a= 1000- (0.16 v²)
a=100/(1000-0.16v²)
if it's wrong please correct me

 

[Tanya Sharma]

He'll need to calculate velocity as a function of distance first, otherwise "velocity as a function of time" information would be useless.

But yes, he does need to calculate "velocity as a function of time".

I believe distance doesn't play a role in calculating terminal velocity ,unless you want to determine if the object achieves terminal velocity before reaching the ground.

 

[NihalSh]

okay , this is how i got a
F= m/a
100/a = (100*10)-(1/2 * 1 * 2 * (0.4²) v²)
100/a= 1000- (0.16 v²)
a=100/(1000-0.16v²)
if it's wrong please correct me

Hey, $F=m*a$.

Tanya already posted about this!...understanding the situation is more important. But here is the result:

taking +ve direction to be downwards, we have:

$$mg-F_{d}=ma$$
$$a=g-\frac{F_{d}}{m}$$
Substitute the correct values, you'll get acceleration as a function of velocity.

 

[NihalSh]

I believe distance doesn't play a role in calculating terminal velocity ,unless you want to determine if the object achieves terminal velocity before reaching the ground.

terminal velocity is just the condition when the object is in equilibrium. $F_{d}=mg$ for the terminal velocity in this question. But the question asks the time of fall, not the time to reach terminal velocity. Time of fall is surely enough, dependent on the distance.

I hope you get the point!

Edit: Yes, distance doesn't play important role here to calculate terminal velocity. But I was talking about velocity in my previous posts. And its time of fall depends on velocity and not terminal velocity. But to calculate final velocity and hence calculate time, we need "velocity as a function of distance"......to reach 100% of terminal velocity you need infinite/very large time, because velocity reaches terminal velocity asymptotically with time.

 

[abdo799]

Hey, $F=m*a$.

Tanya already posted about this!...understanding the situation is more important. But here is the result:

taking +ve direction to be downwards, we have:

$$mg-F_{d}=ma$$
$$a=g-\frac{F_{d}}{m}$$
Substitute the correct values, you'll get acceleration as a function of velocity.

worst mistake I've ever made , sorry , i drank like 5 cups of coffee , didnt sleep , got a test in 2 days

 

[NihalSh]

worst mistake I've ever made , sorry , i drank like 5 cups of coffee , didnt sleep , got a test in 2 days

get some sleep, sleepiness and physics don't go hand in hand!

 

[abdo799]

get some sleep, sleepiness and physics don't go hand in hand!

Will do, thanks

 

[NihalSh]

Will do, thanks

:thumbs: