Mechanics: Determine the Acceleration of an Object in the Presence of Drag
'Determine the Acceleration of an Object in the Presence of Drag'
I've been going through the syllabus for my Mechanics retake, does anybody know what I should know for this/how to figure it out?
Initially I thought it would've been SUVAT but then realised that SUVAT was for constant acceleration
I've been going through the syllabus for my Mechanics retake, does anybody know what I should know for this/how to figure it out?
Initially I thought it would've been SUVAT but then realised that SUVAT was for constant acceleration
Original post by chloeintheskies
'Determine the Acceleration of an Object in the Presence of Drag'
I've been going through the syllabus for my Mechanics retake, does anybody know what I should know for this/how to figure it out?
Initially I thought it would've been SUVAT but then realised that SUVAT was for constant acceleration
I've been going through the syllabus for my Mechanics retake, does anybody know what I should know for this/how to figure it out?
Initially I thought it would've been SUVAT but then realised that SUVAT was for constant acceleration
It's Newton's 2nd Law as well.
F=ma and F is the resultant force.
Is that in the presence of ONLY drag?
In which case if there is no other force and the drag is constant then it's just F=ma where F is drag and a is deceleration.
If there are other forces (eg gravity) then you use the resultant of the two (or more).
For a falling object you have mg downwards and drag F upwards so
(mg -F) = ma
If the drag isn't constant (more than likely), and depends on the velocity of the object, then it's more complex.
Now F = kv (constant x velocity) possibly
Then your equation is -kv = ma and the acceleration depends on velocity.
a = -kv/m
This gives you a differential equation to solve.
a = d2x/dt2 and v = dx/dt
Edit
I should have added that, of course you are correct that you can only use the standard suvat equations for constant (uniform) acceleration.
In the cases above the 3rd, and most likely, example would not be constant acceleration.
Are you asking about a mathematical determination or an experimental determination?
(edited 10 years ago)
Original post by Stonebridge
It's Newton's 2nd Law as well.
F=ma and F is the resultant force.
Is that in the presence of ONLY drag?
In which case if there is no other force and the drag is constant then it's just F=ma where F is drag and a is deceleration.
If there are other forces (eg gravity) then you use the resultant of the two (or more).
For a falling object you have mg downwards and drag F upwards so
(mg -F) = ma
If the drag isn't constant (more than likely), and depends on the velocity of the object, then it's more complex.
Now F = kv (constant x velocity) possibly
Then your equation is -kv = ma and the acceleration depends on velocity.
a = -kv/m
This gives you a differential equation to solve.
a = d2x/dt2 and v = dx/dt
Edit
I should have added that, of course you are correct that you can only use the standard suvat equations for constant (uniform) acceleration.
In the cases above the 3rd, and most likely, example would not be constant acceleration.
Are you asking about a mathematical determination or an experimental determination?
F=ma and F is the resultant force.
Is that in the presence of ONLY drag?
In which case if there is no other force and the drag is constant then it's just F=ma where F is drag and a is deceleration.
If there are other forces (eg gravity) then you use the resultant of the two (or more).
For a falling object you have mg downwards and drag F upwards so
(mg -F) = ma
If the drag isn't constant (more than likely), and depends on the velocity of the object, then it's more complex.
Now F = kv (constant x velocity) possibly
Then your equation is -kv = ma and the acceleration depends on velocity.
a = -kv/m
This gives you a differential equation to solve.
a = d2x/dt2 and v = dx/dt
Edit
I should have added that, of course you are correct that you can only use the standard suvat equations for constant (uniform) acceleration.
In the cases above the 3rd, and most likely, example would not be constant acceleration.
Are you asking about a mathematical determination or an experimental determination?
can you please give me an example for when to use the last equation because in my as ocr physics textbooks it doesnt give me a worked example in detail
Original post by Elhamm
can you please give me an example for when to use the last equation because in my as ocr physics textbooks it doesnt give me a worked example in detail
For physics A Level you won't be expected to set up or solve differential equations like this one.
An example where it would apply would be free fall with drag.
The resultant downwards force on a falling mass M would be made up from
- its weight downwards Mg
-the drag force opposing its motion kv (or kv2)where k is a constant and v the velocity as it falls. This force increases as the object gets faster.
Resultant force Mg - kv
acceleration a is F/m
a = (Mg - kv)/M (using F = kv)
a = g - kv/M
The differential calculus is needed because, as you see here, the acceleration depends on the velocity but velocity depends on acceleration.
d2x/dt2 = g - (k/M) dx/dt
d2x/dt2 + (k/M) dx/dt - g = 0
This is a second order differential equation. You don't need to solve these for A level physics. (You might if you do maths, though. Ask your maths teacher if that is the case.)
The solution is exponential with the speed of the object increasing quickly at first but as it gets faster the acceleration gradually reduces to zero. The object then falls with no acceleration and has reached its terminal velocity.
You should be able to find this if you Google terminal velocity equation and solution to 2nd order differential equations. Remember, this is beyond A Level physics.
(edited 9 years ago)
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