How does the velocity of a ball change without buoyant force acting on it?

[PSN03]

Homework Statement

A small spherical ball (obeying stokes law for viscous force) is thrown up vertically with a speed of 20m/s and reaches back to the observer with a speed of 10m/s. Neglecting the buoyant force on the ball, assuming that the ball never attains it's terminal velocity during the flight, find the time of flight of the ball.

Relevant Equations

Weight of block=mg
Acceleration of the block=ma

So my doubt is at the beginning of the problems hey are saying that the ball obeys stokes law and on the latter part of the question they are saying that no buoyant force is acting then how does the velocity of the ball change in the end?
Also what is the use of specifying 'the ball never attains it's terminal velocity'? What will happen if it does and will the question change?

 

[kuruman]

What does Stokes' law say? Start from there, write the equation of motion and see if you can solve it. The buoyant force is neglected to make the solution simpler. The significance of it is good to know that the ball never attains its terminal velocity will become apparent when you solve for the time required to go up.

 

[PSN03]

What does Stokes' law say? Start from there, write the equation of motion and see if you can solve it. The buoyant force is neglected to make the solution simpler. The significance of it is good to know that the ball never attains its terminal velocity will become apparent when you solve for the time required to go up.

Lemme try it then...I guess I need some time

 

[kuruman]

I would find an expression for the terminal velocity and substitute that in the equations of motion. It simplifies the algebra.

 

[PSN03]

What does Stokes' law say? Start from there, write the equation of motion and see if you can solve it. The buoyant force is neglected to make the solution simpler. The significance of it is good to know that the ball never attains its terminal velocity will become apparent when you solve for the time required to go up.

Ohk so I got this
F=6πnrv
F=weight of ball- buoyant force
Buoyant force=0 ,
Hence
6πnrv=weight of ball
=Mg
Is this correct?

 

[kuruman]

Is the ball accelerating when it is thrown straight up? If so, what is its acceleration? What about the acceleration when the ball is on its way down?

 

[PSN03]

Is the ball accelerating when it is thrown straight up? If so, what is its acceleration? What about the acceleration when the ball is on its way down?

So should I write force as F=ma for upward motion of the body instead of equating it to F=6πnrv?

 

[kuruman]

Yes. Repeat for downward motion because the viscous force changes direction.

 

[PSN03]

Yes. Repeat for downward motion because the viscous force changes direction.

Ohk so the new equation would be
F=weight of body
Ma=Mg

Now my doubt is the RHS side will always be constant, hence the acceleration on the LHS side will also be constant. If everything is constant then what is causing the change in velocity

 

[kuruman]

What happened to the viscous force in the RHS?

 

[PSN03]

What happened to the viscous force in the RHS?

It became zero cause it's given there's no buoyant force.

 

[PSN03]

Ohk so the new equation would be
F=weight of body
Ma=Mg

Now my doubt is the RHS side will always be constant, hence the acceleration on the LHS side will also be constant. If everything is constant then what is causing the change in velocity

Aren't we just equating the sum of all force to M*a. Where are we even using the stokes law?

 

[kuruman]

You are confusing the viscous force which, opposes the motion of an object through a fluid and obeys Stokes' law with the buoyant force which provides a constant force against gravity and obeys Archimedes's principle. Stokes' law allows you to find an expression for the viscous force. I suggest that you look up Stokes' law and viscosity to set your mind straight before proceeding.

 

[PSN03]

You are confusing the viscous force which, opposes the motion of an object through a fluid and obeys Stokes' law with the buoyant force which provides a constant force against gravity and obeys Archimedes's principle. Stokes' law allows you to find an expression for the viscous force. I suggest that you look up Stokes' law and viscosity to set your mind straight before proceeding.

I am really sorry...I misunderstood things
So what I have understood now
Weight- drag force=net force
mg-6πnrv=ma

Now a=dv/dt
mg-kv=mdv/dt Where k=6πnrv
Now when I tried integrating this thing I am ending up with a different answer.
What do to now

 

[kuruman]

Is this for the motion going up or going down? Please write the differential equation in the form
$$m\frac{dv}{dt}=\dots$$
Also,

when I tried integrating this thing I am ending up with a different answer

is not very informative. What did you try? What answer did you get? You say your answer is different, different from what? You must show your work to receive help. Ideally, you should try to use LaTeX. Click on "LaTeX Guide" over "Attach Files" (left side of screen near bottom) to learn how.

 

[PSN03]

Is this for the motion going up or going down? Please write the differential equation in the form
$$m\frac{dv}{dt}=\dots$$
Also,

is not very informative. What did you try? What answer did you get? You say your answer is different, different from what? You must show your work to receive help. Ideally, you should try to use LaTeX. Click on "LaTeX Guide" over "Attach Files" (left side of screen near bottom) to learn how.

My attempt:
-(mg +kv)=m dv/dt (considering upward direction as positive)
-(mg dt + kv dt)=mdv
Integrating the above terms : integrating t from 0 to T and v from U to V where
T is toal time of flight
V=final velocity
U=initial velocity

mg*T+ (integral of viscous force term)=m(V-U)
Now I was wondering if I can write v as dx/dt to get kdx and it's interal would be zero for the whole journey. Am I right in doing this? Cause my final answer is now coming right.

 

[kuruman]

-(mg +kv)=m dv/dt (considering upward direction as positive)

This is the equation for the "going up" part of the motion. There is a separate equation for the "going down" part of the motion because the viscous force changes direction relative to the acceleration of gravity. You need to write two separate equations and do two separate integration. I strongly suggest[/color] that you define as positive the direction of motion in each case otherwise you will most likely get into trouble with signs. Then "v" stands for "speed" not velocity which is good because then the viscous force will always be "-kv".

 

[PSN03]

This is the equation for the "going up" part of the motion. There is a separate equation for the "going down" part of the motion because the viscous force changes direction relative to the acceleration of gravity. You need to write two separate equations and do two separate integration. I strongly suggest[/color] that you define as positive the direction of motion in each case otherwise you will most likely get into trouble with signs. Then "v" stands for "speed" not velocity which is good because then the viscous force will always be "-kv".

I think the problem is solved now. I can do after this I think. Thanks for all your help. Good day