When is the rate of change of Kinetic Energy maximum here?

In summary: Oh yeah...! Since, ##A=dv/dt=-(g+ \frac {kv} m)##, therefore, ##\dot {K}_E## becomes: $$ \dot {K}_E = -mv \cdot (g+ \frac {kv} m)$$ So, again, differentiating this with respect to ##v## gives ##-mg -2kv=0##, giving: $$ |v|= |- \frac {mg} {2k} | \approx 12.2 m/s$$
  • #1
rockinwhiz
28
6
Homework Statement
An experimenter throws a ball of mass ##m=1.0##kg vertically upwards with a velocity ##u=4.0m/s## from the top of a high tower. During the flight of the ball, modulus of the force of air resistance on the ball is given by the equation ##F=kv##, here ##k=0.41kg/s## and ##v## is the speed of the ball. The tower is so high that the ball achieves a constant speed before striking the ground. Find velocity of the ball, when its kinetic energy changes most rapidly. Acceleration due to gravity is ##g=10 m/s^2##
Relevant Equations
##K_E= \frac 1 2 mv^2##
##F= ma##
##a= \frac {dv} {dt}##
Because, ##F=ma=kv##, therefore, ##a=kv/m##. Clearly, the net acceleration ##A=-(g+a)##.
Also, ##A=dv/dt=-(g+ \frac {kv} m )##, so cross multiplying and integrating LHS with respect to ##v## and RHS with respect to ##t## gives me:
$$ v= e^{ \frac {-tk} m } * (u + \frac {gm} k) - \frac {gm} k $$
Now, I'm not sure, but differentiating this with respect to time should gives the time when rate of change of ##K_E## is maximum.
This is where I need help. How do I proceed? And how does the terminal velocity come into play?
FYI, the given answer is: ##v \approx 12.2 m/s##

Thanks!
 
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  • #2
rockinwhiz said:
Homework Statement:: An experimenter throws a ball of mass ##m=1.0##kg vertically upwards with a velocity ##u=4.0m/s## from the top of a high tower. During the flight of the ball, modulus of the force of air resistance on the ball is given by the equation ##F=kv##, here ##k=0.41kg/s## and ##v## is the speed of the ball. The tower is so high that the ball achieves a constant speed before striking the ground. Find velocity of the ball, when its kinetic energy changes most rapidly. Acceleration due to gravity is ##g=10 m/s^2##
Relevant Equations:: ##K_E= \frac 1 2 mv^2##
##F= ma##
##a= \frac {dv} {dt}##

Because, ##F=ma=kv##, therefore, ##a=kv/m##. Clearly, the net acceleration ##A=-(g+a)##.
Also, ##A=dv/dt=-(g+ \frac {kv} m )##, so cross multiplying and integrating LHS with respect to ##v## and RHS with respect to ##t## gives me:
$$ v= e^{ \frac {-tk} m } * (u + \frac {gm} k) - \frac {gm} k $$
Now, I'm not sure, but differentiating this with respect to time should gives the time when rate of change of ##K_E## is maximum.
This is where I need help. How do I proceed? And how does the terminal velocity come into play?
FYI, the given answer is: ##v \approx 12.2 m/s##

Thanks!
You don't have to calculate ##v(t)##. Think about what you have to maximise.
 
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  • #3
PeroK said:
You don't have to calculate ##v(t)##. Think about what you have to maximise.
Rate of change of kinetic energy would be:
$$ \frac {d K_E} {dt} \equiv \frac {d(mv^2/2) } {dt} \equiv mv \cdot \frac {dv} {dt}$$
So, I need to maximize this, most probably. So, should I take take another time derivative?
 
  • #4
rockinwhiz said:
Rate of change of kinetic energy would be:
$$ \frac {d K_E} {dt} \equiv \frac {d(mv^2/2) } {dt} \equiv mv \cdot \frac {dv} {dt}$$
So, I need to maximize this, most probably. So, should I take take another time derivative?
Not a time derivative. Think calculus. How do you maximise a function of a single variable?
 
  • #5
PeroK said:
Not a time derivative. Think calculus. How do you maximise a function of a single variable?
A derivative with respect to ##v## here?
 
  • #6
rockinwhiz said:
A derivative with respect to ##v## here?
Exactly. Note that you need to think about the difference between the upward and downward motion. Is the function the same throughout the whole motion?
 
  • #7
PeroK said:
Exactly. Note that you need to think about the difference between the upward and downward motion. Is the function the same throughout the whole motion?
So, if I'm not wrong, the kinetic energy keeps changing during upward motion. However, during downward motion, eventually the ball will eventually hit the terminal speed, so the kinetic energy will stop changing I guess?
 
  • #8
rockinwhiz said:
So, if I'm not wrong, the kinetic energy keeps changing during upward motion. However, during downward motion, eventually it'll eventually hit the terminal speed.
Yes, you may be able to see the shape of the curve without doing calculus. But, you'll need to use calculus to find the precise ##v## that maximises the function.
 
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  • #9
PeroK said:
Yes, you may be able to see the shape of the curve without doing calculus. But, you'll need to use calculus to find the precise ##v## that maximises the function.
I'll try this in a while and reply. Thanks!
 
  • #10
rockinwhiz said:
Because, ##F=ma=kv##, therefore, ##a=kv/m##. Clearly, the net acceleration ##A=-(g+a)##.

Strange of you to invoke the principle of superposition... (i.e. why not just write ##F_y = -kv_y - mg = ma_y \implies a_y = -(g +\frac{kv_y}{m})##), but fair enough :wink:

rockinwhiz said:
Rate of change of kinetic energy would be:
$$ \frac {d K_E} {dt} \equiv \frac {d(mv^2/2) } {dt} \equiv mv \cdot \frac {dv} {dt}$$
So, I need to maximize this, most probably. So, should I take take another time derivative?

Why not use your expression for ##\frac{dv_y}{dt}##, to eliminate all the time dependency in ##\dot{K}_E##?
 
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  • #11
etotheipi said:
Strange of you to invoke the principle of superposition... (i.e. why not just write ##F_y = -kv_y - mg = ma_y \implies a_y = -(g +\frac{kv_y}{m})##), but fair enough :wink:
Why not use your expression for ##\frac{dv_y}{dt}##, to eliminate all the time dependency in ##\dot{K}_E##?
Oh yeah...! Since, ##A=dv/dt=-(g+ \frac {kv} m)##, therefore, ##\dot {K}_E## becomes:
$$ \dot {K}_E = -mv \cdot (g+ \frac {kv} m)$$

So, again, differentiating this with respect to ##v## gives ##-mg -2kv=0##, giving:
$$ |v|= |- \frac {mg} {2k} | \approx 12.2 m/s$$
Thank you guys so much!
 
  • #12
rockinwhiz said:
Oh yeah...! Since, ##A=dv/dt=-(g+ \frac {kv} m)##, therefore, ##\dot {K}_E## becomes:
$$ \dot {K}_E = -mv \cdot (g+ \frac {kv} m)$$

So, again, differentiating this with respect to ##v## gives ##-mg -2kv=0##, giving:
$$ |v|= |- \frac {mg} {2k} | \approx 12.2 m/s$$
Thank you guys so much!
That's the right answer but you really should have ##g - \frac {kv} m## for the downward motion. Gravity and the resisting force are acting in opposite directions.
 
  • #13
PS there is an ambiguity in the problem. Does KE changing rapidly include negative change during the upward motion?

If so, you need to look at the maximum rate of change for the upward motion. This is clearly at the initial velocity where the air resistance is a maximum. If the ball is thrown up with sufficient speed, then this may be a greater rate of change than the maximum on the way down.

The problem may expect you to check this as well.
 
  • #14
PeroK said:
That's the right answer but you really should have ##g - \frac {kv} m## for the downward motion. Gravity and the resisting force are acting in opposite directions.
Oh yeah I missed that.
 
  • #15
PeroK said:
PS there is an ambiguity in the problem. Does KE changing rapidly include negative change during the upward motion?

If so, you need to look at the maximum rate of change for the upward motion. This is clearly at the initial velocity where the air resistance is a maximum. If the ball is thrown up with sufficient speed, then this may be a greater rate of change than the maximum on the way down.

The problem may expect you to check this as well.
Actually, the book this is from has generally ambiguous questions, so nothing is really specified as such. But I might look into that case too. However, for now, getting the answer is enough and I gained a lot of insights too.
Thanks to all of you.
 
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  • #16
The defining equation for the drag force is ##F = kv## (that is, ##|\vec{F}| = k|\vec{v}|##). You can easily switch this out for ##\vec{F} = -kv \frac{d\vec{r}}{ds}## where ##\frac{d\vec{r}}{ds}## is the tangent vector to the trajectory. Or more simply, ##F_y = -kv_y##.

So just replace every instance of ##v## in your dynamics with ##v_y## and now you don't need to worry about the change in the definition of the force at the turning point.
 
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  • #17
PeroK said:
That's the right answer but you really should have ##g - \frac {kv} m## for the downward motion.
Not if v is positive upwards. The equation doesn't suddenly change at the top of the trajectory (as it does for kv2 drag).
 
  • #18
haruspex said:
Not if v is positive upwards.

That is the notational inconsistency, since the question defines ##v## as the magnitude of the velocity. But it is easy enough to fix it by, like you say, re-interpreting it as the ##y## component of the velocity vector with ##\hat{y}## pointing upward.
haruspex said:
as it does for kv2 drag)
Yeah this is why we must always include the tangent vector in the definition of the drag force, i.e.$$\vec{F} = -kv^2 \vec{t}(s)$$
 
  • #19
haruspex said:
Not if v is positive upwards. The equation doesn't suddenly change at the top of the trajectory (as it does for kv2 drag).
I always take ##g## to be positive.
 
  • #20
PeroK said:
I always take ##g## to be positive.
It's not a question of the sign convention for g.
If we take v as positive up and g as having a positive value then for the upward motion
##m\dot v=-mg-kv##
and for the downward motion
##m\dot v=-mg-kv##
Same equation, but in the first case v is positive so -kv is negative, drag acting downwards; in the second, v is negative so -kv is positive and drag acts upwards.
 

Related to When is the rate of change of Kinetic Energy maximum here?

1. What is Kinetic Energy?

Kinetic Energy is the energy an object possesses due to its motion. It is calculated by the equation KE = 1/2mv^2, where m is the mass of the object and v is its velocity.

2. How is the rate of change of Kinetic Energy calculated?

The rate of change of Kinetic Energy, also known as the power, is calculated by the equation P = Fv, where F is the force acting on the object and v is its velocity.

3. Why is it important to know when the rate of change of Kinetic Energy is maximum?

Knowing when the rate of change of Kinetic Energy is maximum can help in understanding the efficiency of a system or the force required to accelerate an object. It can also be used to optimize the performance of machines and vehicles.

4. How can we determine when the rate of change of Kinetic Energy is maximum?

The rate of change of Kinetic Energy is maximum when the force acting on the object is at its maximum and the velocity is also at its maximum. This can be determined by analyzing the motion of the object and identifying the points where the force and velocity are at their highest.

5. What factors can affect the rate of change of Kinetic Energy?

The rate of change of Kinetic Energy can be affected by factors such as the mass of the object, the force acting on it, and the surface it is moving on. Other factors such as air resistance, friction, and external forces can also impact the rate of change of Kinetic Energy.

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