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

[rockinwhiz]

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!

 

[PeroK]

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.

 

[rockinwhiz]

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?

 

[PeroK]

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?

 

[rockinwhiz]

Not a time derivative. Think calculus. How do you maximise a function of a single variable?

A derivative with respect to $v$ here?

 

[PeroK]

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?

 

[rockinwhiz]

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?

 

[PeroK]

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.

 

[rockinwhiz]

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!

 

[etotheipi]

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

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 $\dot{K}_E$?

 

[rockinwhiz]

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
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!

 

[PeroK]

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.

 

[PeroK]

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.

 

[rockinwhiz]

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.

 

[rockinwhiz]

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.

 

[etotheipi]

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.

 

[haruspex]

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).

 

[etotheipi]

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.

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.$$\vec{F} = -kv^2 \vec{t}(s)$$

 

[PeroK]

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.

 

[haruspex]

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.