Angular Momentum Problem: Torque after fall

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Torque=change in angular momentum

I know how to get to the answer but that's what is confusing me.

To find final velocity I multiply the acceleration by the time the object fell.

Then multiply the velocity by the mass to get momentum.

Now the angular momentum is r x p.

Since the initial angular momentum was 0, this was also the change in angular momentum.

The problem is that to solve this youre not suppose to multiply by the time. You are suppose to do r*a*d*m to get the answer (599.56).

Why don't I need the momentum to find the angular momentum here??

 

[Charles Link]

Torque=rate of change in angular momentum w.r.t. time.
It is also $\vec{r} \times \vec{F}$, where the $\times$ is a vector cross product.
$m \cdot a \cdot d$ should be the correct answer, where $a=9.8 m/sec^2$. I don't get an $r$ in the product.
We, as homework helpers, generally aren't supposed to give the answer, but this one is almost one that involves a definition. Meanwhile, vector cross products may not have been covered yet in your coursework.

The problem gives you a lot of extra detail, but really doesn't give you much direction in how to solve it.

 

[JoeyBob]

Torque=rate of change in angular momentum w.r.t. time.
It is also $\vec{r} \times \vec{F}$, where the $\times$ is a vector cross product.
$m \cdot a \cdot d$ should be the correct answer, where $a=9.8 m/sec^2$. I don't get an $r$ in the product.
We, as homework helpers, generally aren't supposed to give the answer, but this one is almost one that involves a definition. Meanwhile, vector cross products may not have been covered yet in your coursework.

Yeah I meant d.

So torque here would be d x force of gravity.

But isn't torque also the change in angular momentum? Why does calculating the change in angular momentum give the wrong answer?

 

[Charles Link]

The torque multiplied by the time will give the change in angular momentum.
Unless my arithmetic is off, they don't have $h=at^2/2$, where $a=9.8$.
I get that $t \approx 19$ seconds, when it hits the ground.
Maybe they wanted you to simply conclude that at $t=15.8$ seconds, the object hadn't reached the ground yet.
In any case, the problem could be a little more clear to the reader, especially when I think it is supposed to be for a first year student.

 

[haruspex]

torque here would be d x force of gravity.

Yes.

isnt torque also the change in angular momentum?

No. A change in angular momentum is still an angular momentum, just as a change in velocity is dimensionally a velocity, not an acceleration.

Maybe they wanted you to simply conclude that at t=15.8 seconds, the object hadn't reached the ground yet.

It asks for the torque due to gravity, so whether it has hit the ground is irrelevant. I think they are just leaving open the possibility that the torque varies with time ("$\tau(t)$"), and it is up to the student to figure out that it doesn't.