• entries
4
14
• views
1,674

Derivation Unknown

533 views

[ATTACH=CONFIG]61[/ATTACH]

So as I was chugging along on the Rotational Motion WebAssign I was startled to notice a seemingly coincidental relationship between my givens and my answer. But after, procrastinating longer than is healthy, trying it with other numbers, the relationship was consistent.

This pertains to question 2 on the WebAssign. I found that, for a record on a turntable with an initial rpm that slows with a constant angular acceleration until rest in time t in minutes, the number of revolutions the record makes before stopping x = (rpm)(t)/2.

So if, when the turntable is shut off, the record is rotating at 100 rpm and comes to a stop in 24 seconds or 0.4 minutes, the record will make (100)(0.4)/2 = 20 revolutions before coming to rest. You can take a step further to find the angular displacement θ by multiplying by 2π . From this we get θ = (rpm)(t)(π)

I am wondering whether there is a clear derivation and reason for this. After linking it to θ, I think I know why it works and makes sense sorta. It's finding an average. Really it should be Δrpm. It's still a little fuzzy to me.

Can you clear it up?

I'd like to speculate on this... cool find

$x=vt + \frac{1}{2}at^{2}$

change to rotational, If you assume starting at 0 and speeding up, the $\omega_{o}t$ term goes away.

$\theta = \frac{1}{2}\alpha t^{2}$

$(\alpha = \frac{\triangle \omega}{t}=\frac{\omega_{o}}{t})$

sub in...

$\theta = \frac{1}{2}\frac{\omega_{o}}{t}t^{2}$

$\theta = \frac{\omega t}{2}$

Groovy! That makes a lot of sense. Thanks for tying that down!

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
×
• Create New...