Question

Physics

Answer 1

https://www.khanacademy.org/science/high-school-physics/forces-and-newtons-laws-of-motion/newtons-second-law/v/newton-s-second-law-of-motion i hope this helps

Answer 2

what u solve for

Answer 3

net torque * time = L
where L is angular momentum

Fnet = m*a is newton's second law

Answer 4

Multiply to remove the faction then set equal to 0 and solve L=tt

Answer 5

What? That doesn't make any sense.

Answer 6

net torque = a force times distance applied from the AOR (axis of rotation)
L is measured in rotational inertia times angular velocity, units of mR^2 and w^2 where w = v/R
so angular momentum in mv^2
because mR^2*(v^2/R^2) = mv^2

Answer 7

i didnt mean to send that

Answer 8

actually I dont remember how to do this exactly off the top of my head, ill have to look at my notes which ill do after school ^.^

Answer 9

huh

Answer 10

mmh batman has got an idea, but not up the pole for what we're asked -- so this quite a wide topic so imma try and make it as short as possible -- since we are into angular momentum we know that \( L= I \times \omega \) AND \( L_0=I \times \omega \) right?

acoording to the 2nd law of newton we know that \( \sum \tau = I \times \alpha \) as mass is replaced by \( I\), and accleration by \( \alpha \).

to get a little deeper into the formula itself, we can break alpha into its formula:
\[ \sum \tau = I \times \frac{\Delta \omega}{\Delta t}\]
is it all clear till now?

Answer 11

Yes

Answer 12

Great!

so now we can write delta omega as the change of both omegas:
\[ \sum \tau = I \times \frac{ ( \omega-\omega_0)}{\Delta t} \]

that is the same as writing:
\[ \sum \tau = \frac{ I \times\omega- I \times\omega_0}{\Delta t} \]

clear?

Answer 13

another formula comes into the game, and that is \( \Delta L= I \times\omega- I \times\omega_0 \)

knowing that we can put that into the equation we were doing:

\( \sum \tau = \frac{\Delta L}{\Delta t} \Rightarrow \sum \tau = \frac{L-L_0}{\Delta t} \)

any idea what we are going to do now? look at the first reply i made.

Answer 14

We write delta t as t1-t0?

Answer 15

youre actually right!
\[ \sum \tau = \frac{\Delta L}{t-t_0} \]
\( t_0=0 \) right? thats why we write that expression as:
\[ \sum \tau = \frac{\Delta L}{t-\cancel{t_0}} \Rightarrow \sum \tau = \frac{\Delta L}{t} \]

remeber when i said that \( L_0=I \times \omega_0 \) ?

if \( \omega \) is 0, it means that \( L_0 \) is 0 as well

yes?

Answer 16

So we cancel l0 out just ike we did with t0?

Answer 17

yes, well done!

so we get:
\[ \sum \tau = \frac{L-L_0}{t} \Rightarrow \sum \tau = \frac{L-\cancel{L_0}}{t} \]

\[ \sum \tau = \frac{L}{t} \]

and here it is -- we have confirmed that!

Answer 18

Thanks