geerky42
  • geerky42
Physics -> Rotation Help Needed! (see attachment)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
geerky42
  • geerky42
1 Attachment
shubhamsrg
  • shubhamsrg
|dw:1360138607944:dw|
geerky42
  • geerky42
Ok, and?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

shubhamsrg
  • shubhamsrg
where are you stuck ?
shubhamsrg
  • shubhamsrg
|dw:1360138848396:dw|
geerky42
  • geerky42
On torque. I don't know how to find torque.
geerky42
  • geerky42
\(\tau = F_\perp r\) , right?
shubhamsrg
  • shubhamsrg
mg(3) = I (alpha)
geerky42
  • geerky42
I don't understand exactly what this formula means.
shubhamsrg
  • shubhamsrg
I will be around the point where it is pivoted
shubhamsrg
  • shubhamsrg
See its equivalent to F=ma
shubhamsrg
  • shubhamsrg
here its torque = I alpha
shubhamsrg
  • shubhamsrg
Can you calculate moment of inertia about the point where it is pivoted?
geerky42
  • geerky42
\[I = \dfrac{1}{3}m(6^2)\]Right?
shubhamsrg
  • shubhamsrg
RIght.
geerky42
  • geerky42
Ok, just checking. I'm starting to understand a little bit. Thanks.
shubhamsrg
  • shubhamsrg
So mg(3) = I (alpha) simply calculate alpha
geerky42
  • geerky42
Yeah, I got it.
shubhamsrg
  • shubhamsrg
Glad you did.
shubhamsrg
  • shubhamsrg
Its all pretty similar to linear motion, just write net toruqe = I alpha
geerky42
  • geerky42
Ok, One more question. How can you find torque in this system?
1 Attachment
geerky42
  • geerky42
@shubhamsrg
shubhamsrg
  • shubhamsrg
Sorry for the late reply, was busy somewhere.
shubhamsrg
  • shubhamsrg
So whats your progress in this ?
geerky42
  • geerky42
I just need to do the same: find the angular acceleration. I'm not sure what to do with torque. I don't know how to find \(F_\perp\).
shubhamsrg
  • shubhamsrg
|dw:1360140460045:dw|
geerky42
  • geerky42
Can you be a little more clarity?
shubhamsrg
  • shubhamsrg
I remains same, right ?
geerky42
  • geerky42
Right.
geerky42
  • geerky42
But torque isn't, right?
shubhamsrg
  • shubhamsrg
perpendicular distance changes
shubhamsrg
  • shubhamsrg
|dw:1360140731467:dw|
shubhamsrg
  • shubhamsrg
x is the new distance. our eqn will be mgx = I alpha
geerky42
  • geerky42
Ah, thanks!
shubhamsrg
  • shubhamsrg
hmm.

Looking for something else?

Not the answer you are looking for? Search for more explanations.