geerky42
Physics > Rotation
Help Needed! (see attachment)



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geerky42
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shubhamsrg
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dw:1360138607944:dw

geerky42
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Ok, and?

shubhamsrg
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where are you stuck ?

shubhamsrg
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dw:1360138848396:dw

geerky42
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On torque. I don't know how to find torque.

geerky42
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\(\tau = F_\perp r\) , right?

shubhamsrg
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mg(3) = I (alpha)

geerky42
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I don't understand exactly what this formula means.

shubhamsrg
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I will be around the point where it is pivoted

shubhamsrg
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See its equivalent to F=ma

shubhamsrg
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here its torque = I alpha

shubhamsrg
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Can you calculate moment of inertia about the point where it is pivoted?

geerky42
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\[I = \dfrac{1}{3}m(6^2)\]Right?

shubhamsrg
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RIght.

geerky42
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Ok, just checking.
I'm starting to understand a little bit. Thanks.

shubhamsrg
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So
mg(3) = I (alpha)
simply calculate alpha

geerky42
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Yeah, I got it.

shubhamsrg
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Glad you did.

shubhamsrg
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Its all pretty similar to linear motion, just write net toruqe = I alpha

geerky42
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Ok, One more question. How can you find torque in this system?

geerky42
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@shubhamsrg

shubhamsrg
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Sorry for the late reply, was busy somewhere.

shubhamsrg
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So whats your progress in this ?

geerky42
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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
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dw:1360140460045:dw

geerky42
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Can you be a little more clarity?

shubhamsrg
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I remains same, right ?

geerky42
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Right.

geerky42
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But torque isn't, right?

shubhamsrg
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perpendicular distance changes

shubhamsrg
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dw:1360140731467:dw

shubhamsrg
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x is the new distance.
our eqn will be
mgx = I alpha

geerky42
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Ah, thanks!

shubhamsrg
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hmm.