Understanding Negative Torque and Angular Acceleration in Harmonic Movement

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transgalactic said:
we got a rock which has a smaller rock inside, we tilt it by a small angle.

why the torques is negative
why they do
-torque=I*angular accelarations

Hi transgalactic! :smile:

No, the equation given is -∆mgR/2 sinθ = (I1 + I2)θ''

and the left hand side is the torque (not minus the torque) … that's the way θ is defined …

the rock inside pulls down, decreasing θ. :wink:
 
but the this torque is going counter clock wise
it should be taken as positive.
i can't understand what it has to do with

minus the torque) … that's the way θ is defined …

the rock inside pulls down, decreasing θ. :wink:

can you explain this part..
 
transgalactic said:
but the this torque is going counter clock wise

No, the torque of the inside rock about the centre is clockwise. :smile:
 
you are correct
but the problem is if the rock tilts to the left side
the solution still gives a minus sign
despite the fact that the torque is positive
i was told that its because of the sinus graph
?
 
transgalactic said:
you are correct
but the problem is if the rock tilts to the left side
the solution still gives a minus sign
despite the fact that the torque is positive
i was told that its because of the sinus graph
?

I'm not following you :confused:

if the inside rock tilts to the left side, then sinθ is negative, so the equation means the angular acceleration must be positive … which it is! :smile:

yes, the torque is the same sign as the angular acceleration (anti-clockwise and positive) …

it will be in any apparatus …

the equation doesn't say "negative torque = acceleration", it says "torque = acceleration", but the formula for the torque just happens to have a minus in it.
 
i don't use negative sinx
i don't have such thing
i just take the hypotenuse multiply it by sinus theta *Mg
and then look if its goes clock wise or counter clockwise.

if the rock leans to the left and we loo from the point the rock touches the ground
Mg*sin theta=Ma^2

but in another solution it states
-Mg*sin theta=Ma^2

why?

how to get the minus using the method i use??
 
Patient: it hurts when I do that.
Doctor: then don't do that! :biggrin:
transgalactic said:
i don't use negative sinx
i don't have such thing
i just take the hypotenuse multiply it by sinus theta *Mg
and then look if its goes clock wise or counter clockwise.

how to get the minus using the method i use??

erm … don't use your method!

don't "look if its goes clock wise or counter clockwise"!

physics is equations …

if x is negative, then sinx is negative

you can't write "just look at the diagram and you can see it's clockwise" in the middle of an exam proof! :smile:
 
but the angle is a parameter
and if x is negative
then sinx will give us a negative value

we can't say
-sinx

its just adding a minus to the sinus
 
  • #10
can you explain this stuff??
 
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