## geerky42 3 years ago Physics -> Rotation Help Needed! (see attachment)

1. geerky42

2. shubhamsrg

|dw:1360138607944:dw|

3. geerky42

Ok, and?

4. shubhamsrg

where are you stuck ?

5. shubhamsrg

|dw:1360138848396:dw|

6. geerky42

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

7. geerky42

$$\tau = F_\perp r$$ , right?

8. shubhamsrg

mg(3) = I (alpha)

9. geerky42

I don't understand exactly what this formula means.

10. shubhamsrg

I will be around the point where it is pivoted

11. shubhamsrg

See its equivalent to F=ma

12. shubhamsrg

here its torque = I alpha

13. shubhamsrg

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

14. geerky42

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

15. shubhamsrg

RIght.

16. geerky42

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

17. shubhamsrg

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

18. geerky42

Yeah, I got it.

19. shubhamsrg

20. shubhamsrg

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

21. geerky42

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

22. geerky42

@shubhamsrg

23. shubhamsrg

Sorry for the late reply, was busy somewhere.

24. shubhamsrg

So whats your progress in this ?

25. 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 $$F_\perp$$.

26. shubhamsrg

|dw:1360140460045:dw|

27. geerky42

Can you be a little more clarity?

28. shubhamsrg

I remains same, right ?

29. geerky42

Right.

30. geerky42

But torque isn't, right?

31. shubhamsrg

perpendicular distance changes

32. shubhamsrg

|dw:1360140731467:dw|

33. shubhamsrg

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

34. geerky42

Ah, thanks!

35. shubhamsrg

hmm.