## AravindG 4 years ago consider the arrangement shown in fig The string is wrapped around a uniform cylinder which rolls without slipping .The other end of the string is passed over a massless ,frictionless pulley to a falling weight .determine acceleration of falling mass m in terms of only the mass of the cylinder M ,the mass m and g

1. AravindG

|dw:1327757617642:dw|

2. anonymous

i consider that both of the object have the same mass. F = ma F = mg , (gravitational force which pulling the square object) F = (m+m)a ,(grouping thses 2 object into the system) F/2m = a

3. JamesJ

No, that analysis isn't right. The cylinder has a moment of inertia, $I = \frac{Mr^2}{2}$ where $$r$$ is its radius. Hence if the cylinder has velocity $$v$$, then because $$v = \omega r$$ where \omega is the angular velocity. Thus the total kinetic energy is $KE_{linear} + KE_{rotational} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$ $= \frac{1}{2}Mv^2 + \frac{1}{2}Mr^2\frac{v^2}{r^2}$ $= Mv^2$ Now, given that, you can use conservation of energy to find the energy of the cylinder when the weight $$m$$ has fallen a distance $$h$$. From that, use the Work-Energy Theorem and the fact $\Delta U = \Delta KE_{block} + \Delta PE_{m} = Work = \int F \cdot dy = \int ma \ dy$ to find the acceleration $$a$$ on the block.

4. anonymous

james why arent u bringing tension oor gravity into the poicture?

5. anonymous

James is correct, though I believe to correct a small typo, the rotational kinetic energy should be $\frac{1}{2} I \omega^2 = \frac{1}{2}\left(\frac{Mr^2}{2}\right)v^2r^2 = \frac{1}{4}Mv^2$ so the total kinetic energy is $\frac{3}{4}Mv^2$

6. anonymous

@salini we automatically include tension when we equate the velocity of the center of mass of the cylinder with the velocity of the hanging weight, and the F in the integral happens to be the gravitational force.

7. anonymous

when we equate the velocity and acceleration*

8. AravindG

wndful

9. anonymous

And I also had yet another typo, sorry.... $\omega^2 = \frac{v^2}{r^2} \text{ not } v^2r^2$ :)

10. JamesJ

(@jem, thanks)

11. AravindG

i hav a doubt

12. AravindG

u see in eqns of moment of inertia like mr^2,mr^2/2

13. AravindG

what is the r basically??

14. anonymous

r is the radius of the cylinder. The moments of inertia of differently shaped objects are not the same. For a disc and a solid cylinder, I = mr^2/2 (assuming we're rotating about the axis poking through the radial center), whereas for a hoop or a hollow cylinder I = mr^2.

15. anonymous

total energy is k.e+p.e right so y arent u guys including that anywhere ?

16. AravindG

|dw:1327769992982:dw|

17. AravindG

|dw:1327770030252:dw|

18. AravindG

i rotate about axis q then r is??

19. AravindG

|dw:1327770065239:dw|

20. anonymous

@Aravind that's a more complicated moment of inertia, so it's not as simple to write down.

21. anonymous

change in kinetic energy=work done by net force=m*a

22. AravindG

hey actually i did that qn

23. anonymous

no. m*a is not the work done. It is the force.

24. AravindG

pls tell the sides are made of rods

25. anonymous

in ther question they have asked for a in terms of M,m.g but we have to include displacement here

26. AravindG

wat is momentr of inertia?

27. anonymous

You have to do an integral to find that. Post a new question about calculating moment of inertia if you'd like to know how.

28. anonymous

And no we don't, not for the acceleration.

29. AravindG

hey but salini gave me the answr 3mr^2

30. anonymous

when?

31. AravindG

hey we did that remember?

32. anonymous

is athat an equilateral triangle?

33. AravindG

ys

34. AravindG

made of uniform rods of length l

35. AravindG

q is centre of mass

36. anonymous

then use trignometry to find aq

37. AravindG

axis is q

38. AravindG

aq is l/root 3 then??

39. anonymous

3*m*(aq)^2

40. AravindG

that is 3 ml^2

41. AravindG

srry ml^2

42. anonymous

annu

43. AravindG

u gud in maths?

44. AravindG

salini?

45. anonymous

arayila enniku orakam verunu anywaY PARA NOOKATE

46. AravindG
47. anonymous

Using Forces, Tr = Mr^2 / 2 * a / 2r ----- where a is acceleration of block which will be double of cylinder. And T is Tension. => T = Ma/4 Now plugging in equation for block, mg - T = ma mg - Ma/4 = ma Therefore, a = 4mg / ( M + 4m)

48. anonymous

siddhantsharan, why is Tr = Mr^2 / 2 * a / 2r that is I*a/2r why? and how?

49. anonymous

Salini, In the fig. the string is attached to top of the sphere. From rotation dynamics we know that acceleration of the point at the top of sphere = a (COM) + (angular Acceleration) / r = a + a ( pure rolling => a = Angular acceleration / r) = 2a. Where a is acceleration of COM of sphere which is what we consider the sphere's acceleration to be during Force Concept. Now along the string Acceleration has to be equal. Hence, a ( top of sphere) = a (block) Therefore a(block) = 2a. In the answer I have assumed a to be acceleration of block and solved for a. Therefore Acceleration of sphere = a /2 Cheers. :)

50. anonymous

|dw:1327926218152:dw|@ siddhantsharan your 3rd step: acce of sphere = a (COM) + (angular Acceleration) / r

51. anonymous

Sorry. Typo error. Answer is still the same.

52. anonymous

YOu got the concept right?

53. anonymous

Tr = Mr^2 / 2 * a / 2r here, is a/2r the angular acceleration of cylinder if so T=M*a*r/4 not Ma/4

54. anonymous

Wait I'll draw it for clarity.|dw:1327878291024:dw|

55. anonymous

last doubt.....how did u get T*r ?

56. anonymous

|dw:1327880069640:dw| Hence , Force on sphere is Tension. Torque of Tension = r x F = r x T Hence Torque is Tr.