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## ParthKohli one year ago Good question, I guess.

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1. ParthKohli

|dw:1443975149240:dw|

2. ParthKohli

Find out the time $$t_0$$ in which the string rotates by $$\theta = \pi$$. Find $$\vec{d_1}$$ at $$t = \pi \ell / v_0$$. Find the distance travelled by $$1$$ from $$t = 0$$ to $$\dfrac{\pi \ell }{v_o}$$ Find the speed of $$2$$ as a function of time.

3. ParthKohli

|dw:1443975353793:dw|

4. ParthKohli

@ganeshie8

5. ParthKohli

@imqwerty

6. ganeshie8

the center of mass is stationary and that thing is rotating about the center of mass with uniform velocity of $$v_0$$ is it ?

7. ganeshie8

am i interpreting it correctly..

8. ParthKohli

oh, we're given that two masses are connected with a string and at $$t=0$$ the mass 1 is given the velocity $$v_0$$ (the other mass is not). if you think about it, the center of mass will not be stationary. $$v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$$

9. ParthKohli

so $$v_{CM} = \frac{v_0}2 \hat i$$

10. ParthKohli

and I think we're getting very close to the crux of the problem

11. ganeshie8

Ohk.. so it is also translating to the right as it rotates..

12. ParthKohli

exactly...

13. ParthKohli

if we observe the whole setting from the ground, we'll see that these things are accelerating and it'd be very hard to predict anything

14. ParthKohli

in fact even the acceleration is variable as the direction of force varies

15. ganeshie8

so we got velocity of center of mass pretty cheaply next i guess we want to find the angular velocity

16. ganeshie8

$$\omega = \dfrac{2v_0}{l}$$ $$d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t)i+\frac{l}{2}\sin(\omega t)j$$

17. ganeshie8

so $$d_1(\pi l/v_0) = (\dfrac{v_0}{2} + \dfrac{l}{2})i$$

18. ParthKohli

i mean there's one little thing: one part of the velocity helps it to translate, and the other to rotate.

19. ParthKohli

so $$v_0/2$$ is being used in translation and the rest is used for rotation

20. ParthKohli

$\omega = v_o/l$

21. ParthKohli

so I think we're finally in the solution of the problem: try doing calculations in the frame of the center of mass.

22. ganeshie8

|dw:1443977480377:dw|

23. ParthKohli

in the frame of the center of mass|dw:1443977558039:dw|

24. ParthKohli

since tension is perpendicular to the direction of motion, this is uniform circular motion

25. ganeshie8

$$\omega = \dfrac{v_0}{l}$$ $$d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t-\pi/2)i+\frac{l}{2}\sin(\omega t-\pi/2)j$$ so $$d_1(\pi l/v_0) = \dfrac{v_0}{2}i + \dfrac{l}{2}j$$

26. ganeshie8

In frame of reference of center of mass, it should be $$d_1(\pi l/v_0) = 0i + \dfrac{l}{2}j$$

27. ParthKohli

yes, correct!

28. ganeshie8

distance travelled = $$\pi l/2 + \frac{v_0}{2}*\pi l/v_0 = \pi l$$

29. ParthKohli

$s = \int |\vec v| dt$

30. ganeshie8

should get the same answer

31. ganeshie8

right ?

32. ParthKohli

I'm not sure.

33. ganeshie8

|dw:1443981626417:dw| not necessarily, better to evaluate the integral yeah

34. ParthKohli

indeed

35. ParthKohli

hey, could you go back to the question i asked earlier yesterday about the multinomial interpretation of the stars-and-bars problem i understood why it holds but how do i solve a sample problem there

36. ganeshie8

Okay.. so it sems we need to split that into two integrals : $$0 \to 3\pi l/(4v_0)$$ $$3\pi l/(4v_0)\to \pi l/v_0$$

37. ganeshie8

$$\color{red}{*}$$ $$\omega = \dfrac{v_0}{l}$$ $$d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}\color{red}{t}i + \frac{l}{2}\cos(\omega t-\pi/2)i+\frac{l}{2}\sin(\omega t-\pi/2)j$$

38. ganeshie8

distance travelled = $$|d_1(3\pi l/(4v_0)) - d_1(0)|+|d_1(\pi l/v_0) - d_1(3\pi l/(4v_0))|$$

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