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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443975149240:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Find 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443975353793:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1the center of mass is stationary and that thing is rotating about the center of mass with uniform velocity of \(v_0\) is it ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1am i interpreting it correctly..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1oh, 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. \(v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so \(v_{CM} = \frac{v_0}2 \hat i \)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1and I think we're getting very close to the crux of the problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ohk.. so it is also translating to the right as it rotates..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1if 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1in fact even the acceleration is variable as the direction of force varies

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so we got velocity of center of mass pretty cheaply next i guess we want to find the angular velocity

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\omega = \dfrac{2v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t)i+\frac{l}{2}\sin(\omega t)j\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so \(d_1(\pi l/v_0) = (\dfrac{v_0}{2} + \dfrac{l}{2})i\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1i mean there's one little thing: one part of the velocity helps it to translate, and the other to rotate.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so \(v_0/2\) is being used in translation and the rest is used for rotation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\omega = v_o/l \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so I think we're finally in the solution of the problem: try doing calculations in the frame of the center of mass.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443977480377:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1in the frame of the center of massdw:1443977558039:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1since tension is perpendicular to the direction of motion, this is uniform circular motion

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\omega = \dfrac{v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t\pi/2)i+\frac{l}{2}\sin(\omega t\pi/2)j\) so \(d_1(\pi l/v_0) = \dfrac{v_0}{2}i + \dfrac{l}{2}j\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1In frame of reference of center of mass, it should be \(d_1(\pi l/v_0) = 0i + \dfrac{l}{2}j\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1distance travelled = \(\pi l/2 + \frac{v_0}{2}*\pi l/v_0 = \pi l\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[s = \int \vec v dt\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1should get the same answer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443981626417:dw not necessarily, better to evaluate the integral yeah

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1hey, could you go back to the question i asked earlier yesterday about the multinomial interpretation of the starsandbars problem i understood why it holds but how do i solve a sample problem there

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Okay.. so it sems we need to split that into two integrals : \(0 \to 3\pi l/(4v_0)\) \(3\pi l/(4v_0)\to \pi l/v_0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{red}{*}\) \(\omega = \dfrac{v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}\color{red}{t}i + \frac{l}{2}\cos(\omega t\pi/2)i+\frac{l}{2}\sin(\omega t\pi/2)j\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1distance 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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