ParthKohli
  • ParthKohli
Good question, I guess.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
|dw:1443975149240:dw|
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
|dw:1443975353793:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
@ganeshie8
ParthKohli
  • ParthKohli
@imqwerty
ganeshie8
  • 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 ?
ganeshie8
  • ganeshie8
am i interpreting it correctly..
ParthKohli
  • 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. \(v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}\)
ParthKohli
  • ParthKohli
so \(v_{CM} = \frac{v_0}2 \hat i \)
ParthKohli
  • ParthKohli
and I think we're getting very close to the crux of the problem
ganeshie8
  • ganeshie8
Ohk.. so it is also translating to the right as it rotates..
ParthKohli
  • ParthKohli
exactly...
ParthKohli
  • 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
ParthKohli
  • ParthKohli
in fact even the acceleration is variable as the direction of force varies
ganeshie8
  • ganeshie8
so we got velocity of center of mass pretty cheaply next i guess we want to find the angular velocity
ganeshie8
  • ganeshie8
\(\omega = \dfrac{2v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t)i+\frac{l}{2}\sin(\omega t)j\)
ganeshie8
  • ganeshie8
so \(d_1(\pi l/v_0) = (\dfrac{v_0}{2} + \dfrac{l}{2})i\)
ParthKohli
  • ParthKohli
i mean there's one little thing: one part of the velocity helps it to translate, and the other to rotate.
ParthKohli
  • ParthKohli
so \(v_0/2\) is being used in translation and the rest is used for rotation
ParthKohli
  • ParthKohli
\[\omega = v_o/l \]
ParthKohli
  • ParthKohli
so I think we're finally in the solution of the problem: try doing calculations in the frame of the center of mass.
ganeshie8
  • ganeshie8
|dw:1443977480377:dw|
ParthKohli
  • ParthKohli
in the frame of the center of mass|dw:1443977558039:dw|
ParthKohli
  • ParthKohli
since tension is perpendicular to the direction of motion, this is uniform circular motion
ganeshie8
  • 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 \(d_1(\pi l/v_0) = \dfrac{v_0}{2}i + \dfrac{l}{2}j\)
ganeshie8
  • ganeshie8
In frame of reference of center of mass, it should be \(d_1(\pi l/v_0) = 0i + \dfrac{l}{2}j\)
ParthKohli
  • ParthKohli
yes, correct!
ganeshie8
  • ganeshie8
distance travelled = \(\pi l/2 + \frac{v_0}{2}*\pi l/v_0 = \pi l\)
ParthKohli
  • ParthKohli
\[s = \int |\vec v| dt\]
ganeshie8
  • ganeshie8
should get the same answer
ganeshie8
  • ganeshie8
right ?
ParthKohli
  • ParthKohli
I'm not sure.
ganeshie8
  • ganeshie8
|dw:1443981626417:dw| not necessarily, better to evaluate the integral yeah
ParthKohli
  • ParthKohli
indeed
ParthKohli
  • 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
ganeshie8
  • ganeshie8
Okay.. so it sems we need to split that into two integrals : \(0 \to 3\pi l/(4v_0)\) \(3\pi l/(4v_0)\to \pi l/v_0\)
ganeshie8
  • ganeshie8
\(\color{red}{*}\) \(\omega = \dfrac{v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}\color{red}{t}i + \frac{l}{2}\cos(\omega t-\pi/2)i+\frac{l}{2}\sin(\omega t-\pi/2)j\)
ganeshie8
  • ganeshie8
distance travelled = \(|d_1(3\pi l/(4v_0)) - d_1(0)|+|d_1(\pi l/v_0) - d_1(3\pi l/(4v_0))|\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.