Suppose there is a little block attached to a cord passing over apulley through

Question

Suppose there is a little block attached to a cord passing over apulley through a hole in the
show all calculations and work ...THANKS!
Suppose there is a little block attached to a cord passingover a pulley through a hole in the horizontal frictionlesssurface. At first the little block is revolving with an initialangular speed of 6rad/sec at a distance r from the center of thathole. So if this chord is pulled from below until the new radius isr/6, then what will be the new angular velocity?

Show all calculations and work ...THANKS!
w=work andw(i)=initial work r= radius m=mass
Hints given: L= r*p-  = r*m *v(1)
v- =w*r
L(i)=r(i)*m*w(i)*r(i)
Eventually masses will cancel out so left withr(i)^2*w(i)--> solve for w(i)

L(f)=r(f)*m*w(f)*r(f)
Likewise mass will cancel out left r(f)^2*w(f)
Then plug in the values into equation L(i)=L(f) (setthem equal to each other to cancel out masses that way)
Conservation of angular momentum L(i)=L(f) =L(0) -->the angular momentum constant
Solve forW(f)

show all calculations and work ...THANKS!
Suppose there is a little block attached to a cord passingover a pulley through a hole in the horizontal frictionlesssurface. At first the little block is revolving with an initialangular speed of 6rad/sec at a distance r from the center of thathole. So if this chord is pulled from below until the new radius isr/6, then what will be the new angular velocity?

Show all calculations and work ...THANKS!
w=work andw(i)=initial work r= radius m=mass
Hints given: L= r*p-  = r*m *v(1)
v- =w*r
L(i)=r(i)*m*w(i)*r(i)
Eventually masses will cancel out so left withr(i)^2*w(i)--> solve for w(i)

L(f)=r(f)*m*w(f)*r(f)
Likewise mass will cancel out left r(f)^2*w(f)
Then plug in the values into equation L(i)=L(f) (setthem equal to each other to cancel out masses that way)
Conservation of angular momentum L(i)=L(f) =L(0) -->the angular momentum constant
Solve forW(f)

Show all calculations and work ...THANKS!
w=work andw(i)=initial work r= radius m=mass
Hints given: L= r*p-  = r*m *v(1)
v- =w*r
L(i)=r(i)*m*w(i)*r(i)
Eventually masses will cancel out so left withr(i)^2*w(i)--> solve for w(i)

L(f)=r(f)*m*w(f)*r(f)
Likewise mass will cancel out left r(f)^2*w(f)
Then plug in the values into equation L(i)=L(f) (setthem equal to each other to cancel out masses that way)
Conservation of angular momentum L(i)=L(f) =L(0) -->the angular momentum constant
Solve forW(f)

Explanation / Answer

We know that                              v = r So that we have                              1 / r Then we can write                             2 / 1 = r1 /r2 ==> 2 = 1 *r1 / r2                 = 6 * r / r /6                 = 36 rad /s

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