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Circular motion Watch

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    I have no idea what the question means. How can the string length be same but the diagram show different lengths?
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    i think they mean the length from the centre to the first mass =

    the length from the first mass to the second mass ?
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    (Original post by 123PC)
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    I have no idea what the question means. How can the string length be same but the diagram show different lengths?
    2 identical strings. One is attached to the centre of the circle and the inner mass; the other is attached to the inner mass and the outer mass.
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    (Original post by ombtom)
    2 identical strings. One is attached to the centre of the circle and the inner mass; the other is attached to the inner mass and the outer mass.
    But the lengths of the two strings are the same?
    But if the lengths are the same, then shouldn't the tension in both be the same?
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    (Original post by 123PC)
    But the lengths of the two strings are the same?
    But if the lengths are the same, then shouldn't the tension in both be the same?
    As others have pointed out but to explain it further, consider the masses orbiting as in normal cases of circular motion. Since the question specifies that the angular speed is the same for both masses, because they have a same time period, we can write this down as,

    \omega^2= \dfrac{F}{mr}=\dfrac{T}{mr}, where T is the tension in the string. Now you should realise that the tension in the inner string results from its own orbiting mass and the mass attached to the other string that is twice the distance from the center of the orbit. Therefore, the tension in the outer string is independent of the mass attached to the end of the inner string. Writing these down we should get;

    \dfrac{T_1}{mr+m(2r)}=\dfrac{T_2  }{m(2r)}.

    Edit: Even though the strings are of equal lengths, the tension in them is not the same because one of them orbits at twice the distance from the center of the orbit, and from the force equation  F=m\omega^2r, the force is proportional to the distance from the orbit. Notice that we have used this equation rather than F=\dfrac{mv^2}{r} which would give the inverse relationship of force with respect to r. This is because in the question the angular velocity of both masses is constant, but not their tangential velocities.
 
 
 
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