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A bead of mass 'm' is connected at one end of an inextensible massless string. The other end of the string is fixed to a fixed cylinder. An initial impulse J is given to the bead perpendicular to the string. Bead moves on a curve and collides with the cylinder after some time.

Now in this question does the tension remain constant? For getting the answer to this question first we need to see whether velocity of the mass(perpendicular to the string) remains constant or not. If the velocity remains constant, then according to centripetal force we could say that as radius is changing, tension must change as tension=centripetal force. Further, does the work done by string on mass is zero? I think, it must be, because velocity of mass is always perpendicular to the string.

Thanks in advance.

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    $\begingroup$ I tried to get a conceptual doubt cleared, which will also help others on this site, to get a clear understanding of the concept, in accordance with the rules on the meta site. I do not get the reason behind the question being referred to as off-topic. $\endgroup$
    – ksr
    Commented Jan 12, 2016 at 9:47
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    $\begingroup$ I agree with @ksr - the question does not ask "do my homework", but tries to clear up confusion about work done on the mass and tension in the string - conceptual questions about the physics behind the problem. I think it is on topic and voted to reopen. $\endgroup$
    – Floris
    Commented Jan 12, 2016 at 13:42

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The x and y coordinates of the mass can be expressed parametrically in terms of the wrap angle $\theta$ (of the string around the cylinder) as follows: $$x=R\sin\theta+(L-R\theta)\cos\theta$$ $$y=-R\cos\theta+(L-R\theta)sin\theta$$ where x and y are measured from the center of the cylinder, R is the radius of the cylinder, and L is the initial length of string. If we take the time derivatives of these coordinates, we obtain the velocity components of the mass at any time: $$v_x=-(L-R\theta)\frac{d\theta}{dt}\sin\theta$$ $$v_y=(L-R\theta)\frac{d\theta}{dt}\cos\theta$$ Since kinetic energy is conserved in this system, the magnitude of the velocity vector v is constant, and, from the velocity components, we have $$v=(L-R\theta)\frac{d\theta}{dt}$$This equation can be integrated to get the wrap angle as a function of time.

If T represents the magnitude of the tension vector in the string, the components in the x and y directions, respectively, are $T_x=T\cos\theta$ and $T_y=T\sin\theta$. If we take the dot product of the tension vector with the velocity vector (using these components), we find that the tension vector is perpendicular to the velocity vector.

The force balance on the mass gives: $$m\frac{dv_x}{dt}=-T\cos\theta$$ $$m\frac{dv_y}{dt}=-T\sin\theta$$ These equations can be used to determine the tension T: $$T=mv\frac{d\theta}{dt}=m(L-R\theta)\left(\frac{d\theta}{dt}\right)^2=m\frac{v^2}{(L-R\theta)}$$

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    $\begingroup$ Very nicely done. $\endgroup$
    – Floris
    Commented Jan 13, 2016 at 18:30
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    $\begingroup$ Mutual Admiration Society $\endgroup$ Commented Jan 13, 2016 at 20:41
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Go this this site and find the section entitled "skater's spiral". You will get a nice detailed analysis that will show you that since the velocity is at right angles to the string, no work is done on the bead, so the speed is constant. But the radius of curvature of the path is changing as the string gets shorter, so the tension in the string changes.

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    $\begingroup$ This problem can rather easily be solved analytically once one expresses the x and y coordinates of the mass parametrically in terms of the wrap angle. $\endgroup$ Commented Jan 12, 2016 at 13:36
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    $\begingroup$ @ChesterMiller-can you please elaborate? $\endgroup$
    – ksr
    Commented Jan 12, 2016 at 14:17
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    $\begingroup$ The position is a function of the wrap angle and initial length - a simple trig expression. The velocity (which is constant) is obtained as the derivative of the wrap angle - and it has constant magnitude. With that, the equation of motion "appears" cleanly. $\endgroup$
    – Floris
    Commented Jan 12, 2016 at 14:19

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