Mechanics impulse related 
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.
 A: 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:
$$T=mv\frac{d\theta}{dt}=m(L-R\theta)\left(\frac{d\theta}{dt}\right)^2=m\frac{v^2}{(L-R\theta)}$$
A: 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.
