Centripetal force causing tangential acceleration paradox

Question

I was solving this question

Now the general solution for part(a) is that as the tension makes a $180°$ angle with the radius, it wont cause any torque on the mass. So angular momentum is conserved and

$$mr_0v_0=mrv$$

so $$v = \frac{r_0v_0}{r} \tag{1}$$

and hence we can calculate the tension by subsituting from eq(1) in $T=\frac{mv^2}{r}$.

But from eq(1) we can also derive the fact using relation $v=r\omega$ that final angular velocity would be greater than the initial angular velocity. This means that there was an angular acceleration. Ofcourse in the question they have given that we pull the string slowly which means that the angular acceleration must be near about zero, but there will be some angular acceleration. This can only be possible if the string's tension has a tangential component in the rotation of the mass. But that fact doesn't sound right to me. Is it true?

Also note that it is my intuition that says that as the person in the question pulled the string slowly, angular acceleration must be near about zero but it is not written anywhere. Could you verify this too and better prove it?

Answer 1

(a) Because the force is purely radial you have correctly deduced that angular momentum is conserved. It simply doesn't follow that angular velocity is conserved. To have an angular acceleration the mass doesn't require a tangential force (but to gain angular momentum it would need such a force).

(b) Further insight can be gained by considering the work done pulling the string through a small distance, such that the radius of the mass's path changes by a small (negative) increment, d$r$. The work done by the force pulling the string will be equal to the gain in kinetic energy of the mass. So $$\frac{mv^2}r(-dr)=d\left(\tfrac 12 mv^2\right)=mv\ dv,$$ that is $$\frac {dv}v=-\frac {dr}r$$ You'll find that this is exactly the same relationship that you get from angular momentum conservation in the limit as $dr$ approaches zero: $$mvr=m(v+dv)(r+dr).$$