I have a doubt regarding the correct method to solve this question.

A bead of mass $m = 0.5kg$ which can slide along a smooth horizontal rod is attached to one end of a light inextensible string of length $l = 1m$. The other end 'A' of the string is pulled with a constant speed $v_0 = 2m/s$ always directed along the length of the string. The tension in the string at $\theta = 60^\circ$ is .......... (Ans: 24N)


My approach (Incorrect)

From constraint relation, we can conclude that the bead moves along the rod with a speed $v_0 \sec \theta$. Acceleration($a$) is caused by the component of tension($T$) along the rod. So, $$T \cos \theta = m a $$ $$a = \frac{d}{dt}(v_0 \sec \theta) = v_0 \sec \theta \tan \theta \frac{d \theta}{dt}$$ sub-part

$$\cos \theta = x/l \implies \sin \theta \frac{d \theta}{dt} = \frac{-dx/dt}{l} = \frac{v_0 \sec \theta}{l}$$

Solving these 3 equations we get, $T = \frac{m v_0^2}{l \cos^4 \theta} = 32N \Box$

Correct Method

Put the observer at point A. sub-part2

The claim is that relative to A, B undergoes circular motion (can be proved by finding out the direction of resultant velocity vector of B) and tension along with a part of normal reaction, provides the necessary centripetal force. $$N = T \sin \theta$$ $$T - N \cos (90^\circ - \theta) = m v_{B,A}^2/l $$ Solving these we get $T = 24N \Box$

I am unable to spot any mistake in my method. Where is the flaw?

  • $\begingroup$ I had many doubts about the "correct" answer as well. But I also guess that every doubt here starts from the poor form of the text of the problem $\endgroup$
    – basics
    Sep 8 at 12:41
  • $\begingroup$ The angle θ is constant in the data, therefore: dθ/dt=0 $\endgroup$
    – The Tiler
    Sep 8 at 13:20
  • $\begingroup$ @TheTiler No, the angle $\theta$ is not constant. Try visualising the situation. $\endgroup$
    – S Das
    Sep 8 at 14:00
  • $\begingroup$ We need a more clear formulation of the problem. I thought $\theta$ was constant, as well $\endgroup$
    – basics
    Sep 8 at 16:23