# Finding force in a constraint motion problem [closed]

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, $$T \cos \theta = m a$$ $$a = \frac{d}{dt}(v_0 \sec \theta) = v_0 \sec \theta \tan \theta \frac{d \theta}{dt}$$ $$\cos \theta = x/l \implies \sin \theta \frac{d \theta}{dt} = \frac{-dx/dt}{l} = \frac{v_0 \sec \theta}{l}$$

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. 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?

• 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 Sep 8 at 12:41
• The angle θ is constant in the data, therefore: dθ/dt=0 Sep 8 at 13:20
• @TheTiler No, the angle $\theta$ is not constant. Try visualising the situation. Sep 8 at 14:00
• We need a more clear formulation of the problem. I thought $\theta$ was constant, as well Sep 8 at 16:23