# free end of the thread must be lifted vertically upwards so that the block leaves the floor

1 vote
79 views

A light inextensible straight thread of length I = 3.2 m is placed on a
frictionless horizontal floor. A small block of mass m = 3 kg placed on
the floor is attached at one end of the thread. With how much constant
speed $v_0$, the free end of the thread must be lifted vertically upwards so
that the block leaves the floor, when the thread makes an angle $\theta = 30°$
with the floor. Acceleration of free fall is g = 10 m/s$^2$ .

My try : Answer is given as : What answer did you get? I see formulas scattered around, but no answer.

The string pivots about A while the end B (the block) moves in a circular arc. Then the downward vertical components of velocity and acceleration of B are $$v=L\dot \theta \cos\theta, a=L\ddot \theta \cos\theta$$ The constraint is that, while the block remains in contact with the floor, end B moves downwards with constant velocity $v=v_0$. At the instant of losing contact with the floor the vertical velocity of B is still $v_0$ but it is now in free-fall so $a=g$. Therefore at this instant $$\dot v=L(\ddot \theta\cos\theta-\dot \theta^2 \sin\theta)=0$$ $$\ddot\theta\cos\theta=\dot\theta^2\sin\theta$$ $$gL=aL=L^2\ddot\theta\cos\theta=(L\dot\theta)^2\sin\theta$$ $$v_0=L\dot\theta\cos\theta=\sqrt{\frac{gL}{\sin\theta}}\cos\theta$$ I think the square on the cosine is a printing error.