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

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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 :

asked Jun 30 in Physics Problems by koolman (4,116 points) 1 flag
What answer did you get? I see formulas scattered around, but no answer.

1 Answer

1 vote
 
Best answer

Use an inertial frame in which the held end of the string A is at rest.

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.

answered Jun 30 by sammy gerbil (26,166 points)
selected Jun 30 by koolman
I am not quite satisfied with this answer. I suspect that it might be incorrect.
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