# Finding minimum angular velocity of an object attached to two inextensible strings

Could you please explain question (ii) please? (I have no problem with (i))

One end of a light inextensible string of length 0.5m is attached to a fixed point A. The other end of the string is attached to a particle P of weight 6N. Another light inextensible string of length 0.5m connects P to a fixed point B which is 0.8m vertically below A. The particle P moves with constant speed in a horizontal circle with centre at the mid-point of AB. Both strings are taut.

(i) Calculate the speed of P when the tension in the string BP is 2N. (I got it.)

(ii)Show that the angular speed of P must exceed 5 rad s^-1. (I am confused!)

The mark scheme for (ii) says “uses tension in BP = 0 and resolves vertically.”

Since, [0.4T/0.5 = 6], T = 7.5

7.5(0.3/0.5) = (6/g) ω^2 (0.3)

Therefore, ω = 5 rad s–1

I wonder why though because if you see my work below, the right-hand side V (tension in BP=0) must be lower than left-hand side V. But then again, since v = ω r, the one with tension in BP=0 has smaller r. Thus, this might not be a correct reason for this. Does anyone know why the mark scheme said that? Also, should the string on the right-hand side not be taut?

Thank you so much.

• The original title (since corrected by sammy gerbil) was "Finding maximum angular velocity of an object attached to two inextensible strings". Perhaps this was just a typo, or perhaps this is the source of your confusion. This is the minimum angular velocity for which you are solving, not the maximum. An angular velocity less than this minimum means the lower string will be slack. Commented Mar 16, 2017 at 6:27
• Yeah! Exactly! That was one of my confusion. Thank you. Commented Mar 18, 2017 at 9:24

The tension in the lower string BP must be greater than zero, otherwise the string will be slack. This would contradict the condition given in the problem that both strings remain taut. This condition imposes a minimum angular speed on P.

The radius of the circle does not vary. It is fixed because the strings are inextensible and always taut. (Note that there is a 3-4-5 triangle here so it is not necessary to calculate the angle $\theta$.)

Since this explanation is already given in the marking scheme, it is not clear to me what your difficulty is, and what you mean about "the right hand string" - in your diagrams the strings are above and below not right and left.

• Sorry for my obscure meaning. What's I mean is the right part of the lower picture, not the individual strings. But since angle θ is fixed, I understand it now. Thank you! Commented Mar 18, 2017 at 9:29

You want the centre of the horizontal circle of rotation to be at the mid-point of $AB$ and this condition cannot be satisfied unless the plane of rotation is at the midpoint of $AB$.

Go to the extreme when the particle is not moving.
The position of the particle would be below the midpoint of $AB$.

Now have the particle rotating slowly like a conical pendulum with string $BP$ slack.
The plane of rotation would be below the midpoint of $AB$ because the horizontal component of the tension in string $AP$ is sufficient to provide the necessary force for the centripetal acceleration of the particle.

If the particle rotates faster it will rotate in a higher horizontal plane as a greater horizontal force is needed to produce the higher centripetal acceleration.
That force is still only provided by string $AP$.

There will come a speed of rotation when the string $BP$ is straight but with a zero tension and that is the limiting condition you are after because if the speed of rotation is any higher both string will be need to provide the necessary force to produce the centripetal acceleration of the particle.