Hard physics problem about object attached to a thread

Question

A thread of lenght l<AB is tied to the top (point A) of the stem AB. The other end of the thread is attached to an object. The thread is stretched out horizontally and then the object is released. When the tension of the thread is greater than the force of gravity acting on the object k times, the thread breaks and the object falls off the stem to the ground (point B). Find the height of the stem.

Answer 1

It is possible that there is a picture that should accompany the problem.

In its absence I assume that the stem is vertical.

When the thread breaks, the object happens to fall exactly to the foot of the stem -- point B.

I am assuming no drag or friction.

Please draw a picture and enter in there the quantities mentioned below.

Let

m (unknown) be the mass of the object,

h (to be calculated) be the height of A, that is, the distance AB,

l (known) be the length of the thread,

T (unknown) be the tension on the string at the time the thread breaks,

θ (unknown) be the angle the thread forms with the vertical at the time the thread breaks,

v (unknown) be the object's velocity at the time the thread breaks,

t (unknown) be the time it took the object to hit the ground (starting from the point of breakage),

g (known) be gravitational acceleration.

In total there are two forces acting on the object -- gravity, mg, and the tension T.

We are given

T = kmg (1)

The object is subject to centripetal acceleration v²/l (in addition to tangential acceleration).

That centripetal acceleration is caused by the centripetal force, which is

the sum of those two forces projected on the line of the thread.

The projection of gravity has magnitude mgcos(θ), and its direction is opposite the tension T.

Therefore by Newton's Second Law

mv²/l = T - mgcos(θ)

which simplifies into

v²/l = gk - gcos(θ) (2)

Now we use conservation of energy to get another relation between v and θ.

We are told that the object was released from horizontal position.

At that point its kinetic energy was 0, and its potential energy mgh.

At the point where the thread broke, the object was at height h - lcos(θ),

so its potential energy was mg(h - lcos(θ)).

Its kinetic energy was mv²/2.

By conservation of energy

mgh = mv²/2 + mg(h - lcos(θ))

This can be simplified into

v²/l = 2gcos(θ) (3)

Combining (2) and (3)

2gcos(θ) = gk - gcos(θ)

After simplification

cos(θ) = k/3 (4)

Note that equation (4) implies that the thread will break only if k ≤ 3.

Substituting (4) into (3)

v²/l = 2gk/3

Below we will use it in the rewritten form

gl/v² = 3/(2k) (5)

After the thread broke the object started with velocity v, which is perpendicular to the thread.

It then followed a trajectory in the shape of a parabola,

until it hit the ground at the point B.

The parabola is the result of horizontal movement with constant velocity vcos(θ),

and vertical fall, starting with vertical velocity vsin(θ).

In time t the falling object must cover horizontal distance lsin(θ), and

vertical distance h - lcos(θ).

From the equation for movement with constant velocity

lsin(θ) = tvcos(θ)

From that express t as

t = (l/v)tan(θ) (6)

From the equation for vertical fall

h - lcos(θ) = tvsin(θ) + gt²/2

Substituting (6)

h - lcos(θ) = ltan(θ)sin(θ) + g(l/v)²tan²(θ)/2

From (5)

h - lcos(θ) = ltan(θ)sin(θ) + 3ltan²(θ)/(4k)

From that express

h = l(cos(θ) + tan(θ)sin(θ) + (3/(4k))tan²(θ)) (7)

It remains to replace the functions of θ in (7) using (4).

cos(θ) = k/3

tan(θ)sin(θ) = sin²(θ)/cos(θ) = (1 - cos²(θ))/cos(θ) = 1/cos(θ) - cos(θ) = 3/k - k/3

tan²(θ) = sin²(θ)/cos²(θ) = (1 - cos²(θ))/cos²(θ) = 1/cos²(θ) - 1 = 9/k² - 1

Substituting into (7)

h = l(k/3 + 3/k - k/3 + (3/(4k))(9/k² - 1))

= l(3/k + (3/(4k))(9/k² - 1))

= (3l/k)(1 + (1/4)(9/k² - 1))

= (3l/k)(3/4 + 9/(4k²))

= (9l/4k)(1 + 3/k²)