A small mass m swings in a horizontal circle of radius r at the end of a string of length Li and

negligible mass, which makes an angle �i with the vertical. The string is slowly shortened (for
example by pulling it through a hole in its support) until the final length is Lf and the string is
making an angle �f with the vertical. Find an expression for Lf in terms of Li; �i and �f

To find an expression for Lf in terms of Li, �i, and �f, we can use trigonometric identities and the concept of conservation of angular momentum.

Let's start by drawing a diagram of the situation. We have a small mass m swinging in a horizontal circle of radius r at the end of a string of length Li, making an angle �i with the vertical. After the string is shortened, the new length becomes Lf, and the angle with the vertical becomes �f.

Now, let's consider the conservation of angular momentum. When the mass is swinging at angle �i with length Li, its angular momentum is given by:

L_initial = mvr_initial

Here, v is the velocity of the mass and r is the radius of the circle. We can express the velocity v in terms of the length of the string using trigonometry. Since the string is horizontal, the vertical component of the velocity is zero. So, we have:

v_initial * sin(�i) = 0

This implies that v_initial = 0, as sin(�i) is zero when �i is zero.

Now, when the string is shortened and the new length becomes Lf, the angular momentum is given by:

L_final = mvr_final

Again, we can express the velocity v_final in terms of Lf using trigonometry. The vertical component of the velocity is given by v_final * sin(�f).

Using the conservation of angular momentum, we can equate L_initial and L_final:

mvr_initial = mvr_final

Since v_initial is zero, we have:

0 = mvr_final

Simplifying the equation, we get:

r_final = 0

This means that after shortening the string, the mass will stop moving in a circular path and will instead move along a straight line. Hence, Lf = 0.

Therefore, the expression for Lf in terms of Li, �i, and �f is:

Lf = 0

To find an expression for Lf in terms of Li, θi, and θf, we can make use of the conservation of mechanical energy.

The initial mechanical energy of the system, consisting of the mass m and the string, is given by the sum of its potential energy and kinetic energy:

Ei = PEi + KEi

At the highest point of the circle, when the string makes an angle θi with the vertical, the potential energy is given by:

PEi = m * g * (Li * sin(θi))

Since the mass m is moving in a horizontal circle, it has kinetic energy given by:

KEi = (1/2) * m * v^2

where v is the velocity of the mass. The velocity can be related to the length of the string and the angle θi by:

v = ω * r

where ω is the angular velocity, given by:

ω = v / r = (2 * π * f)

where r is the radius of the circle and f is the frequency of the circular motion.

Substituting the expressions for PEi and KEi into the initial mechanical energy equation, we have:

Ei = m * g * (Li * sin(θi)) + (1/2) * m * v^2

Now, consider the final mechanical energy, Ef, when the string has been shortened to a length Lf and makes an angle θf with the vertical. The potential energy at this point is given by:

PEf = m * g * (Lf * sin(θf))

The final kinetic energy is given by:

KEf = (1/2) * m * v^2

Since energy is conserved, we have:

Ei = Ef

Therefore:

m * g * (Li * sin(θi)) + (1/2) * m * v^2 = m * g * (Lf * sin(θf)) + (1/2) * m * v^2

Simplifying the equation, we have:

m * g * (Li * sin(θi)) = m * g * (Lf * sin(θf))

Dividing both sides by m * g:

Li * sin(θi) = Lf * sin(θf)

Finally, rearranging the equation, we have:

Lf = Li * (sin(θi) / sin(θf))

Therefore, the expression for Lf in terms of Li, θi, and θf is:

Lf = Li * (sin(θi) / sin(θf))