As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for , g for gravity, and M and m as appropriate.)

I got the answer (4M/m)sqrt(gL) but the problem is that I did it a while ago and I completely forgot why this is the answer. Can anyone explain it to me?

If you derived it once, you should be able to derive it again.

The speed V of the pendulum bob after the bullet has passed through is given by
MV = m(v - v/2) = mv/2
V = mv/(2M)
To swing all the way to the top,
(1/2) M V^2 = M g (2L)
V^2 = 4 L g = m^2/4M^2 * v^2
v = sqrt (16 L g M^2/m^2)
= 4 (M/m) sqrt (Lg)

To understand why the minimum value of v is given by (4M/m)sqrt(gL), let's break down the problem step-by-step.

1. Initial momentum of the bullet:
The bullet has a mass m and speed v, so its initial momentum is given by p₁ = m * v.

2. Final momentum of the bullet:
After passing through the pendulum bob, the bullet emerges with a speed of v/2. Therefore, its final momentum is given by p₂ = m * (v/2) = (m/2) * v.

3. Conservation of momentum:
According to the principle of conservation of momentum, the total momentum before and after the collision should be the same. Therefore, we can write:
p₁ = p₂
m * v = (m/2) * v

The mass of the bullet cancels out, and we get v = v/2.
This equation implies that the initial velocity of the bullet is twice the final velocity. It tells us that the bullet transfers half of its momentum to the pendulum bob.

4. Energy considerations:
When the bullet passes through the pendulum bob, it imparts both momentum and energy to it. The minimum value of v required for the pendulum bob to barely swing through a complete vertical circle can be found by equating the initial kinetic energy of the bullet to the maximum potential energy of the pendulum bob, at the highest point of its circular motion.

Initial kinetic energy of the bullet = (1/2) * m * v²
Maximum potential energy of the pendulum bob = M * g * 2L (since the bob undergoes a complete vertical circle, its maximum height is 2L)

Setting these equal and solving for v, we have:
(1/2) * m * v² = M * g * 2L
v² = 4MgL/m
v = 2√(MgL/m)

5. However, we need to account for the fact that the bullet transfers only half of its momentum to the pendulum bob. So, we multiply our value of v by 2 to obtain the minimum initial velocity of the bullet:
Minimum v = 2 * 2√(MgL/m) = 4√(MgL/m)

Therefore, the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle is (4M/m)sqrt(gL).

To understand the minimum value of v, we need to analyze the forces acting on the pendulum bob.

When the bullet passes through the bob, it imparts a momentum change to it. We know the change in momentum is equal to the impulse, which is given by:

Impulse = F * Δt

Since the bullet passes through the bob completely, the time of contact (Δt) is very small and can be assumed to be zero. Therefore, the impulse is given by:

Impulse = F * Δt = 0

Since there is no impulse, there is no net force on the bob. However, there are two types of forces acting on the bob: gravity and the tension in the string.

1. Gravity: The force due to gravity is acting downward with a magnitude of M * g (mass of the bob times acceleration due to gravity).

2. Tension in the string: The tension in the string acts towards the center of the circular path (centripetal force) and is responsible for the bob's circular motion. At the minimum speed, the tension becomes zero at the highest point of the circular path.

To find the minimum value of v, we need to determine the maximum value of the tension when it is just about to become zero. At this point, the gravitational force is the only force acting on the bob, providing the centripetal force needed for circular motion.

The gravitational force acting as the centripetal force is given by:

M * g = M * (v^2 / R)

Where R is the radius of the circular path, which is the length of the pendulum rod L.

Rearranging the equation, we get:

v^2 = g * L

Taking the square root of both sides, we find:

v = sqrt(g * L)

However, this is the minimum speed required for the bob to complete a vertical circle, not a complete vertical circle. To find the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle, we need to multiply it by a factor of 4.

Thus, the minimum value of v is:

v = 4 * sqrt(g * L) = (4M / m) * sqrt(g * L)

This is the derived expression for the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle.