Ap physics 1 ratio

Question

Two identical clay spheres of mass m0 traveling with identical velocity v0 collide with and stick to two different vertical rods. The shorter rod has length 2R and mass 2m0, while the longer rod has length 3R and mass 3m0, as shown in the figure. One sphere collides with the end of the short rod, and the rod-sphere system begins to rotate about its pivot with angular momentum L1. The other sphere collides with the center of the long rod, which begins to rotate about its pivot with angular momentum L2. What is the ratio of L1 to L2? Report your answer as L1:L2

Answer 1

Instead of making two calculations, one for each rod,

I will first calculate moment of inertia of a sphere attaching itself

to an arbitrary rod at an arbitrary spot.

Then I will apply the formula to the two cases.

The two cases can be distinguished by two numbers, let's call them p and q.

The number p (which is 2 in one case and 3 in the other) is the multiple of R

giving the length of the rod.

The number q (which is 2 in one case and 1.5 in the other) is the multiple of R

giving the distance of the sphere from the pivot.

Let

p and q be two numbers satisfying (0 < q ≤ p),

m be the mass of the sphere,

pm be the mass of the rod,

pR be the length of the rod,

qR be the distance from the pivot to where the sphere is attached,

v be the velocity with which the sphere strikes the rod.

I will make three assumptions with the justification that without these assumptions

the statement of the problem would be missing information.

Assumption 1: The rod has uniform density.

Assumption 2: The sphere's radius is so small that the sphere can be treated as single point.

Assumption 3: The pivot is at one end of the rod, opposite to where the sphere attaches to the shorter rod.

Once the rod with the sphere starts rotating, the angular momentum is

L = Iω

where

I is the moment of inertial of the combination rod + sphere,

ω is angular velocity.

We need to compute both I and ω.

The moment of inertia I = I_r + I_s, where

I_r is the moment of inertia of the rod and I_s is the moment of inertia of the sphere.

By Assumption 1,

I_r = pm(pR)²/2

By Assumption 2,

I_s = m(qR)²

So

I = pm(pR)²/2 + m(qR)² = mR²(p³/2 + q²)

The angular velocity ω can be obtained from conservation of energy.

Before the collision, the energy of the system consists of kinetic energy of the sphere:

mv²/2

After the collision, the energy of the system is the rotational energy:

Iω²/2

Therefore the two must be equal

Iω²/2 = mv²/2

So

ω = v√(m/I)

Now we can express the angular momentum

L = Iω = Iv√(m/I) = v√(mI) = v√(m²R²(p³/2 + q²)) = vmR√(p³/2 + q²)

Now we can come back to having the two situations characterized by different p and q

Shorter rod: p₁ = 2, q₁ = 2

Longer rod: p₂ = 3, q₂ = 1.5

L₁ : L2 = √(p₁³/2 + q₁²) / √(p₂³/2 + q₂²)

= √((p₁³ + 2q₁²) / (p₂³ + 2q₂²))

= √((2³ + 2×2²) / (3³ + 2×1.5²)) ≈ 0.71