Question

The 5-kg rod is pinned at O and is initially at rest. A 2-g bullet is fired into the rod with a velocity of 500 m/s. The bullet hits the rod and becomes embedded in it. Determine 1. Mass moment of inertia of the rod about an axis that passes through the centroid (G) of the rod. kg-m2 2. Initial angular momentum of the bullet about point O. kg-m2/s 3. The angular velocity of the rod just after the bullet becomes embedded in it. rad/s

Answer 1

(1)The mass moment of initial = 0.6 kg-m², (2)The initial momentum of the bullet about point O =0.96 k8-m²/ 5, (3)the angular velocity of the rod just after the bullet becomes embedded in it is 0.64 rad/s

Step-by-step explanation:

Solution

Recall that,

(1) The initial mass moment is at (G)

where

M = 5kg

The length = 1.2

IG = ML²/L² = 5 * 1.2²/1.2 = 0.6 kg-m²

Therefore

IG = 0.6 kg-m²

(2)The initial angular momentum of the bullet at point O

Now,

Angular momentum

mb * vₓ * l

Mbₓ v cosλ l

which is

( 2* 10 ^⁻³) * 500 * 4/5 * 1.2

where cos θ = 4/5, tan θ = 3/4

V = 500 mg

m = 2g * 10 ^⁻³ kg

L = 1.2

( 2* 10 ^⁻³) * 500 * 4/5 * 1.2

Now,

The angular momentum becomes

= 0.96 k8-m²/ s

(3) The next step is to solve for the angular velocity of the rod

Thus,

I₀ = IG + md²

which is

0.6 + 5 * 0.6²/2

Then,

I₀ = 1.5 kg-m²

By applying the conservation of angular momentum

m V. R = I₀ w

(2 * 10 ^ ⁻³) * (500 cosθ) * 1.2 = 1.5 * w

(2 * 10 ^ ⁻³) * (500 4/5) * 1.2 = 1.5 * w

therefore

w = 0.64 rad/s