Question

Hi there!!


$\sf\huge{\underline{\mathbb{Question!!}}}$

▶A bullet of mass 10 g and speed 500m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embedds into it.

#Hint!
➡ The moment of inertia of the door about vertical axis at one end is $\frac{ML^{2}}{3}$

Figure it out!​

Answer 1

Hey mate....

here's the answer....

Mass of bullet,

m = 10 g = 10 × 10–3 kg

Velocity of bullet,

v = 500 m/s

Thickness of door,

L = 1 m

Radius of door,

r = m / 2

Mass of door,

M = 12 kg

Angular momentum of bullet on the door:

α = mvr

= (10 × 10-3 ) × (500) × (1/2) = 2.5 kg m2 s-1 ...1

Moment of inertia of door:

I = ML2 / 3

= (1/3) × 12 × 12 = 4 kgm2

But α = Iω

∴ ω = α / I

= 2.5 / 4

= 0.625 rad s-1

Hope this helps❤