Physics - Rotation Question with Solution | TestHub

Given, mass of bullet (m) = 10 g = 0.01 kg

Speed of bullet (v) = 500 m s^-1

Width of the door (l) = 1.0 m

Mass of the door (M) = 12 kg

As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door,

​$\text{(r)}=\frac{l}{2}=\frac{1}{2}\text{m}$

Angular momentum transferred by the bullet to the door,

(L) = mv × r

$=\text{0.01}\times \text{500}\times \frac{1}{2}$

= 2.5 J-s

Moment of inertia of the door about the vertical axis at one of its end,

$\text{(I)}=\frac{\text{M}{l}^2}{3}=\frac{\text{12}\times {\text{(1)}}^2}{3}={\text{4 kg-m}}^2$

But angular momentum, (L) = Iω​

∴                        2.5 = 4 × ω

$\text{or }\text{ω}=\frac{\text{2.5}}{4}$

ω = 0.625 rad s^-1