Q. 244.4( 7 Votes )

# A bullet of mass

Answer :

Given:

Mass of Bullet, m = 10g = 10^{-3} kg

Speed of bullet, v = 500m/s

Width of the door, w = 1m

Distance from hinge where bullet hits, r = 0.5 m

Mass of door, M = 12 Kg

We know that angular moment is given by,

α = mvr

where,

α = angular momentum

m = mass of the body

v = velocity of the body

r = radius of the body

by putting the value, we get,

⇒ α = 10 × 10^{-3} kg × 500 ms^{-1} × 0.5 m

⇒ α = 2.5 Kgm^{2}s^{-1} …(i)

Moment of inertia is given by,

I = ML^{2}/3 (For rectangle)

Where,

M = mass of door

L = width of door

I = 1/3 × 12 Kg × (1m)^{2}

I = 4 Kgm^{2}

We also have,

α = Iω

where is α is the angular momentum

l is the linear momentum

ω is the angular velocity

⇒ ω = α /I

By putting the values, we get,

⇒ ω = 2.5/ 4

⇒ ω = 0.625 rad s^{-1}

Note: It is a good practice to memorise the moment of inertia for common shapes like rectangle, circle, triangle, cylinder and sphere.

Rate this question :

A door is hinged Physics - Exemplar

A wheel in uniforPhysics - Exemplar

A uniform sphere Physics - Exemplar

Figure 7.5 shows Physics - Exemplar

Figure 7.4 shows Physics - Exemplar

The net external Physics - Exemplar

A simple pendulumHC Verma - Concepts of Physics Part 1

A uniform cube ofPhysics - Exemplar

A body is in tranHC Verma - Concepts of Physics Part 1

A person sitting HC Verma - Concepts of Physics Part 1