A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m 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 embeds into it.

(Hint: The moment of inertia of the door about the vertical axis at one end is ML²/3.)

Question

(a) Moment of inertia of disc I= I₁

Answer 1

(a)

Moment of inertia of the man-platform system = 7.6 kg m²
Moment of inertia when the man stretches his hands to a distance of 90 cm,

2 × m r²= 2 × 5 × (0.9)²
= 8.1 kg m²
Initial moment of inertia of the system, I_i = 7.6 + 8.1

= 15.7 kg m²
Angular speed, ω_i = 300 rev/min

Angular momentum, L_i = I_iω_i = 15.7 × 30 ...(i)

Moment of inertia when the man folds his hands to a distance of 20 cm

2 × mr²= 2 × 5 (0.2)²

= 0.4 kg m²
Final moment of inertia, I_f = 7.6 + 0.4 = 8 kg m²

Final angular speed = ω_f
Final angular momentum, L_f = I_fω_f = 0.79 ω_f .... (ii)

From the conservation of angular momentum, we have

I_iω_i =  I_fω_f
Therefore,

  ω_f =

(b) Kinetic energy is not conserved in the given process. But, the kinetic energy increases as there is a decrease in the moment of inertia.

The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Answer 2

Given,

Mass of the bullet, m = 10 g = 10 × 10^–3 kg

Velocity of the bullet, v = 500 m/s

Thickness of the door, L = 1 m

Radius of the door, r = m / 2

Mass of the door, M = 12 kg

Angular momentum imparted by the bullet on the door,

α = mvr

= (10 × 10^-3 ) × (500) ×

= 2.5 kg m² s^-1 ...(i)

Moment of inertia of the door is given by,

This is the required angular speed of the door just after the bullet embeds into it.

Answer 3

(a) Moment of inertia of disc I= I₁

Angular speed of disc I = ω₁

Moment of inertia of disc II = I₂
Angular speed of disc II = ω₂
Angular momentum of disc I, L₁ = I₁ω₁
Angular momentum of disc II, L₂ = I₂ω₂
Total initial angular momentum L_i = I₁ω₁ + I₂ω₂
When the two discs are joined together, their moments of inertia get added up.

Moment of inertia of the system of two discs, I = I₁ + I₂
Let ω be the angular speed of the system.

Total final angular momentum, L_T = (I₁ + I₂) ω

Using the law of conservation of angular momentum, we have

L_i = L_T
I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω

∴ ω =

(b) Kinetic energy of disc I, E₁ =

I₁ω₁²
Kinetic energy of disc II, E₂=

I₂ω₂²
Total initial kinetic energy, E_i =

( I₁ω₁² + I₂ω₂²)

When the discs are joined, their moments of inertia get added up.

Moment of inertia of the system, I = I₁ + I₂
Angular speed of the system = ω

Final kinetic energy is given by,

E_f=

( I₁ + I₂) ω²
=

( I₁ + I₂)

=

∴ E_i - E_f =

Solving the equation, we get

=

All the quantities on RHS are positive

Therefore,

E_i - E_f > 0

i.e., E_i > E_f
When the two discs come in contact with each other, there is a frictional force between the two. Hence there would be a loss of Kinetic Energy.

Answer 4

Initial velocity of the solid cylinder, v = 5 m/s

Angle of inclination, θ = 30°

Let the cylinder go up the plane upto a height h.

Using the relation,

If s is the distance up the inclined plane, then as

sin θ =

s =

=

Therefore, time taken to return to the bottom of the inclined plane,

1.53 sec is required for the balloon to return to the bottom.

Answer 5

The given question is illustrated in the figure below:

N_B = Force exerted on the ladder by the floor point B

N_C = Force exerted on the ladder by the floor point C

T = Tension in the rope

BA = CA = 1.6 m

DE = 0. 5 m

BF = 1.2 m

Mass of the weight, m = 40 kg

Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.

ΔABI and ΔAIC are similar triangles.

∴ BI = IC

Hence, I is the mid-point of BC.

DE || BC

BC = 2 × DE = 1 m

AF = BA – BF = 0.4 m … (i)

D is the mid-point of AB.

Hence, we can write,

AD = × BA = 0.8 m ...(ii)

Using equations (i) and (ii), we get

FE = 0.4 m

Hence, F is the mid-point of AD.

FG||DH and F is the mid-point of AD.

Hence, G will also be the mid-point of AH.

ΔAFG and ΔADH are similar

In ΔADH,

AH = (AD² - DH²)^1/2
= (0.8² - 0.25²)^1/2

= 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force.

N_c + N_B = mg = 392 … (iii)

For rotational equilibrium of the ladder, the net moment about A is

-N_B × BI + mg × FG + N_C × CI + T × AG - T × AG = 0

-N_B × 0.5 + 40 × 9.8 × 0.125 + N_C × 0.5 = 0

(N_C - N_B) × 0.5 = 49

N_C - N_B = 98 ...(iv)

Adding equations (iii) and (iv), we get

N_C = 245 N

N_B = 147 N

For rotational equilibrium of the side AB, consider the moment about A,

-N_B × BI + mg × FG + T × AG = 0

-245 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76 = 0

∴ T = 96.7 N.

T is the required tension in the rope.

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