Engineering Mechanics Test 2

Concept:

Impulse (I): It is defined as the integral of force with respect to time. It is a vector quantity. Or it is also defined as a change in the linear moment (P) with respect to time.

I = F  × dt = ΔP 

According to Newton's second law, the Force can be defined as a moment per time.

∴  \({\bf{F}} = \frac{{{\bf{d}}\left( {{\bf{mv}}} \right)}}{{{\bf{dt}}}}\)

This is known as the Impulse momentum equation.

where, m = mass of body, v = relative velocity

Calculation:

Given:

m = 1500 kg, v₁ = 15 m/s, v₂ = 2.6 m/s, t = 0.15 s

Therefore, \(F=\frac{1500\times (15-(-2.6))}{0.15}=\frac{1500\times17.6}{0.15}=1.76\times10^5~N\)

Momentum conservation:

mu = MV

\(\frac{V}{u}=\frac{m}{M}\)

The ratio of kinetic energies of bullet and gun is:

\(\frac{K.E_b}{K.E_g}=\frac{\frac{1}{2}mu^2}{\frac{1}{2}MV^2}=\frac{mu^2}{MV^2}=\frac{m}{M}(\frac{M}{m})^2=\frac{M}{m}=10\)

Concept:

Moment of inertia is the sum of the product of mass of each particle with the square of its distance from the axis of the rotation.

M.O.I for hollow circular cross-section \( = \frac{\pi }{{64}}\left( {d_0^4 - d_i^4} \right)\) 

Where d₀ = outer diameter of cross section

d_i = inner diameter of cross section

Calculation:

d₀ = 8 cm

d_i = 6 cm

\(I = \frac{\pi }{{64}}\left( {{8^4} - {6^4}} \right)\)

= 137.45 cm⁴

Other Important Points:

Formula of moment of inertia for various other figures is given below.

Virtual work is the work done by a real force acting through a virtual displacement or a virtual force acting through a real displacement.

A virtual displacement is any displacement consistent with the constraints ofthe structure, i.e., that satisfy the boundary conditions at the supports.

A virtual force is any system of forces in equilibrium.

The principle of virtual work states, for bodies in equilibrium, for a small arbitrary displacement, the total work done by the system is zero.

In this method, the system is displaced through a small amount about a reference point and the work done by all the forces about the reference point is summed to zero to find the unknown reactions if any.

m = 8t = 8000 kg

ω₀ = 0 (because it starts from rest)

\(\omega = 180\;r.p.m. = \frac{{180 \times 2\pi }}{{60}} = 6\pi \;rad/s\)

t = 3 min = 180 s

I = Mk² = 8000 × (0.6)² = 2880 kg-m²

ω = ω₀ + αt

6π = 0 + α × 180

\(\alpha = \frac{{6\pi }}{{180}} = 0.105\;rad/{s^2}\)

Let initial velocity of man is “u” = 0

V² = u² + 2gh

V² = 0 + 2 × 9.81 × 25

V = 22.15 m/s

Now at a velocity of 22.15 m/s man starts dives in to a swimming pool and goes up to x metre at which its velocity = 0

So, V₁ = 0

u₁ = 22.15 m/s

a₁ = -a (because body starts moving upward)

\(v_1^2 = u_1^2 + 2{{\rm{a}}_1}{\rm{x}}\)

0 = (22.15)² + 2(-a) × X

\({\rm{a}} = \frac{{ - {{\left( {22.15} \right)}^2}}}{{ - 2{\rm{x}}}} = \frac{{245.31}}{{\rm{x}}}\)

Now, the net force acting on the man must be equal to the product of mass and acceleration

F_r – W = ma

\(7300-650 = \left( {\frac{{650}}{{9.81}}} \right) \times \frac{{245.31}}{{\rm{x}}}\)

Concept:

It is assumed hitting the ball with a bat as a 1D collision. During a collision, the momentum is conserved.

Momentum lost by bat = Momentum gained by bat

M_bat u_bat - M_bat V_bat = M_ball V_ball - M_ball u_ball

Impulse on the ball will be change in momentum

I = d(mv)

Average force exerted by bat on the ball will be

F = I/Δt

Calculation:

Given:

M_ball = 160 gms = 0.16 kg; M_bat = 1.5 kg; u_ball = 90 kmph = 25 m/s;

V_ball = 144 kmph = 40 m/s; u_bat = 54 kmph = 15 m/s; Δt = 3 ms = 0.003 s

Conservation of momentum:

1.5 × 15 + 0.16 × 25 = 0.16 × 40 + 1.5 × V_bat

⇒ V_bat = 13.4 m/s = 48.24 kmph (Option 4)

Impulse = change in momentum = M_ball (u_ball – (V_ball))

⇒ I = 0.16 (25 – (- 40)) = 10.4 kg m/s (Option 2)

Average force exerted is given by

F_avg = I/Δt = 10.4/0.003 = 3.46 kN (Option3)

Mistake point:

While calculating impulse, directions of velocities are also needed to be considered because directions play a key role in kinematic analysis.

If direction not considered, I = 0.16 × 15 = 2.4 kg.m/s

Concept:

If the person is sitting in train and observing train velocity, he will observe the train is at rest as the relative velocity of train w.r.t person is zero.

V_TP = V_T – V_P = V_T – V_T = 0

If the person is outside the train and stationery, he will observe the actual speed of train.

V_TP = V_T – V_P = V_T – 0 = V_T

If the person is sitting in another vehicle and observing the train, the observed train velocity will be different.

V_TP = V_T – V_P ⇒ V_T = V_TP + V_P

Hence, the vectors V_T, V_P and V_TP constitute a triangle.

Calculation:

Given u_T = 54 kmph = 15 m/s; u_B = u_P = 0; a_T = 0.2 m/s²; a_P = 0.4 m/s²;

At t = 25 sec,

V_T = u_T + a_T t = 15 + 0.2 × 25 = 20 m/s;

V_P = u_P + a_P t = 0 + 0.4 × 25 = 10 m/s;

Now both the velocities’ directions are different, let’s solve in each direction separately,

We will assume, East direction as positive x – axis and North direction as positive Y – axis.

V_TX = 20 × cos 30° = 17.32 m/s;

V_TY = 20 × sin 30° = 10 m/s;

V_PX = 10 × cos 0° = 10 m/s

V_PY = 10 × sin 0° = 0 m/s;

V_TPX = V_TX – V_PX = 7.32 m/s;

V_TPY = V_TY – V_PY = 10 – 0 = 10 m/s;

Since Both the east component and north component are positive, the direction of Relative velocity vector lies in 1^st quadrant,

∴ Passenger observes the train moving in north east direction (Option 2)

For angle,

\(\tan \theta = \frac{{{V_{TPY}}}}{{{V_{TPX}}}} = \frac{{10}}{{7.32}} = 1.366\) 

⇒ θ = 53.8°

As the angle is measured from East direction, so θ = 53.8° (Option 3)

This problem can also be solved using Velocity triangle approach.

Mistake points:

The angle calculated (θ) is the angle with East direction (+ve x-axis)

If the angle is measured from north direction, then the answer will be 90 - θ degrees.

Concept:

The figure of two objects P and Q as per given data in question is,

Conservation of linear momentum (P)

If there is no external force acting on system then initial momentum of system is equal to final momentum is called conservation of liner momentum.

\({{\rm{m}}_P}{\rm{\;}}{{\rm{U}}_P} + {\rm{\;}}{{\rm{m}}_Q}{\rm{\;}}{{\rm{U}}_Q} = {{\rm{m}}_P}{\rm{\;}}{{\rm{V}}_P} + {{\rm{m}}_Q}{\rm{\;}}{{\rm{V}}_Q}\)

And 

Coefficient of restitution (e)

It is the ration of final relative velocity after collision to the initial relative velocity before collision.

\({\rm{e}} = \frac{{\left( {{{\rm{V}}_Q} - {{\rm{V}}_P}} \right)}}{{\left( {{{\rm{U}}_P} - {{\rm{U}}_Q}} \right)}}\)

Calculation:

Given:

U_P = 8 m/s, U_Q = 6 m/s, m_P = 3 kg, m_Q = 5 kg, e = 0.7

m_P × 8 + m_Q × 6 = m_P­V_P + M_QV_R

⇒ 3 × 8 + 5 × 6 = 3 V_P + 5 V_R

3 V_P + 5 V_R = 54 ------ (1)

For Coefficient of restitution (e)

\(0.7 = \frac{{{V_Q} - {V_P}}}{{8 - 6}}\)

\({V_Q} - {V_P} = 1.4\)

5V_R + 3V_P = 54 

3V_Q - 3V_P = 1.4 × 3

By adding both equations we will get, 

8 V_Q = 58.2

∴ V_Q = 7.275 m/s

V_P = 8.875 m/s

Concept:

\({V_y} = \frac{{dy}}{{dt}}\)

\({a_y} = \frac{{{d^2}y}}{{dt}}\)

Calculation:

Given

y = x³-3x+150

From, \(\overrightarrow {{V_0}} = 3\hat i - 14\hat j\)

V_x = 3 m/s (Constant)

\({V_y} = \frac{{dy}}{{dt}} = 3{x^2}{V_x} - 3{V_x}\)

\({V_y} = 3{x^2}\left( 3 \right) - 3\left( 3 \right)\)

\({V_y} = 9{x^2} - 9\)      (1)

\({a_y} = 18x\)      (2)

Substitute x = 5 in (1) and (2)

V_y = 216 m/s

a_y = 90 m/s²