Engineering Mechanics Test 1

Concept:

x = 6t² – t³

\(V = \frac{{dx}}{{dt}} = 12t - 3{t^2}\)

 Velocity will be maximum, when

\(\frac{{dv}}{{dt}} = \frac{{d_x^2}}{{d{t^2}}} = 0\)

∴ 12 – 6t = 0

∴ t = 2 seconds

\(\therefore V = \frac{{dx}}{{dt}} = 12t - 3{t^2}\) 

V_max = 12(2) – 3(2)²

∴ V_max = 12 m/s

Concept:

Principle of virtual work: (unit-load Method)

Developed by Bernoulli:

To find Δ at point A due to loads P₁, P₂, P₃. Remove all loads, apply virtual load P’ on point A.

For simplicity P’ = 1

It creates internal load u on representative element.

Now remove this load, apply P₁, P₂, P₃ due to which pt. A will be displaced by Δ

∴ External virtual work = 1.Δ

Internal virtual work = u.dL

P^' = 1 = external virtual unit load in direction of Δ.

u = internal virtual load acting on element in direction of a dL.

Δ = external displacement caused by real loads.

dL = internal deformation caused by real loads.

BG=Centre of pontoon – Centre of immersed portion=0.3-0.45*0.3=1.65

Metacentric height=I/∀ -BG

I = π*r⁴ = π*2.5⁴

∀ = π*r*r*h = π*2.5*2.5*6

Metacentric height=-0.61.

Concept:

Impulse (I) = change in momentum

Calculation:

Given:

V₁ = 20 m/s, V₂ = -20 m/s, M = 150 g = 0.15 kg

Now,

Impulse (I) = change in momentum

I = m (V₂ – V₁)

I = 0.15 (-20 - 20)

Concept:

Conservation of momentum:

20 m = m v₁ + m v₂ ; v₁, v₂ → velocity of smooth and rough after impact

\(e = \frac{{velocity\;of\;separation}}{{velocity\;of\;approach}}\)

\(e = \frac{{{v_2} - {v_1}}}{{20}}\)

Perfectly elastic, e = 1

Calculation:

e = 1 ⇒ v₂ – v₁ = 20     

∴ v₂ = v₁ + 20

or  v₁ = v₂ – 20

Also, v₁ + v₂ = 20

∴ v₂ = 20 m/s, v₁ = 0 m/s

Friction retardation on rough block, a = μg

∴ Distance travelled = v² – u² = 2as

⇒ - 400 = - 2μgs

s = 200/μg

∴ s = 101.936 m

Note:

When perfect elastic impact occurs between two identical mass out of which one is at rest, then velocities are exchanged.

Concept:

Angular momentum of the system about its center is conserved.

Calculation:

Given:

R = 1m, M = 0.6 kg, m₁ = m₂ = 65g = 0.065 kg

r₁ = 0.4 m, r₂ = 0.6 m, v₁ = v₂ = 25 m/s

Initially, whole system is at rest, therefore

Initial angular momentum (L_i) = 0

After fixing of guns, let system start rotating with angular velocity ‘ω’.

Then,

Final angular momentum (L_f)

L_f = m₁ u₁ r₁ + m₂ u₂ r₂ + I_ω

\({{\rm{L}}_{\rm{f}}}{\rm{\;}} = {m_1}{v_1}\left( {{r_1} + {r_2}} \right) + \frac{1}{2}M{R^2}.\omega \)

\({L_f} = 0.065 \times 25 \times \left( {0.4 + 0.6} \right) + \frac{1}{2} \times 0.6 \times {1^2} \cdot \omega \)

L_f = 1.625 + 0.3 ω

Conservation of angular momentum,

L_i = L_f

0 = 1.625 + 0.3 ω

a = h/4 = 10/4 = 2.5 m

Explanation:

A Wound Watch Spring and Its Energy Form

• A wound watch spring is a tightly coiled strip of metal that stores energy in the form of mechanical potential energy when it is wound. This energy is later released to power the movement of the watch. When the spring is wound, it undergoes deformation, and the energy required to deform it is stored in the spring as potential energy. This is an example of elastic potential energy, which falls under the broader category of mechanical potential energy.
• When you wind a watch spring, you apply a force to twist or coil the spring further than its relaxed state. This action stores energy in the spring due to its elastic properties, as the metal resists deformation and "wants" to return to its natural shape. As the watch operates, the spring gradually unwinds, releasing the stored energy to drive the gears and hands of the watch.

Correct Option Analysis: The correct option is:

Option 4: Mechanical potential energy

Explanation:

Centre of gravity of right circular cone lies at a distance of h/4 from the base and at a distance of 3h/4 from the apex.

Similarly, the centre of gravity for different shapes are:

Concept:

Impulse \((I) = \mathop \smallint \limits_0^t F.dt\)

Calculation:

Given:

F = 600 - 2 × 10⁵ t

Now,

When F = 0

600 – 2 × 10⁵ t = 0

⇒ t = 3 × 10^-3 s

Now,

\(I = \mathop \smallint \limits_0^t F.dt = \mathop \smallint \limits_0^t \left( {600 - 2 × {{10}^5}t} \right)dt = 600t - {10^5}{t^2}\)

I = 600 (3 × 10^-3) – 10⁵ (3 × 10^-3)²

∴ I = 1.8 - 0.9

∴ I = 0.9 N.s

Concept:

Rate of change of velocity is known as acceleration. Its unit is m/s2. It is a vector quantity.

a = change in velocity/time

Equations of motion:

Concept:

• Average velocity = total displacement/ total time duration
• Equation of motion:
• v = u + at
• v² = u² + 2as
• s = ut + 1/2 at2

Calculation:

Given:

Time interval = 5 s & 1 s, Initial velocity u = 0, and, Acceleration a = 2 m/s²

​When an object starts from rest, then the total distance covered in time 1 sec and 5 sec is,

s = ut + 1/2 at2

Object is at rest, so, u = 0 m/s.

Total time taken, t = t₂ - t₁ = 5 - 1 = 4 sec

Average velocity = total displacement/ total time duration

Average velocity = 24/4 = 6 m/s

Average velocity between time 1 s and 5 s = 6 m/s

A fielder pulling his hand backward while catching a ball is an application of newton’s second law of motion.

Newton's Second Law of motion states that the rate of change of momentum of an object is proportional to the applied force in the direction of the force. ie., F=ma. Where F is the force applied, m is the mass of the body, and a is the acceleration produced.

Newton's Third Law of Motion states that 'To every action there is an equal and opposite reaction'.

A fielder pulls his hand backward; while catching a cricket ball coming with a great speed, to reduce the momentum of the ball with a little delay. According to Newton's Second Law of Motion; rate of change of momentum is directly proportional to the force applied in the direction.