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

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

• This option correctly identifies the type of energy stored in a wound watch spring. The energy stored in the spring is mechanical in nature because it results from the deformation of the spring due to the applied force. Additionally, it is potential energy because it is stored and has the potential to do work as the spring unwinds. This energy is neither kinetic (as the spring is stationary when wound) nor a combination of kinetic and potential energy. Furthermore, the term "spring energy" is not a standard term used in physics, making "mechanical potential energy" the most accurate description.

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:

When a block rests on an inclined plane, its weight can be resolved into two components:

• Component parallel to the inclined plane
• Component perpendicular to the inclined plane

The component of the gravitational force acting parallel to the inclined plane is given by:

Concept:

The kinetic energy of a body moving with a certain velocity is given by:

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:

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

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.

s₂ - s₁ = 24 m

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.

Explanation:

• Momentum is conserved in all collisions.
• In elastic collision, kinetic energy is also conserved.
• In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.

Perfectly elastic collision:

If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If law of conservation of momentum holds good during collision while that of kinetic energy is not.

Coefficient of restitution (e)

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Concept:

The friction force is given by:

f = μN

where μ is the coefficient of friction between the surfaces in contact, N is the normal force perpendicular to friction force.

Calculation:

Given:

μ = 0.1, m = 1 kg, F = 0.8 N

Now, we know that

From the FBD as shown below

Normal reaction, N = mg = 1 × 9.81 = 9.81 N

Limiting friction force between the block and the surface, f = μN =  0.1 × 9.81 = 0.98 N

But the applied force is 0.8 N which is less than the limiting friction force.

∴ The friction force for the given case is 0.8 N.

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:

Calculating net horizontal and vertical forces,

The magnitude of resultant force is, F = \(\sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_V}} \right)}^2} +{{\left( {{\rm{\Sigma }}{{\rm{F}}_H}} \right)}^2}} \)

Calculation:

Given:

ΣF_v = 40 – 80 = -40 N

ΣF_H = 20 – 60 = -40 N

Resultant force:

F = \(\sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_V}} \right)}^2} + \;{{\left( {{\rm{\Sigma }}{{\rm{F}}_H}} \right)}^2}} \)

F = \(\sqrt {{{\left( { - 40} \right)}^2} + \;{{\left( { - 40} \right)}^2}}\) = \(\sqrt {3200}\) = 40\(\sqrt 2 \) N

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

Concept:

The CG of a semicircular plate of  r radius, from its base, is

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:

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:

If s = f(t)

Then, First derivative with respect to time represents the velocity

Concept:

Moment of inertia of circular plate,

Calculation:

The ratio of the moment of inertia of a circular plate to that of a square plate is, Which is less than 1.

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:

Equation of motion:

• The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering the force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move in a straight line.

There are three equations of motion:

where, v = final velocity, u = initial velocity, s = distance travelled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

Calculation:

Given:

Part-I:

When the ball will reach the highest point then the final velocity will be zero.

Initial velocity = u m/sec, final velocity = 0 m/sec, acceleration = -g m/sec²

applying 1st equation of motion

To find the resultant of two equal forces P making an angle θ, we use the formula for the resultant of two forces:

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

Moment of inertia for a circular section about an axis perpendicular to the section is given by , where d is the diameter of the circle. Therefore, the correct answer is Option C.

The application of mechanics to issues involving common engineering aspects is known as engineering mechanics. The purpose of the Engineering Mechanics course is to expose students to mechanics problems as they are applied to realistic circumstances.