Engineering Mechanics MCQ Quiz - Objective Question with Answer for Engineering Mechanics - Download Free PDF

Explanation:

Angular velocity: 

The rate of change of angular position of an object in a rotational/circular motion is called angular velocity.

The SI unit of angular velocity is radian per second.

It is given by the formula:

ω = dθ/dt

where ω = angular velocity in radian per second, dθ = small angular displacement in radian, and dt = small time in second.

Relation of angular velocity with linear velocity:

V = ωr

where, V = linear velocity of the object, ω = angular velocity of the object, and r = radius of the circle in which the body is moving.

Concept:

From Newton's Second Law of motion

F = ma

and distance traveled S 

\(s = ut + \frac{1}{2}a{t^2}\)

Claculation:

Given: m = 5 kg, F= 12 N, t = 2s

a = F/m = 12/5 = 2.4 m/s²

Here u = 0 (body is at rest)

\(s = 0 + \frac{1}{2} \times 2.4 \times {2^2} = 4.8\;m\)

Concept:

Law of Parallelogram of forces

This law is used to determine the resultant of two coplanar forces acting at a point.

It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of parallelogram which passes through that common point.”

Let two forces P and Q, acting at the point O be represented, in magnitude and direction, by the directed line OA and OB inclined at an angle θ with each other.

Then if the parallelogram OACB be completed, the resultant force R will be represented by the diagonal OC.

\({\rm{R}} = \sqrt {{\rm{P}}^2 + {\rm{Q}}^2 + 2{{\rm{P}}}{{\rm{Q}}}\cos {\rm{\theta }}}\)

Explanation:

Coulomb Law of Friction

• Coulomb’s law of sliding friction can represent the most fundamental and simplest model of friction between dry contacting surfaces.

• When sliding takes place, the coulomb law states that the tangential friction force is proportional to the magnitude of the normal contact force.
• It is independent of relative tangential velocity.

Important Points

• When contacting bodies slide or tend to slide relative to each other tangential forces are generated.
• These forces are usually referred to as friction forces.
• Three basic principles have been established namely:
  • The friction force acts in a direction opposite to that of the relative motion between the two contacting bodies.
  • The friction force is proportional to the normal load on the contact.
  • The friction force is independent of a normal area of contact.

Explanation:

Coulomb Law of Friction

• Coulomb’s law of sliding friction can represent the most fundamental and simplest model of friction between dry contacting surfaces.

• When sliding takes place, the coulomb law states that the tangential friction force is proportional to the magnitude of the normal contact force.
• It is independent of relative tangential velocity.

Important Points

• When contacting bodies slide or tend to slide relative to each other tangential forces are generated.
• These forces are usually referred to as friction forces.
• Three basic principles have been established namely:
  • The friction force acts in a direction opposite to that of the relative motion between the two contacting bodies.
  • The friction force is proportional to the normal load on the contact.
  • The friction force is independent of a normal area of contact.

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:

• v = u + at
• v2 – u2 = 2as
• \(s = ut + \frac{1}{2}a{t^2}\)

Calculation:

Given:

u = 36 km/h = 10 m/s; S = 125 m ; v = 54 km/h = 15 m/s,  t = ?

v2 – u2 = 2as

\(a = \frac{{{v^2} - {u^2}}}{{2s}} = \frac{{{{15}^2} - {{10}^2}}}{{2 \times 125}} = 0.5\;m/s^2\)

v = u + at

\(t = \frac{{v - u}}{a} = \frac{{15 - 10}}{0.5} = 10\;sec\)

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.

\(s_2 - s_1 = \frac12 a(t_2^2-t_1^2)\)

\(s_2 - s_1 = \frac12 \times 2(5^2-1^2)\)

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)

\(e = \frac{{Relative\;velocity\;after\;collision}}{{Relative\;velocity\;before\;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\)

• 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:

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,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

Concept:

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

\(\bar y = {4r\over 3 \pi}\)

Calculation:

Given:

r = 33 cm

\(\bar y = {4r\over 3 \pi}={4\times 33\over3\times{22\over 7}}\)

y̅ = 14 cm

∴ the C.G. of a semicircular plate of 66 cm diameter, from its base, is 14 cm.

Additional Information

C.G. of the various plain lamina are shown below in the table. Here x̅  & y̅  represent the distance of C.G. from x and y-axis respectively.

Circle
Semicircle
Triangle
Cone
Rectangle
Quarter Circle
Solid hemisphere

Concept:

If s = f(t)

Then, First derivative with respect to time represents the velocity

\(v=\frac{ds}{dt}\)

Acceleration is given by

\(a=\left( \frac{{{d}^{2}}S}{d{{t}^{2}}} \right)\)

Where s is the displacement

Calculation:

Given:

s = 2t3 – t2 - 1 and t = 1 sec.

\(\frac{ds}{dt}=6{{t}^{2}}-2t\)

\(\frac{{{d}^{2}}s}{d{{t}^{2}}}=12t-2\)

Concept:

Moment of inertia of circular plate,

\({{\rm{I}}_{{\rm{xx}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}\)

Moment of inertia of Square plate,

\({{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{d}}\times{{\rm{d}}^3}}}{{12}}\)

Calculation:

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

Important Points 

The following table shows the Second moment of inertia of different shapes

Shape	Figure	Moment of Inertia
Rectangle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{d}}^3}}}{{12}}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{d}}{{\rm{b}}^3}}}{{12}}\)
Triangle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{h}}{{\rm{b}}^3}}}{{36}}\)
Circle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}\)
Semicircle		\({{\rm{I}}_{{\rm{xx}}}} = {\rm{\;}}0.11{{\rm{R}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{8}{{\rm{R}}^4}\)
Quarter circle		\({{\rm{I}}_{{\rm{xx}}}} = 0.055{{\rm{R}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = 0.055{{\rm{R}}^4}\)

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:

v = u + at

v2 = u2 + 2as

\(s =ut + \frac{1}{2}at^2\)

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

v = u + at

0 = u - gt₁

\(t_1=\frac{u}{g}\)

Part-II:

Initial velocity will be zero as the ball is at the highest point.

applying 1st equation of motion

v = u + at

3u = 0 + gt₂

\(t_2=\frac{3u}{g}\)

Therefore total time is:

t = t₁ + t₂

\(t=\frac{u}{g}+\frac{3u}{g}=\frac{4u}{g}\)