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Test: Curvilinear Motion - Civil Engineering (CE) MCQ


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7 Questions MCQ Test Engineering Mechanics - Test: Curvilinear Motion

Test: Curvilinear Motion for Civil Engineering (CE) 2024 is part of Engineering Mechanics preparation. The Test: Curvilinear Motion questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Curvilinear Motion MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Curvilinear Motion below.
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Test: Curvilinear Motion - Question 1

Three identical cars A, B, and C are moving at the same speed on three bridges. Car A moves on a plane bridge, B on a convex bridge, and C on a concave bridge. FA, FB, and FC are the normal forces extorted by the cars on the bridges, then we can say that

Detailed Solution for Test: Curvilinear Motion - Question 1

When a body moves along a curved path a net force acts on the body that is directed towards the center of curvature. This force is called the centripetal Force, FC
FC = mV2/r
where, m is body mass, V is magnitude of the velocity, r is radius of curvature.

Calculation:
Given:
Case 1:
The car A is running on a plane bridge as shown -

The normal force FA = mg 
Case 2:-
The car B is running on a convex bridge as shown - 

Now this is a case of plane circular motion and in this case, the net centripetal force acts towards the center of curvature.
So the normal force

Case 3:
The car C is running on a concave bridge as shown - 

In this case, there is also a net centripetal force acting towards the center of curvature.
So the normal force FC

Hence, by looking at all three equations,
FC > FA > FB 
we can say that the contact force FC is the maximum among all three forces.

Test: Curvilinear Motion - Question 2

The radial component of velocity and acceleration in curvilinear motion are 

Detailed Solution for Test: Curvilinear Motion - Question 2

Velocity:
There is the only a tangential component of velocity in the curvilinear motion.

Let the unit vector in the tangential direction be denoted as 

Here, v is nothing but the derivative of a position vector

Here, the first term represents the tangential component of acceleration and the second term represents the radial component of acceleration

So, the radial component of acceleration = r˙θ˙
And the velocity component in a radial direction is 0.

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Test: Curvilinear Motion - Question 3

The speed of a shaft increases uniformly from 300 rpm to 800 rpm in 10 s. The angular acceleration is

Detailed Solution for Test: Curvilinear Motion - Question 3

Angular acceleration is the rate of change of angular velocity

Calculation:
Given

Test: Curvilinear Motion - Question 4

The radial component of velocity and acceleration in curvilinear motion are 

Detailed Solution for Test: Curvilinear Motion - Question 4

Radial and Transverse co-ordinates:

  • In this system, the position of the particle is defined by the polar coordinates r and θ
  • Further, the velocity and acceleration of the particle are resolved into components along and perpendicular to the position vector  These components are called:
    • Radial components denoted by
    • Transverse components denoted by 

The unit vectors of  are represented by  respectively. Here A1A2 and B1B2 are perpendicular to the x-axis and are drawn from the endpoints A1 and B1 of the unit vectors  respectivel
Furthur: 
That gives: A1A2 = Sin θ and B1B2 = Cos θ, PA2 = Cos θ and PB2 = Sin θ
Applying the triangle law of addition of vectors

Velocity Component: When the above identities are differentiated with respect to time t.


In terms of the position vector , the velocity vector is defined as

Acceleration:

Substituting the values for  in acceleration.

The radial component of acceleration,

Test: Curvilinear Motion - Question 5

A body moves with a speed of 10 m/s in the curved path of 25 m radius of curvature. If the tangential acceleration is 3 m/s2, then total acceleration for the body will be:

Detailed Solution for Test: Curvilinear Motion - Question 5

Centripetal Acceleration (ac): 

  • 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 = v2/r
Where v = velocity of the object and r = radius
Tangential acceleration (at):

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


Where α = angular acceleration and r = radius
CALCULATION:
Given – v = 10 m/s, r = 25 m and at = 3 m/s2

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


Centripetal Acceleration (ac):

Hence, net acceleration 

Test: Curvilinear Motion - Question 6

A car travels on a horizontal circular track of radius 9 m, starting from rest at a constant tangential acceleration of 3 m/s2. What is the resultant acceleration of the car, 2 sec after starting? 

Detailed Solution for Test: Curvilinear Motion - Question 6

Tangential acceleration = r × angular acceleration

Now,

Now,
Normal acceleration 

∴ Resultant acceleration 

Test: Curvilinear Motion - Question 7

The midpoint of a rigid link of a mechanism moves as a translation along a straight line, from rest, with a constant acceleration of 5 m/s2. The distance covered by the said midpoint in 5 s of motion is

Detailed Solution for Test: Curvilinear Motion - Question 7

Distance covered by translation, S = u × t + (1/2)at2
u is the initial motion, t is the time and a is the acceleration
Calculation:
Given u = 0, a = 5 m/s2 and t = 5 s

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