Past Year Questions: Translational And Rotational Motion

# Past Year Questions: Translational And Rotational Motion

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## 15 Questions MCQ Test | Past Year Questions: Translational And Rotational Motion

Past Year Questions: Translational And Rotational Motion for Mechanical Engineering 2022 is part of Mechanical Engineering preparation. The Past Year Questions: Translational And Rotational Motion questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Past Year Questions: Translational And Rotational Motion MCQs are made for Mechanical Engineering 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Past Year Questions: Translational And Rotational Motion below.
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Past Year Questions: Translational And Rotational Motion - Question 1

### AB and CD are two uniform and identical bars of mass 10 kg each, as shown. The hinges at/A and B are frictionless. The assembly is released from rest and motion occurs in the vertical plane. At the instant that the hinge B passes the point B, the angle between the two bars will be [1996]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 1

θ = 60° will remain same as weight mg of the rod CD passes through hinge point B so that there will be no torque in rod CD with respect to B.

Past Year Questions: Translational And Rotational Motion - Question 2

### A car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is [1998]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 2

For t1 =5 secs, S1 = 450 m
and for t2 = 5 secs  S2 = 700 m
From the relation S = ut + 1/2 at2
450 = 5u + 1/2 × a × 25 [since initial velocity = u]
or 450 = 5u + 25a/2     (i)
Also after first 5 seconds, velocity of car,
v = u + at
= u + 5a.
After first 5 seconds, u + 5a will be the initial velocity for next 5 seconds, therefore
700 = (4 + 5a)5 + 1/2a × 25
= 20 + 25a + 25a/2
From equation (i), we have
700 = (5u + 25a/2) + 25a = 450 + 25a, a = 10 m/sec2.

Past Year Questions: Translational And Rotational Motion - Question 3

### The time variation of the position of a particle in rectilinear motion is given by x = 2t3 + t2 + 2t. If v is the velocity and a is the acceleration of the particle in consistent units, the motion started with [2005]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 3

Given x= 2t3 + t2 + 2t
Velocity : v = dx/dt = 6t2 + 2t + 2
At the start of motion, t = o
V = 0 + 0 + 2
V = 2 m/s
Acceleration; a = dv/dt = 12t + 2
at t = 0
a = 0 + 2
∴ a = 2 m/s2

Past Year Questions: Translational And Rotational Motion - Question 4

A simple pendulum of length of 5 m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of 5 m/s. The net force on the bob at the mean position is

[2005]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 4

The net force at the mean position will be zero.

Past Year Questions: Translational And Rotational Motion - Question 5

A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of 2L/3 from the hinge so that the rod swings to the right. The horizontal reaction at the hinge is

[2009]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 5

∴ R × L = P × L/3

Past Year Questions: Translational And Rotational Motion - Question 6

A and B are the end points of a diameter of a disc rolling along a straight line with a counter clockwise angular velocity as shown in figure. Referring to the velocity vectors  and  shown in the figure

[1990]

Past Year Questions: Translational And Rotational Motion - Question 7

The cylinder shown below rolls without slipping. Toward which of the following points is the acceleration of the point of contact on the cylinder directed?

[1993]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 7

Acceleration of the point of contact A acts towardsgeometric centre.

Past Year Questions: Translational And Rotational Motion - Question 8

A stone of mass m at the end of a string of length l is whirled in a vertical circle at a constant speed. The tension is the string will be maximum when the stone is

[1994]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 8

T1 > T2
Maximum tension will be maximum at the bottom

Past Year Questions: Translational And Rotational Motion - Question 9

A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane, the velocity of point A on the rod is 1 m/s. The angular velocity of the rod at the instant is

[1996]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 9

VA = rA ω
1 = 1 cos 60° × ω

Past Year Questions: Translational And Rotational Motion - Question 10

As shown in figure, a person A is standing at the centre of a rotating platform facing person B who is riding a bicycle, heading East. They relevant speeds and distances are shown in given figure person, a bicycle, heading East. At the instant under consideration, what is the apparent velocity of B as seen by A?

[1999]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 10

Relative velocity = 8

Past Year Questions: Translational And Rotational Motion - Question 11

A rigid body shown in the fig. (a) has a mass of 10 kg. It rotates with a uniform angular velocity w. A balancing mass of 20 kg is attached as shown in fig. (b). The percentage increase in mass moment of inertia as a result of this addition is

[2004]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 11

I1 = 10 × (200)2 = 4 × 105
I2 = 10 × (200)2 + 20 × (100)2 = 6 × 105
∴ % increase in moment of inertia

Past Year Questions: Translational And Rotational Motion - Question 12

A circular disk of a radius R rolls without slipping at a velocity V. The magnitude of the velocity at point P (see figure) is:

[2008]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 12

The instantaneous centre of rotation of the disk is 0, i.e. its point of contact with ground every other point on the disk is rotating with respect to this center with an angular velocity.
ω = v/R

⇒ Velocity of P = ω. OP = v/R .√3R =√3v

Past Year Questions: Translational And Rotational Motion - Question 13

A solid disk of radius r rolls without slipping on a horizontal floor with angular velocity ω and angular acceleration α. The magnitude of the acceleration of the point of contact on the disc is

[2012]

Past Year Questions: Translational And Rotational Motion - Question 14

A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is

[2012]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 14

K.E. of fly wheel = 1/2 Iω2
where,

∴ K.E. = 1/2 × 0.4 × 62.832 = 790 J

Past Year Questions: Translational And Rotational Motion - Question 15

A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on motor with respect to center of gravity (CG) of the mobile and the rest of the dimension of the mobile phone are shown. The mobile is kept on a flat horizontal surface.

Given in addition that the eccentric mass = 2 grams, eccentricity = 2.19 mm, mass of the mobile = 90 grams, g= 9.81 m/s2. Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately

[2015]

Detailed Solution for Past Year Questions: Translational And Rotational Motion - Question 15

When lifted from ground at Q
Reaction = 0
∴ taking moments about ‘p’ and equating to 0
.09 × .06 = mr ω2 × .09
9.01 × .09 × .06 = .002 × 2.19 × 10–3 × ω2 × .09