System of Particles & Rotational Motion Test

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System of Particles & Rotational Motion Test - 6

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Question 1
1 / -0

Direction of linear velocity of a particle rotating in a circular motion is :

A
Along the tangent

B
Along the radius away from the centre

C
Along the axis of rotation

D
Along the radius of the centre

Solution

Unlike linear motion, where velocity and acceleration are directed along the line of motion, in circular motion the direction of velocity is always tangent to the circle. This means that as the object moves in a circle, the direction of the velocity is always changing. A linear velocity of a particle performing circular motion, which is directed along the tangent to the circular path at given point on the circular path at that instant is called instantaneous velocity. It is also called as tangential velocity.
Question 2
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A ball is under the effect of circular motion about a perpendicular axis with respect to the reference direction. The angular position (in radians) is given by the following function θ = t² - 0.4t + 2 .The angular position when angular velocity is zero is

A
1.96 radians

B
0.99 radians

C
1.06 radians

D
2.96 radians
Question 3
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Let H and E respectively be the angular speeds of the hour hand of a watch and that of the earth around its own axis. Compare the angular speeds of the earth and hour hand of a watch.

A
H = E

B
^ωH = 2E

C
^ωH = (1/2)E

D
^ωH = 4E

Solution

Given,
Angular velocity of the hour hand-H
Earth’s speed around its own axis-E
So, time taken by hour hand to complete a rotation is=T_H=12hr
Time taken by earth to complete a rotation =T_E=24 hours
So,
ω_E/ω_H =(2π/T_E )/2π/T_H=T_H/T_E=1/2
⇒ ω_E/ω_H=1/2
⇒ 2 ω_E= ω_H
Question 4
1 / -0

The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by . The linear velocity of the particle

A
1.84 m/s

B
1.80 m/s

C
1.82 m/s

D
1.88 m/s

Solution

The answer is option A
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
Question 5
1 / -0

All the points of a rigid body rotating about the given axis have same :

A
Linear acceleration

B
Angular velocity

C
Linear velocity

D
Angular momentum

Solution

Thus, it is the velocity of a reference point fixed to the body. During purely translational motion (motion with no rotation), all points on a rigid body move with the same velocity. However, when motion involves rotation, the instantaneous velocity of any two points on the body will generally not be the same.
Question 6
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A mixer grinder rotates clockwise, its angular velocity will be :

A
zero

B
negative

C
uniform but not zero

D
positive

Solution

Clock wise direction means inward field so angular velocity is negative. Anticlockwise means outward field so angular velocity is positive.
Question 7
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Which of the following is not a unit of angular displacement?

A
Degrees

B
revolutions

C
metre

D
radian

Solution

Angular displacement is measured in units of radians. 2 pi radians equals 360 degrees. The angular displacement is not a length (i.e. not measured in meters or feet), so an angular displacement is different than a linear displacement.
Question 8
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In an equilateral triangle of length 6 cm , three masses m₁= 40g , m₂= 60g and m₃= 60g are located at the vertices. The moment of inertia of the system about an axis along the altitude of the triangle passing through m₁is

A
1080 g cm²

B
1020 g cm²

C
1000 g cm²

D
1100 g cm²

Solution

Axis is along AD so Moment of inertia of first mass is zero and 2nd and 3rd masses are at distance of 6/2 from axis, so by using simply I=mr² we get, I=60×36/4+60×36/4=1080 g cm²
Question 9
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A CD accelerates uniformly from rest to 200 spins per minute in 8 seconds. If the rate of change of speed is constant, determine the instantaneous acceleration in rad/s² .

A
200.2 π rad/s²

B
25 rad/ s²

C
20.9 rad/ s²

D
2.6 rad/ s²
Question 10
1 / -0

A boy is playing with a tire of radius 0.5m. He accelerates it from 5rpm to 25 rpm in 15 seconds. The linear acceleration of tire is

A
70 m/s²

B
7 m/s²

C
10 m/s²

D
0.07 m/s²

Solution

r=0.5m, t=15s
n₁=5rpm=5/60 rps
n₂=25rpm=25/60rps
… ω₁=2π(5) rad/s
… ω₂=2π(25) rad/s
As,
a=r∝ [∝=ω₂- ω₁/t]
so, a=0.5[2π(n₂-n₁)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s²
a≈0.7 m/s²
Question 11
1 / -0

The angular momentum of a body is 31.4 Js and its rate of revolution is 10 cycle per second. The momentum of inertia of the body about the axis of rotation is

A
0.5 kgm²

B
0.15 kgm²

C
1.5 kgm²

D
2.5 kgm²

Solution

F = w/2π
w = 2πf
w = 2π × 10
w = 62.8
L = Iw
I = L/w
I = 31.4/62.8
I = 0.5 kgm2
Hence A
Question 12
1 / -0

A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly accelerates until it rotates at 0.8 rev/min. This occurs in 11s. Once it has reached this rate , it rotates at 0.8 rev/min uniformly . Determine the angular acceleration

A
7.55 x 10^-3 rad/s²

B
6.7 x 10^-3 rad/s²

C
0.083 rad/s²

D
0.083 rad/s

Solution

First we have to find ‘v’ by using the relation v = rw
Now, v = 0.167m/s
Then we find ‘a’ by using the relation a = v/t
So, a = 0.0152m/s².
Now, we can find angular acceleration by using the formula α = a/r
Finally, α = 0.0076 rad/sec²
Question 13
1 / -0

A demo CD, 8 cm in diameter, spins 200 rev/min. Determine its linear velocity in cm/s for a point 3 cm from the center.

A
62.8 cm/s

B
20/6 cm/s

C
20.9 cm/s

D
40 π/6 cm/s
Question 14
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The Equations of Rotational Motion are

A

B

C

D

Solution
Question 15
1 / -0

Rotational analogue of force is :

A
Moment of Inertia

B
Radius of gyration

C
Gyration

D
Torque

Solution

The equation τ = mr2α is the rotational analog of Newton's second law (F=ma), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass (or inertia).
Question 16
1 / -0

A wheel rotates with a constant acceleration of 2.0 rad/s² . If the wheel starts from rest, the number of revolutions it makes in the first ten seconds will be approximately:

A
8

B
32

C
24

D
16

Solution
Question 17
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The angular velocity of a second hand of a clock is :

A
π/ 30 rad/s

B
π/5 rad/s

C
2π rad/s

D
π /3 rad/s

Solution

Time taken to complete one rotation is 60 seconds.
One rotation involves 360 degree = 2π
Formula of angular velocity = theta/time taken
Hence, omega = 2π/60
= π/30 rad/s
Question 18
1 / -0

A car is decelerating uniformly from velocity v₁ to velocity v₂ while getting displaced by s. If the diameter of the wheel is d, the angular acceleration of the wheel is

A

B

C

D
Question 19
1 / -0

A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and center of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB respectively. Then

A
ω_B = ω

B
ω_A< ω_B

C
ω_A = ω

D
ω_A = ω_B

Solution

So, “I” gets cancelled.
Question 20
1 / -0

If n is the frequency of rotation of a body, its angular velocity is:

A
1/n

B
2π/n

C
n/2π

D
2πn