System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 8

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Question 1
4 / -1

A person standing on a rotating platform with his hands lowered outstretches his arms. The angular momentum of the person

A
becomes zero

B
decreases

C
remains constant

D
inreases

Solution

Yes because there is absence of any external force or torque so angular momentum will remain constant
here outstretching the hands means internal forces are working.. so moment of inertia increases in this case and to make angular momentum constant when angular velocity decreases.
Question 2
4 / -1

The mass of an electron is 9 x 10^-31 kg. It revolves around the nucleus of an atom in a circular orbit of radius 4.5 Å, with a speed of 8 x 10⁵m/s. The angular momentum of electron is

A
2.24x 10^-4kgm²s^-1

B
3.24 x 10^-34kgm²s^-1

C
4.24 x10^-34kgm²s^-1

D
2.24 x 10^-34kgm²s^-1

Solution

M=9x10^-31Kg,r=4.5x10^-10
V=8x10⁵m/s
L=mvr
=9x10^-31x 4.5 x 10^-10x 8 x 10⁵= 3.24 x 10^-34 kgm²s^-1
Question 3
4 / -1

The earth rotates about an axis passing through its north and south poles with a period of one day. If it is struck by meteorites, then

A
moment of inertia decreases and angular velocity increases

B
moment of inertia decreases and angular velocity decreases

C
moment of inertia increases and angular velocity decreases

D
moment of inertia increases and angular velocity increases

Solution

Because angular momentum remains constant as the moment of inertia is increasing here so the angular velocity will decrease.
Question 4
4 / -1

Relation between torque and angular momentum is similar to the relation between

A
force and linear momentum

B
energy and displacement

C
acceleration and velocity

D
mass and moment of inertia

Solution

Torque is the rate of change of angular momentum. Force in linear motion corresponds to torque in rotary motion. Thus, just as Newton’s second law can be written as F = dP/dt (F for linear motion), it can be written as T = dL/dt for rotary motion.
Question 5
4 / -1

The dimensions of angular momentum are

A
[M¹L²T^-1]

B
[M¹L^-1T^-1]

C
[M¹L^-1T^-2]

D
[M¹L¹T^-1]

Solution

We know that angular momentum,
L = r x P
Thus its dimension is [r][P] = ML²T^-1.
Question 6
4 / -1

An earth satellite is moving around the earth in a circular orbit. In such case, what is conserved?

A
force

B
velocity

C
angular momentum

D
linear momentum

Solution

An earth satellite is moving in a circular motion due to change in the direction of velocity. Its angular momentum about centre is conserved as there is no external force acting on the system.
Question 7
4 / -1

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad/sec². It's net acceleration in m/s² at the end of 2.0 s is a approximately.

A
7.0

B
6.0

C
3.0

D
8.0

Solution

A uniform circular disc of radius 50 cm at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be shown by the figure given below:

Therefore,
Angular acceleration = α = 2 rad/s²
Angular speed, ω = αt = 4 rad/s
Centripetal acceleration, a_{c = ω}²r
4² x 0.5
=16 x 0.5
= 8m/s²
Linear acceleration at the end of 2 s is,
a_t = αt = 2 x 0.5 = 1 m/s²
Therefore, the net acceleration at the end of 2.0 sec is given by
Question 8
4 / -1

kgm²s^-1 is the unit of

A
force

B
angular momentum

C
tourqe

D
moment of inertia

Solution

L = Iw
L is the angular momentum, I is the MOI and w is the rotational velocity.
[I] = [ML²T⁰], [w] = [M⁰L⁰T^-1]
[L] = [MLT^-1]
Question 9
4 / -1

Angular momentum is equal to____.(All the symbols used have their usual meanings)

A
L = I²ω

B
L = Iω

C
L = ma

D
L = I/α

Solution

Angular momentum is equal to L = Iω. It is the formula.
Question 10
4 / -1

In case of a projectile, the angular momentum is minimum

A
at some location other than those mentioned above

B
at the starting point

C
at the highest point

D
on return to the ground

Solution

In case of a projectile, the angular momentum is minimum at the starting point.

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System of Particles and Rotational Motion Test - 8

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System of Particles and Rotational Motion Test - 8

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Question 1

becomes zero

decreases

remains constant

inreases

Solution

Question 2

2.24 x 10-4 kgm2s-1

3.24 x 10-34 kgm2s-1

4.24 x10-34 kgm2s-1

2.24 x 10-34 kgm2s-1

Solution

Question 3

moment of inertia decreases and angular velocity increases

moment of inertia decreases and angular velocity decreases

moment of inertia increases and angular velocity decreases

moment of inertia increases and angular velocity increases

Solution

Question 4

force and linear momentum

energy and displacement

acceleration and velocity

mass and moment of inertia

Solution

Question 5

[M1L2T-1]

[M1L-1T-1]

[M1L-1T-2]

[M1L1T-1]

Solution

Question 6

force

velocity

angular momentum

linear momentum

Solution

Question 7

7.0

6.0

3.0

8.0

Solution

Question 8

force

angular momentum

tourqe

moment of inertia

Solution

Question 9

L = I2ω

L = Iω

L = ma

L = I/α

Solution

Question 10

at some location other than those mentioned above

at the starting point

at the highest point

on return to the ground

Solution

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2.24x 10^-4kgm²s^-1

3.24 x 10^-34kgm²s^-1

4.24 x10^-34kgm²s^-1

2.24 x 10^-34kgm²s^-1

[M¹L²T^-1]

[M¹L^-1T^-1]

[M¹L^-1T^-2]

[M¹L¹T^-1]

L = I²ω