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QUESTION: 1

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

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

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 10^{5}m/s. The angular momentum of electron is

Solution:

M=9x10^{-31}Kg,r=4.5x10^{-10}

V=8x10^{5}m/s

L=mvr

=9x10^{-31 }x 4.5 x 10^{-10 }x 8 x 10^{5 }= 3.24 x 10^{-34} kgm^{2}s^{-1}

QUESTION: 3

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

Solution:

Because angular momentum remains constant as the moment of inertia is increasing here so the angular velocity will decrease.

QUESTION: 4

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

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

The dimensions of angular momentum are

Solution:

We know that angular momentum,

L = r x P

Thus its dimension is [r][P] = ML^{2}T^{-1}.

QUESTION: 6

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

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

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/sec^{2}. It's net acceleration in m/s^{2} at the end of 2.0 s is a approximately.

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/s^{2}

Angular speed, ω = αt = 4 rad/s

Centripetal acceleration, a_{c = ω}^{2}r

4^{2} x 0.5

=16 x 0.5

= 8m/s^{2}

Linear acceleration at the end of 2 s is,

a_{t} = αt = 2 x 0.5 = 1 m/s^{2}

Therefore, the net acceleration at the end of 2.0 sec is given by

QUESTION: 8

kgm^{2}s^{-1} is the unit of

Solution:

L = Iw

L is the angular momentum, I is the MOI and w is the rotational velocity.

[I] = [ML^{2}T^{0}], [w] = [M^{0}L^{0}T^{-1}]

[L] = [MLT^{-1}]

QUESTION: 9

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

Solution:

Angular momentum is equal to L = Iω. It is the formula.

QUESTION: 10

In case of a projectile, the angular momentum is minimum

Solution:

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

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