Test: Angular Momentum - JEE MCQ

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.

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

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

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.

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

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.

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 

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]

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

At the starting point of the projectile, the perpendicular distance from the point of rotation is zero as it's on the ground itself. Since one factor of the product is zero, the angular momentum at the starting point will be zero too.