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Solution: We can find the angular momentum by using the formula, and the moment of inertia of a solid disc (ignoring the hole that is present in the middle). The angular momentum will be: L = I ω. L = (12MR2)ω. So, L = 12(0.0200kg)(0.0600m)2(160.0radians/s) L = 0.00576 kg. m2 /s. The angular momentum of this DVD will be 0.00576 kg. m2 /s.
Angular Momentum. Torque and angular momentum are closely related to each other. Angular momentum is the rotational analogue of linear momentum ‘p’ and is denoted by ‘l’. It is a vector product. Angular momentum of the particle is. l = r × p. l = r p sinθ, where θ is the angle between r and p. Relation between Torque and Angular Momentum
Angular momentum can be defined as the vector product of the angular velocity of a particle and its moment of inertia. When a particle of mass m shows linear momentum (p) at a position (r) then the angular momentum with respect to its original point O is defined as the product of linear momentum and the change in position.
SI Unit of angular momentum kgm2/sAngular momentum =moment of inertia×Angular velocity....(1)Dimensional formula of moment of inertia=M1L2T0Dimensional formula of Angular velocity =M0L0T−1Putting these values in above eq. (1)So dimensional formula of angular momentum =M1L2T−1. Was this answer helpful?
The angular momentum as electron is a given orbital is calculated using, Orbital angular momentum = √ l (l + 1) h 2 π. Now for d-electron, l=2
A particle performs uniform circular motion with an angular momentum L.If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes:
If $$ \omega_1$$ is the angular valocity when the moment of inertia is $$ l_1 $$ and $$ \omega_2 $$ is the angular velocity when the moment of inertia is $$ i_2 $$ , then from the law of conservation of angular momentum it follows that $$ l_1,\omega_1=l_2 \omega _2 $$ Examples: 1.A spinning ballet dancer uses the principle of conservtion of ...
An electron in an excited state of L i 2 + ion has angular momentum 3 h 2 π. The de Broglie wavelength of the electron in this state in p π α 0 (where α 0 is the Bohr radius). The value of p is
The angular momentum of the electron in the nth orbit= n h 2 π where orbit number = n = 4 (Given) Angular momentum of the electron in the fourth orbit= 4 h 2 π = 2 h π
Angular momentum in 3 rd orbit P 1 = n h 2 π = 3 2 h π Angular momentum in 3 d -orbital ( l = 2 for d-orbitals) P 2 = h 2 π √ l ( l + 1 ) = h 2 π × √ 6
