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A particle of mass 2kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. The height of the table from the ground is 0.8m If the angular speed of the particle is 12rads1. The magnitude of its angular momentum about a point on the ground right under the center of the circle is


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Solution

Step 1. Given Data,

The mass of a particle =2kg
The radius of the circular path of motion of the body, R=0.6m
Height =0.8m
The angular speed of the particle, ω=12rads1

Step 2. Formula Used,

The angular momentum of an object having mass (m) and linear velocity (v) is:

|L|=m|v||r|sin(θ)

|L| is the magnitude of a vector quantity of the angular momentum of the body,

|r| is the magnitude of the radius of the circular path of motion of the body,

|v|is the magnitude of another vector quantity, the velocity with which the body is travelling,

θ is the angle between the radius vector and the velocity vector.

L= r×p

p=mv

L= Angular Momentum

m = mass of the particle

p= linear momentum

v is the velocity of the particle

From Pythagoras theorem,

H=P2+B2

H is the resultant radius vector, B is the base, and P is the height.

Step 3. Calculating the magnitude of its angular momentum,

We consider Has the resultant radius vector and we can denote it as rperpendicular

The perpendicular distance between the point and mass rperpendicular is given by,

rperpendicular=|r|=(0.6)2+(0.8)2=1m

The angle between the resultant radius vector and the linear velocity vector is θ=90.
The angular momentum is,

|L|=m|v||r|sin(θ)

|L|=m|ωR||r|sin(90o)
L=2×12×0.6×1×1=14.4kgm2s1
Thus the magnitude of angular momentum, L=14.4kgm2s1.

Therefore, the magnitude of its angular momentum about a point on the ground right under the center of the circle is14.4kgm2s-1.


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