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A particle of mass m=5 is moving with uniform speed v=3√2 in the XY plane along the line y=x+4. The magnitude of angular momentum about origin is: ans: 60 first of all, how can a body undergoing linear motion have angular momentum. secondly, they have used L=mvr and taken r as 4/√2. How?
A particle of mass m=5 is moving with uniform speed v=3√2 in the XY plane along the line y=x+4. The magnitude of angular momentum about origin is:
ans: 60
first of all, how can a body undergoing linear motion have angular momentum.
secondly, they have used L=mvr and taken r as 4/√2. How?
the body is undergoing rotational motion about the origin i.e. it is rotating in a circle having its centre as the origin and the radius is equal to the distance of that point from the origin. the distance of the given line from origin is 4/square root2 now the angular momentum can be calculated easily by putting the values
the body is undergoing rotational motion about the origin i.e. it is rotating in a circle having its centre as the origin and the radius is equal to the distance of that point from the origin. the distance of the given line from origin is 4/square root2
now the angular momentum can be calculated easily by putting the values
see one can calculate da angular momentum undergoing linear motion from da origin......nw u draw da coordinate axes and draw da line u ll find dat da prependicular distance of da origin to the line is 4/V2......
y = x+4 (given) x-y+4 = 0 perpendicular distance from origen = 4/sqrt2=2sqrt2 m now angular momentam = mvr (r is the perpendicular distance of line of motion of particle & point about which we have to calculate angular momentam) =5*(3sqrt2)2sqrt2 =5*3*4 =60
y = x+4 (given)
x-y+4 = 0
perpendicular distance from origen = 4/sqrt2=2sqrt2 m
now angular momentam = mvr (r is the perpendicular distance of line of motion of particle & point about which we have to calculate angular momentam)
=5*(3sqrt2)2sqrt2
=5*3*4
=60
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