Question

A particle of mass 0.350 kg is attached to the 100-cm mark of a meterstick of mass 0.150 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 6.00 rad/s.

(a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark.

(b) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark.

Answer 1

(a)

angular momentum L = (I + mparticle*r^2)*w

I = moment of inertia of stick = (1/12)*mstick*l^2 = (1/12)*0.15*1^2 = 0.0125 kg m^2

l = 100 cm = 1 meter

r = distace of particel from pivot point = 50 cm = 0.5 m

L = (0.0125 + 0.35*0.5^2)*6

L = 0.6 kg m^2/s

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(b)

angular momentum L = (I + mparticle*r^2)*w

I = moment of inertia of stick = (1/3)*mstick*l^2 = (1/3)*0.15*1^2 = 0.05 kg m^2

l = 100 cm = 1 meter

r = distace of particel from pivot point = 100 cm = 1 m

L = (0.05 + 0.35*1^2)*6

L = 2.4 kg m^2/s

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DONE please check the answer. any doubts post in comment box