A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of ma
ID: 1480244 • Letter: A
Question
A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of mass 0.200 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.
kg · m2/s
(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.
kg · m2/s
Explanation / Answer
Givne 1 meter stick 100 cm, mass of meter stick m = 0.2 kg,
mass of particle M = 0.5 kg
angular speed of 6.00 rad/s
Angular momentum L = I*W I moment of inertia, W angular speed
a) axis passing through the centre I = ml^2 / 12
total momet of inertia of the system is I = ml^2/12 +Mr^2
= 0.2*1^2/12 + 0.5*0.5^2
= 0.1417 kg m2
Angular momentum L = IW = 0.1417*6
= 0.8502 kg m2/s
b) at one end of the meter stick moment of inertia is
I = ml^2/3 + Mr^2
= 0.2*1^2/3 + 0.5*1^2
= 0.567 kg m2
L = I*W = 0.567* 6.00
= 3.402 kg m2/s
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