A particle of mass 0.300 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.300 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 2.00 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.

(Answers should be in kg*m^2/s)

Explanation / Answer

A)

I = 1/12 m1L2 + m2 0.52

I = 1/12 0.2*12 + 0.3*0.52

I = 0.916kg m2

L = Iw

L = 0.916*2

L = 0.183 kg m2 /s

(b)

I = 1/3 m1L2 + m2 R2

I = 1/3*0.2*12 + 0.3*12

I = 0.366 kg m2

L = Iw

L = 0.366*1

L = 0.733 kg m2 /s

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