A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of mass 0.175 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.
__________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

I = 1/12 m1L2 + m2 0.52

I = 1/12 0.175*12 + 0.5*0.52

I = 0.1395 kg m2

L = Iw

L = 0.1395*2

L = 0.279 kg m2 /s

L = 0.279 J-sec

(b)

I = 1/3 m1L2 + m2 R2

I = 1/3 0.175*12 + 0.5*12

I = 0.558 kg m2

L = Iw

L = 0.558*2

L = 1.116 kg m2 /s

L = 1.116 J-sec

I = 1/12 m1L2 + m2 0.52

I = 1/12 0.175*12 + 0.5*0.52

I = 0.1395 kg m2

L = Iw

L = 0.1395*2

L = 0.279 kg m2 /s

L = 0.279 J-sec

(b)

I = 1/3 m1L2 + m2 R2

I = 1/3 0.175*12 + 0.5*12

I = 0.558 kg m2

L = Iw

L = 0.558*2

L = 1.116 kg m2 /s

L = 1.116 J-sec

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