Four identical particles of mass 0.337 kg each are placed at thevertices of a 3.
ID: 1743165 • Letter: F
Question
Four identical particles of mass 0.337 kg each are placed at thevertices of a 3.91 m x 3.91 m square and held there by fourmassless rods, which form the sides of the square. What is therotational inertia of this rigid body about an axis that(a) passes through the midpoints of opposite sidesand lies in the plane of the square, (b) passesthrough the midpoint of one of the sides and is perpendicular tothe plane of the square, and (c) lies in the planeof the square and passes through two diagonally oppositeparticles?
(a)
Number
Units
(b)
Number
Units
(c)
Number
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(a)
Number
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radrad/srad/s^2kg·m^2N·mkg·m^2/srad^2/s^2Explanation / Answer
a) I = 4 m * (L / 2)2 = mL2 where L is the length of a side -all 4 masses L / 2 from axis of rotation b) Length of diagonal from midpoint to corner d =( L2 + (L / 2)2 ) andd2 = 5 L2 / 4 The other 2 particles are L / 2 fromthe axis of rotation I = 2 m [ 5 L2 / 4 +L2 / 4 ] = 3 m L2 c) Distance from diagonal to corner d = [(L / 2)2 + [(L / 2 )2] d2 =L2 / 2 I = 2 m * L2 /2 = mL2 c) Distance from diagonal to corner d = [(L / 2)2 + [(L / 2 )2] d2 =L2 / 2 I = 2 m * L2 /2 = mL2Related Questions
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