Question

A block of 0.500 kg weight is fixed to the 200 cm mark of a 0.300 kg weight wooden
pole. The rod runs circularly on the surface of a horizontal frictionless table with 5.00
rad/s angular velocity. What is the magnitude of the system's angular momentum when
the rod is rotated through the 100 cm mark on an axis perpendicular to the table?
(a) 1.50 kg.m2/s
(b) 2.50 kg.m
· m/s
(c) 3.00 kg . m2/s
(d) 3.50 kg.m2/s​

Answer 1

Answer:

A 1.50 kg particle moves in the xy plane with a velocity of v = (4.20 i – 3.60 j ) m/s. Determine the angular momentum of the particle when its position vector is r = (1.50 i + 2.20 j ) m.

L = r X p L = (m) (r X v)

i j k

1.5 2.2 0

4.2 -3.6 0

1.5(-3.6) = -5.4 4.2(2.2) = 9.2

+ ( 2.2*0 – (-5.4) * 0 ) i

- ( 1.5*0 – 6.3 * 0 ) j

+ ( 1.5*-3.6 – 4.2 * 2.2 ) k

1.5 (-14.6) = -22

m(r x v) = -22.0 k (kg m2/s) or Joule-sec

Ch 11.3 #25

A particle of mass 0.4 kg is attached to the 100-cm mark of a meter stick of mass 0.1 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4 rad/sec. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50-cm mark and (b) perpendicular to the table through the 0-cm mark.

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