Preparation problems
13-19. In the Olympic hammer throw, a contestant whirls
a 7.3-kg steel ball on the end of a 1.2-m cable. If the contestant's
arms reach an additional 90 cm from his axis of rotation, and if the
speed of the ball just prior to release is 27 m/s, what is the magnitude
of its angular momentum?
Solution:
We can use the definition of angular momentum from our constants and equation
sheet:
Since r and p are perpendicular, this simplifies
to
We know r = 2.1 m, m = 7.3 kg, and v = 27 m/s, giving us
13-21. A gymnast of rotational inertia 62 kg-m2 is
tumbling head over heels. If her angular momentum is 470 kg-m2/s,
what is her angular speed?
Solution:
We can use the definition of angular momentum from our constants and equation
sheet:
We can then simply solve for w as
Follow-up problems
13-11. A force
F = 2.0 i - 8.0j N is applied at the point x = 1.2 m, y = 3.6 m. Find the
resulting torque about the origin.
Solution:
We can use our definition of torque:
This gives us
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[(1.2 m)(-8.0 N) - (2.0 N)(3.6 m)] |
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k
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13-16. Find the cross product of the vectors
A = i - 2j + 3k and
B = 2i + 4j + 2k.
Solution:
We simply must perform the cross product
This gives us
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[(-2)(2)-(3)(4)] |
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i
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- [(1)(2)-(3)(2)] |
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j
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+ [(1)(4)-(-2)(2)] |
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k
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Solutions translated from TEX by TTH, version 1.57.
