Physic- A hammer thrower rotates
2008-12-15 2:32 am
A hammer thrower rotates faster and faster until he releases the hammer. the last revolution takes place at a constant angular speed and last 4.0s. During the last revolution , the hammer of mass 6.0kg is at a distance of 2.0m from the central vertical axis of rotation through the man's body. Find the magnitude of the force exerted by the man's hands on the hammer. F=?
回答 (1)
2008-12-15 3:01 am
✔ 最佳答案
The force exerted by the man's hand on the hammer accounts for the centripetal force of the hammer.Now, mass of the hammer, m = 6.0 kg
Radius of the circular path, r = 2.0 m
Angular frequency, ω = 2π / T
= 2π / 4.0
= 1.57 rads-1
Now, we have
F = mrω2
F = (6.0)(2.0)(1.57)2
F = 29.6 N
參考: Myself~~~
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