While a person is walking, his arms swing through approximately a 45.0 ∘ angle in 0.480 s . As a reasonable approximation, we can assume that the arm moves with constant speed during each swing. A typical arm is 70.0 cm long, measured from the shoulder joint.
1.What is the acceleration (m/s^2) of a 1.70 gram drop of blood in the fingertips at the bottom of the swing?
2. Find the force (N) that the blood vessel must exert on the drop of blood in part (a).
3. What force (N) would the blood vessel exert if the arm were not swinging?
angular velocity ω = Θ / t,
and Θ = 45º = π/4 radians, so
ω = π/4 rads / 0.480s = 1.6362 rad/s
A) At the bottom of the swing of the arm, the blood is drawn
downward by both gravitational acceleration and radial
acceleration, so
a = g + ω²r = 9.8m/s² + (1.6362rad/s)² * 0.700m = 11.6740
m/s²
B) F = ma = 0.0017kg * 11.6740m/s² = 0.01984 N ( here a is net
acceleration at bottom point)
C) F' = mg = 0.00170kg * 9.8m/s² = 0.01666 N ( sice it is not
swinging so angular velocity = 0)
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