Question

While a person is walking, his arms (with typical lengths 70 cm measured from the shoulder joint) swing through approximately a 45? angle in 0.5 s. As a reasonable approximation, we can assume that the arm moves with constant speed during each swing. Find the magnitude of the force that the blood vessel must exert on the drop of blood (In N). Find the direction of the force that the blood vessel must exert on the drop of blood. What force would the blood vessel exert if the arm were not swinging? (In N)

Answer 1

1.)

angular velocity of arm = w = d/dt

given, d = 45 deg = pi/4

dt = 1/2 = 0.5 sec.

So, w = (pi/4)*0.5= pi/2

now, velocity(v) = w*R

given, R = length of arm = 70 cm = 0.70 m

So, v = 0.70*pi/2 = 1.10 m/sec.

now, radial acceleration is given by,

ar = v^2/R = 1.10/0.70

ar = 1.73 rad/s^2

From figure,

By force balance in vertical direction,

Fnety = m*ay

From figure,

F - m*g = m*ar

F = m*(g+ar)

here, m = mass of a the drop of blood = 1 gm = 1*10^-3 kg (from reference book)

So, F = (1*10^-3)*(9.81 + 1.73)

F = 1.15*10^-2 N

2.)

For equilibrium of drop,

direction of  the force on the drop of blood will be in upward direction.

3.)

when arm is at rest,

radial acceleration = ar = 0

So, by force balance,

F - m*g = 0

F = m*g = (1*10^-3)*9.81

F = 9.81*10^-3 N

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