# Physics: terminal velocity of a sphere; stress in the bones of the arm during the swing

1. Find an equation for the terminal velocity of a sphere falling through a viscous fluid (like water or blood). How does the terminal velocity scale with the size of the falling object? Assume that the density of that object is much larger than that of the fluid through which it falls (neglect buoyancy).

a. V ~ L^-3

b. V ~ L^0

c. V ~ L^1

d. V ~ L^2

e. V ~ L^3

2. A person is bowling with a 8 kg bowling ball. Their arm is 1 m long. If they pull the ball back 0.7 m (hint: call this A) and use no muscle effort to accelerate their bowl (gravity provides the driving force), what is the stress in the bones of their arm during the swing, just as the arm passes the straight up-and-down position. Treat the arm as a simple pendulum with one long bone, and assume that the cross-sectional area of this bone is 2 cm2. Neglect the mass of the arm itself.

a. 4.8 x 106 dyn/cm2

b. 2.7 x 106 dyn/cm2

c. 3.9 x 106 dyn/cm2

d. 6.0 x 105 dyn/cm2

e. 5.8 x 106 dyn/cm2.

https://brainmass.com/physics/velocity/stress-bones-arm-during-swing-282883

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Please refer to the attachment.

1 Find an equation for the terminal velocity of a sphere falling through a viscous fluid (like water or blood). How does the terminal velocity scale with the size of the falling object? Assume that the density of that object is much larger than that of the fluid through which it falls (neglect buoyancy).

a. V ~ L-3

b. V ~ L0

c. V ~ L1

d. V ~ L2

e. V ~ L3

Solution:

Fd

mg

The sphere is subjected to two forces i) weight mg down wards and ii) viscous drag upwards (third force due to buoyancy has been neglected). As per Stoke's formula, viscous drag is given by : Fd = 6Πηrv where η = coefficient of viscosity, r = radius of the sphere and v = velocity

The sphere reaches the terminal velocity (final constant velocity) when the viscous drag force equals the weight mg.

6Πηrv = mg .........(1)

Mass m = (4Π/3)r3ρ where ρ is the density of the sphere

Substituting for m in (1):

6Πηrv = (4Π/3)r3ρg

v = (2ρg/9η)r2

Hence, v is directly proportional to square of the radius r (d is correct).

2. A person is bowling with a 8 kg bowling ball. Their arm is 1 m long. If they pull the ball back 0.7 m (hint: call this A) and use no muscle effort to accelerate their bowl (gravity provides the driving force), what is the stress in the bones of their arm during the swing, just as the arm passes the straight up-and-down position. Treat the arm as a simple pendulum with one long bone, and assume that the cross-sectional area of this bone is 2 cm2. Neglect the mass of the arm itself.

a. 4.8 x 106 dyn/cm2

b. 2.7 x 106 dyn/cm2

c. 3.9 x 106 dyn/cm2

d. 6.0 x 105 dyn/cm2

e. 5.8 x 106 dyn/cm2

Solution:

O

θ 1m

1m

C A

B

0.7m

Angle θ = 0.7/1 = 0.7 radians = 0.7 x 180/Π degrees = 40O

CO = cos40O = 0.76 m

Height through which the ball is raised = h = BC = BO - CO = 1 - 0.76 = 0.24 m

Potential energy rise of the ball in position A = mgh

This PE is converted into kinetic energy when the ball is at its lowest position. Hence,

½ mv2 = mgh where v is the velocity of the ball at the lowest position

v2 = 2gh = 2 x 9.8 x 0.24 = 4.7 or v = 2.17 m/s

In the straight vertical position of the arm (i.e. at the lowest position of the ball), there are two forces acting on the ball i) weight of the ball acting vertically downwards, ii) centrifugal force acting vertically down wards.

Weight of the ball = 8x9.8 = 78.4 N

Centrifugal force = mv2/R = 8x4.7/1 = 37.6 N

Total force = 78.4 + 37.6 = 116 N

An equal and opposite tension is developed in the arm.

Stress in the arm = Tension in the arm/Area of cross section = 116/(2x10-4) = 5.8x105 N/m2

Or 5.8x106 dynes/cm2.

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