Problem 6/3 An object moving in a liquid experiences a linear drag force: vector

Question

Problem 6/3 An object moving in a liquid experiences a linear drag force: vector D = (bu , direction opposite the motion), where b is a constant called the drag coefficient. For a sphere of radius .R, the drag constant can be computed as b = 6 Pi n R, where n is the viscosity of the liquid. Part A Find an algebraic expression for ux (t), the x-component of velocity as a function of time, for a spherical particle of radius R and mass m that is sho horizontally with initial speed uo through a liquid of viscosity n. Express your answer in terms of the variables uo, n, R, t, m, at d appropriate constants. Ux (t) = Part B Water at 20 Degree C has viscosity n = 1.0 x 10^3 Ns/m2. Suppose a 4.0-cm-diameter. 40g ball is shot horizontally into a tank of 20 degree C water. How long will it take for the horizontal speed to decrease to 50 % of its initial value? Express your answer using two significant figures. t=

Explanation / Answer

Part A

Let velocity at any time t be vx

Drag coefficient experienced by the particle = b = 6(pi)nR

Drag force experienced = - b*vx

This is the only horizontal force being experienced by the particle. Hence, acceleration :

a = -b*vx / m

or, dvx/dt = -b * vx / m

or, dvx/vx = -b * dt / m

Integtate both sides from 0 to t, we get :

ln[vx(t)] - ln[v(0)] = -(b/m) * [t - 0]

or, ln[vx(t)/v(0)] = -bt/m

Hence, vx(t) = v(0) * e-bt/m

or, vx(t) = v(0) * e-6(pi)nRt/m

Part B

n = 10-3 Ns/m2

R = 0.02 m

m = 0.040 kg

And v(t) = 0.5v0

Using the reslut of Part A,

we have :

vx(t) = v(0) * e-6(pi)nRt/m

0.5v(0) = v (0) * e-6(pi)nRt/m

Putting in the values of n, R and m and solving, we get :

t = 73.72 seconds

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