6.8. A marble with a mass m of 3.00 g is released from rest in a bottle of syrup. Its terminal speed vt is 1.79 cm/s. (a) Presuming the resistive force R = −bv, what is the value of b (N s/m)? (b) What is the strength of the resistive force (N) when the marble reaches terminal speed? (c) How long in milliseconds does it take for it to reach a speed of 0.600vt?

a. 1.642 Ns/m b. 0.0294 N c. 5 × 10⁵ ms

Explanation:

a. Presuming the resistive force R = −bv, what is the value of b (N s/m)?

Using the equation of motion on the object,

W + R = ma where W = weight of the mass = mg where m =mass of marble = 3.00 g = 0.003 kg and g = acceleration due to gravity = 9.8 m/s² , R = resistive force = -bv where v = velocity and a = acceleration of marble.

So,

mg - bv = ma

At terminal speed, a = 0,

So, mg - bvt =m(0)

mg - bvt = 0

mg = bvt

b = mg/vt since terminal speed vt = 1.79 cm/s = 0.0179 m/s

So, b = 0.003 kg × 9.8 m/s²/0.0179 m/s

b = 0.0294 kgm/s² ÷ 0.0179 m/s

b = 1.642 Ns/m

b. What is the strength of the resistive force (N) when the marble reaches terminal speed?

Since the resistive force R = -bv, at terminal speed, vt

R = -bvt

R = -1.642 Ns/m × 0.0179 m/s

R = -0.0294 N

So, its strength is 0.0294 N

(c) How long in milliseconds does it take for it to reach a speed of 0.600vt?

Using mg - bv = ma where a = dv/dt,

mg - bv = mdv/dt

g - bv/m = dv/dt

separating the variables, we have

dv/(g - bv/m) = dt

Integrating, we have

∫dv/(g - bv/m) = ∫dt

(-b/m)/(-b/m) × ∫dv/ (g - bv/m) = ∫dt

1/(-b/m) ∫(-b/m)dv/(g - bv/m) = ∫dt

1/(-b/m) ㏑(g - bv/m) = t + C

㏑(g - bv/m) = -m/bt - mC/b

㏑(g - bv/m) = -m/bt + C'  (C' = -mC/b)

taking antilogarithm of both sides, we have

g - bv/m = exp(-m/bt + C')

g - bv/m = exp(-m/bt)expC'

g - bv/m = Aexp(-m/bt) (A = expC')

bv/m = g - Aexp(-m/bt)

v = mg/b - (Am/b)exp(-m/bt)

when t = 0, v = 0 (since the marble starts from rest)

0 = mg/b - (Am/b)exp(-m/b(0))

0 = mg/b - (Am/b)exp(0))

-mgb = -Am/b

A = g

v = mg/b - (mg/b)exp(-m/bt)

when v = 0.600vt = 0.600 × 0.0179 m/s = 0.01074 m/s

mg/b = 0.003 kg × 9.8 m/s²/1.642 Ns/m = 0.0179 m/s and m/b = 0.003 kg/1.642 Ns/m = 0.00183/s

So,v = mg/b - (mg/b)exp(-m/bt)

0.01074 m/s = 0.0179 m/s - (0.0179 m/s)exp[(-0.00183/s)t]

0.01074 m/s - 0.0179 m/s = - (0.0179 m/s)exp[(-0.00183/s)t]

-0.00716 m/s = - (0.0179 m/s)exp[(-0.00183/s)t]

exp[(-0.00183/s)t] = -0.00716 m/s/-0.0179 m/s

exp[(-0.00183/s)t] = 0.4

taking natural logarithm of both sides, we have

(-0.00183/s)t = ㏑(0.4)

(-0.00183/s)t = -0.9163

t = -0.9163/-0.00183

t = 500 s

t = 500 × 1000 ms

t = 5 × 100000

t = 5 × 10⁵ ms

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm