Question

A small, spherical bead of mass 3.00 g is released from rest at t 5 0 from a point under the surface of a viscous liquid. The terminal speed is observed to be vT 5 2.00 cm/s. Find (a) the value of the constant b that appears in Equation 6.2, (b) the time t at which the bead reaches 0.632vT, and (c) the value of the resistive force when the bead reaches terminal speed.

Answer 1

The acceleration of the bead in the viscous liquid at terminal speed is zero, from which the value of the constant b, can be determined.

• (a) The constant b is approximately 1.4715 kg/s.

• (b) The time at which the bead reaches  0.632·$v_T$, is approximately 2.038 × 10⁻³ seconds.

• (c) The value of the resistive force when the bead reaches terminal velocity is 0.02943 N.

Reasons:

The mass of the bead, m = 3.00 g

Time at which the mass is released, t = 0

Terminal velocity of $v_T$ = 2.00 cm/s = 0.02 m/s

(a) The equation 6.2, obtained online is presented as follows;

$\overrightarrow R = -b \cdot \overrightarrow v$

Where;

$\overrightarrow R$ = The resistive force = m·a

$\overrightarrow v$ = The relative velocity of the object

a = The acceleration of the object

We get;

m·g - b·v = m·a

At the terminal velocity, $v_T$, we have;

a = 0

Therefore;

m·g - b·$v_T$ = m × a = 0

m·g = b·$v_T$

$b= \dfrac{m \cdot g}{v_T}$

Therefore;

$b= \dfrac{0.003 \, kg \times 9.81 \, m/s^2}{0.02 \, m/s} = 1.4715 \, kg/s$

b = 1.4715 kg/s.

(b) From, m·g - b·v = m·a, and $a = \dfrac{dv}{dt}$, we have;

$m \cdot \dfrac{dv}{dt}= m \cdot g - b\cdot v$

Which gives;

$\right. \dfrac{dv}{dt}= \mathbf{g - \dfrac{b}{m} \cdot v}$

The expression for v obtained by differential equation is expressed as follows;

$v = \dfrac{m \cdot g}{b} \cdot \left(1 - e^{-b \cdot t/m} \right)$

Which gives;

$v = \mathbf{v_T \cdot \left(1 - e^{-b \cdot t/m} \right)}$

Therefore, when the bead reaches 0.632·$v_T$, we get;

$1 - e^{-b \cdot t/m}$ = 0.632

$e^{-b \cdot t/m}$ = 1 - 0.632 = 0.368

$t= \dfrac{ln(0.368)}{-\dfrac{1.4715}{0.003} } \approx 2.038 \times 10^{-3}$

The time at which the bead reaches  0.632·$v_T$, t ≈ 2.038 × 10⁻³ s.

(c) The value of the resistive force at terminal velocity, $v_T$, is $\overrightarrow F_R = -b \cdot v_T$

Which gives;

$\overrightarrow F_R$ = 1.4715 kg/s × 0.02 m/s = 0.02943 N.

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