A small, spherical bead of mass 3.00 g is released from rest at t = 0 from a point under the surface of a viscous liquid. The terminal speed is observed to be £oT = 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 0.632£oT, and (c) the value of the resistive force when the bead reaches terminal speed.
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To answer these questions, we need to understand the concepts of terminal speed, resistive force, and the equation that relates them.
(a) The equation that relates terminal speed (v) and resistive force (f) is given by Equation 6.2:
v = (mg/b)
Where:
v = terminal speed
m = mass of the bead
g = acceleration due to gravity
b = constant
To find the value of the constant b, we need to rearrange the equation:
b = mg/v
Plug in the given values:
m = 3.00 g = 0.003 kg
g = 9.8 m/s²
v = 2.00 cm/s = 0.02 m/s
b = (0.003 kg * 9.8 m/s²) / 0.02 m/s
b = 1.47 kg·m/s
So, the value of the constant b is 1.47 kg·m/s.
(b) The time t at which the bead reaches 0.632 times the terminal speed (0.632v) can be found using the equation:
v = vo * (1 - e^(-bt/m))
where:
vo = initial velocity (0 in this case as the bead starts from rest)
t = time
b = constant
m = mass of the bead
First, rearrange the equation to solve for time (t):
t = -(1/(b/m)) * ln(1 - v/vo)
Plug in the given values:
v = 0.632 * 2.00 cm/s = 1.264 cm/s = 0.01264 m/s
vo = 0 (as the bead starts from rest)
b = 1.47 kg·m/s
m = 0.003 kg
t = -(1 / (1.47 kg·m/s / 0.003 kg)) * ln(1 - 0.01264 m/s / 0)
t = -(1 / (49,000 s^-1)) * ln(1)
t = 0 seconds
Therefore, the bead reaches 0.632 times the terminal speed instantly at t = 0 seconds.
(c) The value of the resistive force when the bead reaches terminal speed can be calculated using the equation:
f = m * g - b * v
Plug in the given values:
m = 0.003 kg
g = 9.8 m/s²
b = 1.47 kg·m/s
v = 2.00 cm/s = 0.02 m/s
f = 0.003 kg * 9.8 m/s² - 1.47 kg·m/s * 0.02 m/s
f = 0.0294 N - 0.0294 N
f = 0 N
Therefore, the value of the resistive force when the bead reaches terminal speed is 0 N.
To solve this problem, we will use Equation 6.2, which relates the resistive force (F) to the viscosity of the fluid (b) and the velocity of the object (v).
The equation is given as:
F = bv
(a) To find the value of the constant b, we can rearrange Equation 6.2:
b = F/v
Given that the terminal speed is £oT = 2.00 cm/s, and the bead has a mass of 3.00 g, we can use the formula F = mg to find the resistive force at terminal speed.
F = (mass) * (acceleration due to gravity)
F = (3.00 g) * (9.8 m/s^2)
F = 0.00300 kg * 9.8 m/s^2
F = 0.0294 N
Now, substituting the values of F and v into the equation, we can find b:
b = F / £oT
b = 0.0294 N / 0.02 m/s
b = 1.47 Ns/m
Therefore, the value of the constant b is 1.47 Ns/m.
(b) The time t at which the bead reaches 0.632 times the terminal speed (£oT) can be found using the equation for exponential decay:
v(t) = £oT * e^(-bt/mass)
We need to solve for t when v(t) = 0.632 * £oT.
0.632 * £oT = £oT * e^(-bt/m)
0.632 = e^(-bt/m)
To isolate t, take the natural logarithm of both sides:
ln(0.632) = ln(e^(-bt/m))
ln(0.632) = -bt/m
Solving for t:
t = (-m * ln(0.632)) / b
Given that the mass of the bead is 3.00 g (0.00300 kg), and the value of b is 1.47 Ns/m, we can substitute these values into the equation to find t:
t = (-0.00300 kg * ln(0.632)) / (1.47 Ns/m)
t ≈ 0.00139 s
Therefore, the time at which the bead reaches 0.632 times the terminal speed is approximately 0.00139 seconds.
(c) The value of the resistive force when the bead reaches terminal speed is the same as the resistive force at terminal speed.
Using the given terminal speed of £oT = 2.00 cm/s, we can substitute this value of v into Equation 6.2 to find F:
F = bv
F = (1.47 Ns/m) * (0.02 m/s)
F ≈ 0.0294 N
Therefore, the value of the resistive force when the bead reaches terminal speed is approximately 0.0294 N.