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For a steel ball bearing (diameter 2 mm and density 7.8
Chapter 2, Problem 2.10(choose chapter or problem)
For a steel ball bearing (diameter 2 mm and density \(7.8 \mathrm{~g} / \mathrm{cm}^{3}\)) dropped in glycerin (density \(1.3 \mathrm{~g} / \mathrm{cm}^{3}\) and viscosity \(12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) at STP), the dominant drag force is the linear drag given by (2.82) of Problem 2.2.
(a) Find the characteristic time \(\tau\) and the terminal speed \(v_{\text {ter }}\). [In finding the latter, you should include the buoyant force of Archimedes. This just adds a third force on the right side of Equation (2.25).] How long after it is dropped from rest will the ball bearing have reached 95% of its terminal speed?
(b) Use (2.82) and (2.84) (with \(\kappa=1 / 4\) since the ball bearing is a sphere) to compute the ratio \(f_{\text {quad }} / f_{\text {lin }}\) at the terminal speed. Was it a good approximation to neglect \(f_{\text {quad }}\)?
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QUESTION:
For a steel ball bearing (diameter 2 mm and density \(7.8 \mathrm{~g} / \mathrm{cm}^{3}\)) dropped in glycerin (density \(1.3 \mathrm{~g} / \mathrm{cm}^{3}\) and viscosity \(12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) at STP), the dominant drag force is the linear drag given by (2.82) of Problem 2.2.
(a) Find the characteristic time \(\tau\) and the terminal speed \(v_{\text {ter }}\). [In finding the latter, you should include the buoyant force of Archimedes. This just adds a third force on the right side of Equation (2.25).] How long after it is dropped from rest will the ball bearing have reached 95% of its terminal speed?
(b) Use (2.82) and (2.84) (with \(\kappa=1 / 4\) since the ball bearing is a sphere) to compute the ratio \(f_{\text {quad }} / f_{\text {lin }}\) at the terminal speed. Was it a good approximation to neglect \(f_{\text {quad }}\)?
ANSWER:Step 1 of 5
Consider the given data as follows.
The diameter of the bearing is \(D = 2\;{\rm{mm}} = 0.2\;{\rm{cm}}\).
The density of the bearing is \({\rho _b} = 7.8\;\frac{{\rm{g}}}{{{\rm{c}}{{\rm{m}}^3}}}\).
The density of \(glycerin is {\rho _g} = 1.3\;\frac{{\rm{g}}}{{{\rm{c}}{{\rm{m}}^3}}} \).
The viscosity of glycerin is,
\(\eta = 12\;\frac{{{\rm{N}} \cdot {\rm{s}}}}{{{{\rm{m}}^2}}}\)
\( = 12\;\frac{{{\rm{kg}} \cdot {\rm{m}} \cdot {\rm{s}}}}{{{{\rm{s}}^2}{{\rm{m}}^2}}}\)
\( = 12\;\frac{{{\rm{kg}}}}{{{\rm{m}} \cdot {\rm{s}}}} \)
Solve further:
\(\eta = 12 \times {10^3} \times {10^{ - 2}}\;\frac{{\rm{g}}}{{{\rm{cm}} \cdot {\rm{s}}}} \)
\( = 120\;\frac{{\rm{g}}}{{{\rm{cm}} \cdot {\rm{s}}}} \)
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