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• Notes

$\begin{array}{l}{\text { Under some circumstances, a star collapse into an }} \\ {\text { extremely dense object made mostly of neurrons and called a }} \\ {\text { neutron star. The density of a neutron star is roughly } 10^{14} \text { times as }} \\ {\text { great as that of ordinary solid mater. Suppose we represent the star }} \\ {\text { as a uniform, solid, rigid sphere, both before and after the collapse. }} \\ {\text { The star's initial radius was } 7.0 \times 10^{5} \mathrm{km} \text { (comparable to our }} \\ {\text { sun); its final radius is } 16 \mathrm{km} \text { . If the original star rotated once in } 30} \\ {\text { days, find the angular speed of the neutron star. }}\end{array}$

$I=\frac{2}{5} M R^{2}$

$\omega_{2}=\omega_{1}\left(\frac{R_{1}}{R_{2}}\right)^{2}$

$=\left(\frac{2 \pi \ ra d}{30 d(86400 s/ d)}\right)\left(\frac{7 \times 10^{5} km}{16 k m}\right)^{2}$

$=4.6 * 10^{3} \mathrm{rad} / \mathrm{s}$

$k=\frac{1}{2} I \omega^{2}=\frac{1}{2} L \omega$

$\begin{array}{l}{\text { A solid wood door } 1.00 \mathrm{m} \text { wide and } 2.00 \mathrm{m} \text { high is }} \\ {\text { hinged along one side and has a total mass of } 40.0 \mathrm{kg} \text { . Initially open }} \\ {\text { and at rest, the door is struck at its center by a handful of sticky mud }} \\ {\text { with mass } 0.500 \mathrm{kg} \text { , traveling perpendicular to the door at } 12.0 \mathrm{m} / \mathrm{s}} \\ {\text { just before impact. Find the final angular speed of the door. Does }} \\ {\text { the mud make a significant contribution to the moment of inertia? }}\end{array}$

$l \rightarrow \text { width }$

$m v\left(\frac{2}{2}\right)$

$I_{\text {door }}=\frac{1}{3} Ml^{2}, \quad I_{\text {mvd }}= m (\frac{l}{2})^2$

$\omega=\frac{L}{I} \rightarrow L=\omega I$

$\frac{mv (\frac{1}{2})}{\left(\frac{1}{3} M l^{2}\right)+M\left(\frac{l}{2}\right)^{2}}=\frac{0.5 * 12 * 0.5}{\frac{1}{3}(40)(1)^{2}+(0.5)(0.5)^{2}}=0.223 \ rad/s$

$\begin{array}{l}{\text { A thin uniform rod has a length of } 0.500 \mathrm{m} \text { and is rotating }} \\ {\text { in a circle on a frictionless table. The axis of rotation is perpendicular }} \\ {\text { to the length of the rod at one end and is stationary. The rod has an }} \\ {\text { angular velocity of } 0.400 \mathrm{rad} / \mathrm{s} \text { and a moment of inertia about the }} \\ {\text { axis of } 3.00 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2} . \text { A bug initially standing on the rod at }}\end{array}$

$\begin{array}{l}{\text { the axis of rotation decides to crawl out to the other end of the rod. }} \\ {\text { When the bug has reached the end of the rod and sits there, its tan- }} \\ {\text { gential speed is } 0.160 \mathrm{m} / \mathrm{s} \text { . The bug can be treated as a point mass. }} \\ {\text { (a) What is the mass of the rod?(b) What is the mass of the bug? }}\end{array}$

$I=\frac{1}{3} M L^{2}$

$I_{1} \omega_{1}=I_{2} \omega_{2}$

$M=\frac{3 I}{L^{2}}=\frac{3\left(3*10^{-3} kg \cdot m^{2}\right)}{(0.5m)^{2}}=0.036 \mathrm{kg}$

$\begin{array}{l}{L_{1}=L_{2}} \\ {I_{1} \omega_{1}=I_{2}\omega_{2}}\end{array} \quad ]\longrightarrow \quad \omega_{2}=\frac{V}{r}=\frac{0.16}{0.5}=0.32 \mathrm{rad} / \mathrm{s}$

$∴\left(3*10^{-3}\right)(0.4)=\left(3*10^{-3}+m_{\text {bvg }}(0.5)^{2}\right)(0.32)$

$\longrightarrow m_{bvg}=\frac{3*10^{-3}(0.4- 0.32)}{0.32* 0.52}=3* 10^{-3} \mathrm{kg}$