横浜国立大学　物理H15~19

目次

H15年度　H16年度　H17年度　H18年度　H19年度

平成15年度

(1)斜面に対する垂直抗力Nは、
　$$N=mgcos\theta ＋m\alpha sin\theta$$

$$\therefore \underset{―}{g\frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta }\leqq \alpha \leqq g\frac{sin\theta +\mu cos\theta }{cos\theta -\mu sin\theta }}$$

(2) $$\begin{array}{r}N=mgcos\theta +m\alpha sin\theta \\ N\leqq 0\mathrm{で}\mathrm{離}\mathrm{れ}\mathrm{る}\mathrm{の}\mathrm{で}\\ 0=mgcos\theta -m\beta sin\theta \\ \beta =\frac{cos\theta }{sin\theta }g=\underset{―}{\frac{g}{tan\theta }}\end{array}$$

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平成16年度

(1) $$\begin{array}{r}1:M_1\frac{d^2{Z}_1}{d{t}^2}=T-M_{1}g\\ 2:M_2\frac{d^2{Z}_2}{d{t}^2}=T-M_{2}g\end{array}$$ (2) $$\begin{array}{r}{Z}_1+Z_2=C(\mathrm{一}\mathrm{定}）\\ \therefore \frac{d^2{Z}_1}{d{t}^2}+\frac{d^2{Z}_2}{d{t}^2}=0\\ (1)\mathrm{か}\mathrm{ら}\mathrm{値}\mathrm{を}\mathrm{そ}\mathrm{れ}\mathrm{ぞ}\mathrm{れ}\mathrm{代}\mathrm{入}\mathrm{し}\\ 0=\frac{T-M_{1}g}{M_1}+\frac{T-M_{2}g}{M_2}\\ T=\frac{2M_1{M}_2}{M_1+M_2}g:(1)1\mathrm{へ}\mathrm{代}\mathrm{入}\\ M_1\frac{d^2{z}_1}{d{t}^2}=2\frac{M_1-M_2}{M_1+2}g-M_{1}g\\ \frac{d^2{z}_1}{d{t}^2}=\frac{M_2-M_1}{M_1+2}g\\ \frac{d^2{z}_2}{d{t}^2}=-\frac{d^2{Z}_1}{d{t}^2}=\frac{M_1-M_2}{M_1+M_2}g\\ \therefore \underset{―}{\mathrm{等}\mathrm{加}\mathrm{速}\mathrm{度}\mathrm{運}\mathrm{動}}\end{array}$$ (3) $$\begin{array}{r}{M}_1\frac{d^2{Z}_1}{d{t}^2}=T-M_{1}g-v_1{\gamma }_1\mathrm{な}\mathrm{の}\mathrm{で}(v_i=\frac{d^2{Z}_i}{d{t}^2})\\ [kg\cdot m/s^2]=[m/s]\cdot {\gamma }_j\\ \therefore \underset{―}{{\gamma }_j\mathrm{の}\mathrm{単}\mathrm{位}：kg/s}\end{array}$$ (4) $$\begin{array}{r}1:M_1\frac{d{v}_1}{dt}=T-M_{1}g-v_1{\gamma }_1\\ 2:M_2\frac{d{v}_2}{dt}=T-M_{2}g-v_2{\gamma }_2\\ v_1=-v_2\mathrm{よ}\mathrm{り}\\ 3:-M_1\frac{d{v}_2}{dt}=T-M_{1}g-v_2{\gamma }_1\\ 2-3\\ (M_1+M_2)\frac{d{v}_2}{dt}=(M_1-M_2)g-(r_1+r_2)v_2\\ \mathrm{最}\mathrm{終}\mathrm{的}\mathrm{に}\mathrm{右}\mathrm{辺}=0\mathrm{と}\mathrm{な}\mathrm{る}\mathrm{の}\mathrm{で}\\ v_2=\frac{M_1-M_2}{r_1+r_2}g\\ \to \mathrm{終}\mathrm{端}\mathrm{速}\mathrm{度}\underset{―}{V=\frac{|M_1-M_2|}{r_1+r_2}g}\end{array}$$

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平成17年度

(1)垂直抗力Nは、 $$\begin{array}{r}N=Fcos\theta +mg\\ \therefore \mu (Fcos\theta +mg)\geqq Fsin\theta \mathrm{で}\mathrm{滑}\mathrm{り}\mathrm{出}\mathrm{さ}\mathrm{な}\mathrm{い}\mathrm{の}\mathrm{で}\\ \underset{―}{\mu \geqq \frac{Fsin\theta }{Fcos\theta +mg}}\end{array}$$ (2) mgは無視できて $$\begin{array}{r}\underset{―}{\mu \geqq tan\theta }\end{array}$$ (3) 問図のように力を加えて、滑り出す直前の角度を測ることによって\ 最大静止摩擦係数μを導くことができる。この手法は傾斜法と呼ばれる。 (4) $$\begin{array}{r}ma=Fsin\theta -{\mu }^{\prime }(Fcos\theta +mg)\\ \underset{―}{a=\frac{F(sin\theta -{\mu }^{\prime }cos\theta )}{m}-{\mu }^{\prime }g}\end{array}$$

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平成18年度

(1)
$$\begin{array}{r}m\frac{dv}{dt}=-a\\ v=-\frac{a}{m}t+v_o\\ v=0\mathrm{で}\mathrm{静}\mathrm{止}\mathrm{す}\mathrm{る}\mathrm{の}\mathrm{で}\\ 0=-\frac{a}{m}T+v_o\\ \therefore \underset{―}{T=v_o\frac{m}{a}}\\ x=\int vdt={[-\frac{a}{2m}{t}^2+v_{o}t]}_0^{v_o\frac{m}{a}}\\ \therefore \underset{―}{d=v_o^2\frac{m}{a}-\frac{m}{2a}{v}_o^2=\frac{m}{2a}{v}_o^2}\end{array}$$

(2) $$\begin{array}{r}m\frac{dv}{dt}=-bv\\ \int \frac{1}{v}dv=-\frac{b}{m}\int dt\\ v=Aexp(-\frac{b}{m}t)\\ t=0\mathrm{で}v=v_o\mathrm{よ}\mathrm{り}\\ A=v_o\\ \therefore v=v_{o}exp(-\frac{b}{m}t)\\ 0=v_{o}exp(-\frac{b}{m}T)\\ \therefore \underset{―}{T=\mathrm{\infty }}\\ x={[-\frac{m}{b}{v}_{o}exp(-\frac{b}{m}t)]}_0^{\mathrm{\infty }}\\ \underset{―}{d=0+\frac{m}{b}{v}_o=\frac{m}{b}{v}_o}\end{array}$$

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平成19年度

(1) $$\begin{array}{r}\underset{―}{L=I\omega }\end{array}$$ (2) $$\begin{array}{r}\underset{―}{E=\frac{1}{2}I{\omega }^2}\end{array}$$ (3) $$\begin{array}{r}I={\int }_0^2\pi {\int }_0^1\frac{1}{\pi \mathrm{・}{1}^2}\mathrm{・}{r}^2\mathrm{・}rd\theta dr\\ I=\frac{1}{\pi }\mathrm{・}2\pi {\left[\frac{1}{4}{r}^4\right]}_o^1=\underset{―}{\frac{1}{2}}\end{array}$$ (4) $$\begin{array}{r}I=\frac{d\omega (t)}{dt}=N\\ \frac{1}{2}\frac{d\omega (t)}{dt}=1\\ \frac{d\omega (t)}{dt}=2\\ \omega (t)=2t+\omega \\ 2\omega =2t+\omega \\ \therefore \underset{―}{t=\frac{1}{2}\omega }\end{array}$$