Physics: A mass travels in a straight line on a horizontal plane. It is attached at one end to a fixed vertical wall through a spring (spring constant ) while at the other end, it is connected through an inextensible cable to a pendulum of mass m. The lower mass m will oscillate in the vertical plane ABC and there is no friction. Using the position x of the mass and the angle theta of the mass m, as the configuration variables, write the two equations of motion.

(Li) Lithium and (F) Fluorine would create a binary ionic compound which would be lithium fluoride which would be (LiF).They would create an ionic compound because lithium is a metal and fluorine is a nonmetal. Metal + Nonmetal = binary ionic compound....

A mass travels in a straight line on a horizontal

Ответ:

kookycookiefanx

20.01.2022

Physics

The equations are as below:

$m\ddot{\theta}(l-l_t+x)+2m\dot{\theta}\dot{x}=mgsin\theta\\\ddot{x}=\dfrac{m\dot{\theta}}{m+M}(l-l_t+x)+\dfrac{k}{m+M}(l_o-x)-\dfrac{mg}{m+M}cos\theta$

Explanation:

As the complete question is not given, the complete question is attached herewith.

From the given data

The mass of the block is M

The length of the spring at any given time is l

The length of the spring initially is lo

The length of the platform is lr

So the coordinates in y and z direction are given as

$y=(l-(l_t-x))cos\theta$

$\dot{y}=-\dot{\theta}(l-l_t+x)sin\theta+\dot{x}cos\theta$

$z=(l-(l_t-x))sin\theta$

$\dot{z}=\dot{\theta}(l-l_t+x)cos\theta+\dot{x}sin\theta$

Now the Lagrange is given as

$L=\dfrac{1}{2}m\dot{y}^2+\dfrac{1}{2}m\dot{z}^2+\dfrac{1}{2}M\dot{x}^2-mg(l-(l_t-x))cos\theta-\dfrac{1}{2}k(l_o-x)^2$

$L=\dfrac{1}{2}m(-\dot{\theta}(l-l_t+x)sin\theta+\dot{x}cos\theta)^2+\dfrac{1}{2}m(\dot{\theta}(l-l_t+x)cos\theta+\dot{x}sin\theta)^2+\dfrac{1}{2}M\dot{x}^2-mg(l-(l_t-x))cos\theta-\dfrac{1}{2}k(l_o-x)^2\\L=\dfrac{1}{2}m(\dot{x}^2+\dot{\theta}^2(l-l_t+x)^2)+\dfrac{1}{2}M\dot{x}^2-mg(l-l_t+x)cos\theta-\dfrac{1}{2}k(l_o-x)^2\\\\$

Now The value is given as

$\dfrac{dL}{d\dot{\theta}}=m\dot{\theta}(l-l_t+x)^2\\\dfrac{dL}{d\theta}=mg(l-l_t+x)sin\theta$

So the 1st equation becomes

$\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{\theta}}\right)=\dfrac{\partial L}{\partial \ddot{\theta}}\\\dfrac{d}{dt}\left(m\dot{\theta}(l-l_t+x)^2\right)=mg(l-l_t+x)sin\theta\\m\ddot{\theta}(l-l_t+x)+2m\dot{\theta}\dot{x}=mgsin\theta$

Now the second equation is as

$\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{x}}\right)=\dfrac{\partial L}{\partial \ddot{x}}\\\ddot{x}=\dfrac{m\dot{\theta}}{m+M}(l-l_t+x)+\dfrac{k}{m+M}(l_o-x)-\dfrac{mg}{m+M}cos\theta$

A mass travels in a straight line on a horizontal plane. It is attached at one end to a fixed verti

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