System of Particles and Rotational Motion Test - 64

Given :    $$m_1 = 1kg$$             $$m_2 = 2kg$$                 $$m_3 = 3kg$$

$$ M{ a }_{ Max }\times h=Mg\times p$$

The deflection of the plate can be noticed by going to a co- rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance $$\xi$$ from the axis is $$a=\xi\beta$$ and the moment of the forces exerted by the section between $$x$$ and $$l$$ is

$$\displaystyle N=\rho lh\beta\int_x^l{\xi^2d\xi}=\frac{1}{3}\rho lh\beta(l^3-x^3)$$.

From the fundamental equation

$$\displaystyle EI\frac{d^2y}{dx^2}=\frac{1}{3}\rho lh\beta(l^3-x^3)$$.

The moment of inertia $$\displaystyle I=\int_{\displaystyle-\frac{h}{2}}^{\displaystyle+\frac{h}{2}}{z^2ldz}=\frac{lh^3}{12}$$

Note that the neutral surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and $$z$$ is perpendicular to it.

$$\displaystyle\frac{d^2y}{dx^2}=\frac{4\rho\beta}{Eh^2}(l^3-x^3)$$. Integrating

$$\displaystyle\frac{dy}{dx}=\frac{4\rho\beta}{Eh^2}\left(l^3x-\frac{x^4}{4}\right)+c_1$$

Since $$\displaystyle\frac{dy}{dx}=0$$, for $$x=0$$, $$c_1=0$$. Integrating again,

$$\displaystyle y=\frac{4\rho\beta}{Eh^2}\left(\frac{l^3x^2}{2}-\frac{x^5}{20}\right)+c_2$$

$$c_2=0$$ because $$y=0$$ for $$x=0$$

Thus $$\displaystyle\lambda=y(x=l)=\frac{9\rho\beta l^5}{5Eh^2}$$

Let velocity of the particle at the base of the inclined plane be  $$v$$

The moment of inertia about a rectangular lamina with area density $$\rho$$ is $$\int \int \rho r^2 dA$$. Here r is the distance from each point to the axis of rotation and $$dA$$ is the differential of area.
Take the center of the rectangle to be at $$(0,0)$$ so the vertices are $$(L/2, B/2)$$, etc. Then the distance from (x,y) to the axis of rotation (passing through $$(0,0)$$ perpendicular to the plate) is $$x^2+ y^2$$ and the moment of inertia is
$$\displaystyle\int^{L/2}_{-L/2}\int^{B/2}_{B/2}\rho(x^2+y^2)dxdy$$
or
$$\displaystyle\int^{L/2}_{-L/2}\rho(Bx^2+\frac{1}{12}B^3)$$
or
$$\displaystyle\frac{1}{12}\rho(L^3B+LB^3)$$
Now as $$M=\rho LB$$, thus we get moment of inertia as $$\displaystyle\frac{M}{12}(L^2+B^2)$$

Consider the frame made using four rods as shown.
We have the moment of inertia of rod AB about the center of mass as $$\dfrac{Ml^2}{12}+M(\dfrac{l}{2})^2=\dfrac{Ml^2}{3}$$
Thus we get the moment of inertia of all the four rods as $$\dfrac{4}{3}Ml^2$$
Now using perpendicular axis theorem we see the the moment of inertia about perpendicular medians is given as $$2I=\dfrac{4}{3}Ml^2$$
Thus we get $$I=\dfrac{2}{3}Ml^2$$
Thus none of the above is the correct answer.

Velocity of center of mass is given by:
$$v = \dfrac{v_{1} m_{1} + v_{2} m_{2}}{m_{1} + m_{2}} = i + j$$
Acceleration of center of mass is given by:
$$a = \dfrac{a_{1} m_{1} + a_{2} m_{2}}{m_{1} + m_{2}} = \dfrac{1}{2}(i + j)$$
Acceleration and velocity both have the same direction, hence the body moves along the same line (i+j).
Hence it moves in a straight line.
Option D is correct.

correct option is (D)

 $$\textbf{HINT:}$$ Moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section . Spinning figure skaters can reduce their moment of inertia by pulling in their arms, allowing them to spin faster due to conservation of angular momentum

$$\textbf{Step1:}$$  finding momentum of inertia of upper ring about  axis

Moment of inertia of the upper ring about the axis $$\mathrm{I}_{1}=\dfrac{1}{2} \mathrm{mr}^{2}$$

Now, Momentum of inertia of one the lower ring

Moment of inertia of the one of the lower ring about the axis $$\mathrm{I}_{2}=\dfrac{1}{2} \mathrm{mr}^{2}+$$ $$\mathrm{mr}^{2}=\dfrac{3}{2} \mathrm{mr}^{2}$$

$$\textbf{Step 2:}$$ finding momentum  of inertia of  of second lower ring

 

Similarly, moment of inertia of the second lower ring about the axis $$\mathrm{I}_{3}=$$ $$\dfrac{3}{2} \mathrm{mr}^{2}$$

$$\textbf{Step 3:}$$: fining total momentum

Total moment of inertia of the system about the axis $$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=\dfrac{7}{2} \mathrm{mr}^{2}$$ 

Radius of gyration $$\mathrm{k}=\sqrt{\dfrac{\mathrm{I}}{3 \mathrm{~m}}}=\sqrt{\dfrac{7}{6}} \mathrm{r}$$

If KE = 0, then all velocities have to be zero(if non-massless particles),

$$\displaystyle L=\frac{l}{2} \sec 45^{\circ} =\frac{l}{\sqrt{2}}$$
$$\displaystyle m=Lx =\frac {lx}{\sqrt{2}}$$
$$\displaystyle

I=2 \left[ \frac{mL^{2}}{3} \sin^{2} 45^{\circ} \right] =2 \left[

\left( \frac{lx}{3\sqrt{2}}\right) \left( \frac{l}{\sqrt{2}}\right)^{2}

\left( \frac{1}{2} \right) \right]$$
$$\displaystyle =\frac{xl^{3}}{6\sqrt{2}}$$