CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 11 - MCQExams.com

Correct answer: Option B

Hint: In this case, the train's length is half the circumference of the circular track.

Step 1:Train’s linear momentum

$$\rho  = mv$$ is the formula for calculating a body's linear momentum, where m is the mass and v is the velocity.

The linear density of an item is equal to $$\lambda  = \dfrac{m}{l}$$ where $$m$$ is the object's mass and $$l$$ is the object's length.

To get an equation for the train's linear momentum, consider a small part of it.

Consider a small element that is at an angle $$\theta $$ to the X-axis. An angle $$d\theta $$ is subtended by this element. This is seen in the diagram.

The small element's length will be $$dl =Rd\theta$$

The small element's mass can therefore be represented as  dm =$$\lambda dl$$        -------(1)

We obtain by substituting for $$\lambda  = \dfrac{M}{{\pi R}}$$ and $$dl = Rd\theta $$ in equation (1):

$$dm = \dfrac{M}{{\pi R}} \times Rd\theta $$$$ = \dfrac{M}{\pi }d\theta $$

Because the vertical component of the velocity $$vcos\theta $$ at A and B is oriented in the positive and negative Y directions, they will cancel each other out. As a result, only the horizontal component $$v\sin \theta $$ is important.

We may now write $$dp = dm\left( {vsin\theta } \right)$$ for the small element's linear momentum.         --------(2)

When we replace $$dm = \dfrac{M}{\pi }d\theta $$ in equation (2), we obtain

$$dp = \dfrac{M}{\pi }(v\sin \theta )d\theta $$                    ---------(3)

The linear momentum of the small element is expressed in equation (3).

Step 2: Finding total linear momentum

Integrating equation (3) will give us the train's total linear momentum.

$$P = \mathop \smallint \limits_0^\pi  d{p_x} = \mathop \smallint \limits_0^\pi  \dfrac{M}{\pi }\left( {v\sin \theta } \right)d\theta $$             

$$P = \dfrac{{Mv}}{\pi }\int\limits_0^\pi  {sin\theta d\theta } $$$$=  - \dfrac{{Mv}}{\pi }[cos\theta ]_0^\pi $$              

Applying limits,

$$P =  - \dfrac{{Mv}}{\pi }[ - 1 - ( + 1)]$$$$ = \dfrac{{2Mv}}{\pi }$$                      

As a result, the train's linear momentum is calculated as $$P = \dfrac{{2Mv}}{\pi }$$

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

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