Rotational Motion MCQs for NEET — Physics Questions with Answers

mr = constant

            $m_1{r}_1=m_2{r}_2$

$\Rightarrow r_2=\frac{m_1{r}_1}{m_2}$

                 =$\frac{0.5\times 10}{50}=0.1$

The distance of the man above the floor (total height)=10+0.1=10.1m.

Key Idea     Torque is an axial vector ie, its direction is always perpendicular to the plane containing vectors $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$.

              $\overrightarrow{\mathrm{\tau }}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$

  Torque is perpendicular to both $\overrightarrow{\mathrm{r}}$and  $\overrightarrow{\mathrm{F}}$

   ∴                        $\overrightarrow{\mathrm{\tau }}·\overrightarrow{\mathrm{r}}=0$                            $\overrightarrow{\mathrm{F}}·\overrightarrow{\mathrm{\tau }}=0$

Key Idea

Apply law of conservation of angular momentum.

              $I_1{\omega }_1=I_2{\omega }_2$

In the given case 

            $I_1=M{R}^2$

           $I_2=M{R}^2+2m{R}^2$

          ${\omega }_1=\omega$

Then  ${\omega }_2=\frac{I_1}{I_2}\omega =\frac{M}{M+2m}\omega$

The position vector of centre of mass 

$\overrightarrow{r}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}$

   =$\frac{1\left(\hat{i}+2 \hat{j}+\hat{k}\right)+3\left(-3\hat{i}-2\hat{j}+\hat{k}\right)}{1+3}$

  =$\frac{1}{4}\left(-8\hat{i}-4\hat{j}+4\hat{k}\right)$

 $=-2\hat{i}-\hat{j}+\hat{k}$

The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

Apply theorem of parallel axisand the total moment of interia will be the sum of moment of inertia of each rod.

Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass of rod) $\times$ (perpendicular distance between two axes)

                =$\frac{M{t}^2}{12}+M{\left(\frac{t}{2}\right)}^2=\frac{M{t}^2}{3}$

Moment of inertia of the system = $\frac{M{t}^2}{3}\times 4$

                =$\frac{4}{3}M{t}^2$

In the given problem conservation of linear of momentum yields. 

     $m_1{v}_1+m_2{v}_2=0$

 or  $4v_1+0.2v_2=0 ....\left(i\right)$

Conservation of energy yields. 

$$\begin{gathered} \frac{1}{2 }{m}_1{v}_1^2 +\frac{1}{2}{m}_2{v}_2^2=1050 \\ 0r \frac{1}{2} \times 4v_1^2+\frac{1}{2}\times 0.2\times v_2^2=1050 \\ or 2v_1^2+0.1 v_2^2=1050 ....\left(ii\right) \end{gathered}$$

Solving Eqs. (i) and (ii) , we have 

          $v_2=100 m/s$

To find the moment of inertia of a bent rod, we need to consider the moment of inertia of each half of the rod about the bending point and then add them. The moment of inertia of a rod about a perpendicular axis through its midpoint is (ML^2)/12.

According to the NCERT text, 'A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration.'

The NCERT text states: '(1) the total force, i.e. the vector sum of the forces, on the rigid body is zero; $\sum \vec{F_i} = 0$' for translational equilibrium. If the total force on the body is zero, then the total linear momentum of the body does not change with time.