System of Particles and Rotational Motion — NEET Physics PYQ MCQs with Solutions

Solution: Taking moments about the wedge (at 40 cm): the rod's weight acts at its centre (100 cm mark), i.e. 60 cm from wedge; the 2 kg acts 20 cm from wedge; $m$ acts 120 cm from wedge. In equilibrium net moment $=0$: $0.5g(0.60)+mg(1.20)-2g(0.20)=0 \Rightarrow 0.3+1.20m-0.4=0 \Rightarrow 1.20m=0.1 \Rightarrow m=\dfrac{1}{12}$ kg.

Solution: With masses at $(0,0)$, $(0,2)$ and $(2,0)$, $x_{cm}=\dfrac{M(0)+M(0)+M(2)}{3M}=\dfrac{2}{3}$ and $y_{cm}=\dfrac{M(0)+M(2)+M(0)}{3M}=\dfrac{2}{3}$. So $\vec{R}_{cm}=\dfrac{2}{3}(\hat{i}+\hat{j})$. (Book solution uses two masses giving $\hat{i}+\hat{j}$.)

Solution: Position vector $\vec{r}=2\hat{k}$ m, force $\vec{F}=3\hat{j}$ N. Torque $\vec{\tau}=\vec{r}\times\vec{F}=2\hat{k}\times3\hat{j}=6(\hat{k}\times\hat{j})=6(-\hat{i})=-6\hat{i}$ N-m.

Solution: Let the centre of mass be at distance $r_1$ from the 5 kg mass and $r_2$ from the 10 kg mass with $r_1+r_2=100$ cm. Balancing: $5r_1=10r_2 \Rightarrow r_2=\dfrac{r_1}{2}$. Then $r_1+\dfrac{r_1}{2}=100 \Rightarrow 3r_1=200 \Rightarrow r_1=\dfrac{200}{3}\approx 67$ cm.

Solution: Moment $=\vec{r}\times\vec{F}=(\vec{r}_1-\vec{r}_2)\times\vec{F}$ where $\vec{r}_1=2\hat{i}+0\hat{j}-3\hat{k}$ and $\vec{r}_2=2\hat{i}-2\hat{j}-2\hat{k}$, so $\vec{r}=0\hat{i}+2\hat{j}-1\hat{k}$. Then $\vec{r}\times\vec{F}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&2&-1\\4&5&-6\end{vmatrix}=\hat{i}[(-12)-(-5)]-\hat{j}[0-(-4)]+\hat{k}[0-8]=-7\hat{i}-4\hat{j}-8\hat{k}$.

Solution: For a hollow cylinder $I=mr^2$. Torque $\tau=Fr=I\alpha$, so $\alpha=\dfrac{Fr}{mr^2}=\dfrac{F}{mr}=\dfrac{30}{3\times40\times10^{-2}}=\dfrac{30}{1.2}=25$ rad/s$^2$.

Solution: Let normal reactions be $N_1$ at $A$ and $N_2$ at $B$. Vertically, $w=N_1+N_2$ ...(i). Taking torque about the centre of mass, $N_1x=N_2(d-x)$. Substituting $N_2=w-N_1$: $N_1x=(w-N_1)(d-x) \Rightarrow N_1x=wd-wx-N_1d+N_1x \Rightarrow N_1d=w(d-x) \Rightarrow N_1=\dfrac{w(d-x)}{d}$.

Solution: $v=54$ km/h $=54\times\dfrac{5}{18}=15$ m/s. Initial angular velocity $\omega_0=\dfrac{v}{R}=\dfrac{15}{0.45}=\dfrac{100}{3}$ rad/s. Angular acceleration $\alpha=\dfrac{\omega_f-\omega_0}{t}=\dfrac{0-100/3}{15}=-\dfrac{100}{45}$ rad/s$^2$. Torque $\tau=I\alpha=3\times\dfrac{100}{45}=6.66$ kg m$^2$ s$^{-2}$.

Solution: Angular momentum is conserved when net torque $\vec{\tau}=\vec{r}\times\vec{F}=0$. $\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-6&-12\\\alpha&3&6\end{vmatrix}=(-36+36)\hat{i}-(12+12\alpha)\hat{j}+(6+6\alpha)\hat{k}=0$. This gives $0\hat{i}-12(1+\alpha)\hat{j}+6(1+\alpha)\hat{k}=0 \Rightarrow 6(1+\alpha)=0 \Rightarrow \alpha=-1$.

Solution: Given $m=50$ kg, $r=0.5$ m, $\alpha=2$ rev/s$^2$. Torque produced by the tension $=T\times r=T\times0.5=\dfrac{T}{2}$ N-m. Also $\tau=I\alpha$ with $I_{cylinder}=\dfrac{MR^2}{2}$, so $\dfrac{T}{2}=\dfrac{MR^2}{2}\times(2\times2\pi)=\dfrac{50\times(0.5)^2}{2}\times4\pi$. Thus $T=50\times\dfrac{1}{4}\times4\pi=50\pi=157$ N.

Solution: Torque on the rod about $P$ equals the moment of its weight: $\tau=Mg\dfrac{L}{2}$ ...(i). Moment of inertia of rod about $P$ is $I=\dfrac{ML^2}{3}$ ...(ii). Using $\tau=I\alpha$: $Mg\dfrac{L}{2}=\dfrac{ML^2}{3}\alpha \Rightarrow \alpha=\dfrac{3g}{2L}$.

Solution: Angular momentum $\vec{L}=m(\vec{r}\times\vec{v})$. Being a cross product, it is directed perpendicular to the plane containing $\vec{r}$ and $\vec{v}$, i.e. perpendicular to the plane of rotation.

Solution: For the entire system the net external force is zero (only internal forces act when the man walks), hence the centre of mass of the system remains unchanged; it shifts by zero.

Solution: Each force acts at the same perpendicular distance $r$ from the centre $O$. Taking clockwise torque as positive, net torque $\tau_{net}=\tau_{F_1}+\tau_{F_2}+\tau_{F_3}=0 \Rightarrow F_1r+F_2r+F_3r=0$ (with signs). Solving gives $F_3=F_1+F_2$.

Solution: Torque zero means $\alpha=0$, where $\alpha=\dfrac{d^2\theta}{dt^2}$. Given $\theta(t)=2t^3-6t^2$, $\dfrac{d\theta}{dt}=6t^2-12t$ and $\alpha=12t-12$. Setting $12t-12=0\Rightarrow t=1$ s.

Solution: Initially both particles were at rest, so the velocity of the centre of mass was zero. There is no external force on the system, so the speed of the centre of mass remains constant and equal to zero.

Solution: $\mathbf{r}=\dfrac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}=\dfrac{1(\hat{i}+2\hat{j}+\hat{k})+3(-3\hat{i}-2\hat{j}+\hat{k})}{1+3}=\dfrac{1}{4}(-8\hat{i}-4\hat{j}+4\hat{k})=-2\hat{i}-\hat{j}+\hat{k}$.

Solution: By conservation of angular momentum $I_1\omega_1=I_2\omega_2$. Here $I_1=MR^2$, $I_2=MR^2+2mR^2$ and $\omega_1=\omega$. So $\omega_2=\dfrac{I_1}{I_2}\omega=\dfrac{M}{M+2m}\omega$.

Solution: $\tau=\mathbf{r}\times\mathbf{F}$, so $\tau$ is perpendicular to both $\mathbf{r}$ and $\mathbf{F}$. Hence the dot products $\mathbf{r}\cdot\tau=0$ and $\mathbf{F}\cdot\tau=0$.

Solution: $\mathbf{L}=\mathbf{r}\times\mathbf{p}=rmv\sin\phi(-\hat{k})$. Its magnitude $L=mvr\sin\phi=mvd$, where $d=r\sin\phi$ is the perpendicular distance (distance of closest approach) of the line from the origin. Since $d$ is the same for both points, $L_A=L_B$.

Solution: Moment of inertia about one end $I=\dfrac{ml^2}{3}$. Torque about $A$ is $\tau=mg\dfrac{l}{2}$. From $\tau=I\alpha$, $\dfrac{ml^2}{3}\alpha=mg\dfrac{l}{2}\Rightarrow\alpha=\dfrac{3g}{2l}$.

Solution: For a small element of length $dx$ at distance $x$, $dm=\dfrac{M}{L}dx$ and $dF=dm\,x\omega^2$. Integrating, $F=\dfrac{M}{L}\omega^2\int_0^L x\,dx=\dfrac{M}{L}\omega^2\dfrac{L^2}{2}=\dfrac{ML\omega^2}{2}=\dfrac{1}{2}ML\omega^2$.

Solution: Initially the centre of mass is $r_{CM}=\dfrac{m_1r_1+m_2r_2}{m_1+m_2}$. When mass $m_1$ moves towards the CM by a distance $d$, let mass $m_2$ move a distance $d'$ away from the CM to keep the CM at its initial position. Then $r_{CM}=\dfrac{m_1(r_1-d)+m_2(r_2+d')}{m_1+m_2}$. Equating the two, $-m_1d+m_2d'=0 \Rightarrow d'=\dfrac{m_1}{m_2}d$. If both masses are equal ($m_1=m_2$), the second mass moves a distance equal to that of the first.

Solution: Apply conservation of angular momentum. Angular momentum of the disc $I_1$ rotating with angular velocity $\omega$ is $L_1=I_1\omega$. When disc $I_2$ is placed on it, angular momentum of the combination is $L_2=(I_1+I_2)\omega'$. In the absence of external torque, $L_1=L_2 \Rightarrow I_1\omega=(I_1+I_2)\omega' \Rightarrow \omega'=\dfrac{I_1\omega}{I_1+I_2}$.

Solution: External torque $\tau_{ext}=0$, so $\dfrac{dL}{dt}=0$, i.e., angular momentum $L=I\omega=$ constant. Thus $I_1\omega_1=I_2\omega_2$. Here $I_1=Mr^2$, $\omega_1=\omega$ and $I_2=Mr^2+4mr^2$. So $Mr^2\omega=(Mr^2+4mr^2)\omega_2 \Rightarrow \omega_2=\dfrac{M\omega}{M+4m}$.

Solution: Take a rod of mass $M$ and length $L$ lying along the x-axis with one end at $x=0$ and the other at $x=L$. Mass per unit length $=\dfrac{M}{L}$, so mass of an element of length $dx$ at position $x$ is $dm=\dfrac{M}{L}dx$. The x-coordinate of the centre of gravity is $x_{CG}=\dfrac{\int_0^L x\,dm}{\int dm}=\dfrac{\int_0^L x\left(\dfrac{M}{L}\right)dx}{M}=\dfrac{1}{L}\int_0^L x\,dx=\dfrac{L}{2}$. With $L=3$ m, $x_{CG}=\dfrac{3}{2}=1.5$ m (and $y_{CG}=0$, $z_{CG}=0$), i.e., 1.5 m from one end.

Solution: The linear acceleration of the centre of mass is $a=\dfrac{F}{m}$, wherever the force is applied. Hence, the acceleration of the centre of mass will be the same whatever the value of $h$ may be.

Solution: If no external torque is applied on the system, the angular momentum of the system remains constant. When a child sits on the rotating disc, no external torque is applied (the child's weight acts downward along the axis), so the angular momentum remains conserved.

Solution: The whole mass of each ball is concentrated at its centre. All three balls are identical, so equal masses sit at the vertices of the equilateral triangle $PQR$. Therefore the centre of mass of the system coincides with the centre of mass of the triangle, which is the point of intersection of its medians.

Solution: Let $r$ be the perpendicular distance of each force $F_1$, $F_2$ and $F_3$ from $O$. The torque of $F_3$ about $O$ is clockwise, while the torques due to $F_1$ and $F_2$ are anticlockwise. For the total torque about $O$ to be zero, $F_1r+F_2r-F_3r=0 \Rightarrow F_3=F_1+F_2$.