Rotational Motion Practice Questions

Moment of inertia for the square plate through 0 , perpendicular to the plate is $I_{n n^{'}}=\;\frac{1}{12} m (a^{2+}a^2) =\;\frac{m a^2}{6}$Also, $D O=\;\frac{D B}{2}=\;{{\sqrt{2}} a}/{2}=\;{a}/{{\sqrt{2}}}$According to parallel axis theorem$\; {\text{I}}_{m m^{'}}=I_{n n^{'}}+m (\;{a}/{{\sqrt{2}}}) ^2$ $\;=\;\frac{m a^2}{6}+\;\frac{m a^2}{2}$ $\;=\;\frac{m a^{2+}3 m a^2}{6}$ $\;=\;\frac{2}{3} m a^2$

By perpendicular axes theorem, $I_x=I_x+I_y \;\;$ or,$\;\; I_z=2 I_y$ ( as $I_x=I_y$ by symmetry of the figure) $\therefore I_{E F}=\;\frac{I_z}{2}$....(i) By perpendicular axes theorem, $I_z=I_{A C}+I_{B D}=2 I_{A C}$ (As $I_{A C}=I_{B D}$ by symmetry of the figure) $\therefore I_{A C}=\;\frac{I_z}{2}$....(ii) From $(i)$ and $(i i)$, we get $I_{E F}=I_{A C}$.

A uniform body of radius ${\text{R}}$, mass ${\text{M}}$ and moment of inertia $I$ rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is$a=\;\frac{g sin \;\theta }{1+\;\frac{I}{M R^2}}$

We know that $\vec{\tau _c}=\;\frac{d \vec{L_c}}{d t}$ where $\vec{\tau _c}$ torque about the center of mass of the body and $\vec{L_c}=$ Angular momentum about the center of mass of the body. Given that $\vec{L_a}=$ constant $\therefore \;\frac{d \vec{L_c}}{d t}=0$ $\Rightarrow \vec{\tau _c}=0$ [as $\vec{\tau _c}=\;\frac{d \vec{L}_c}{d t}$]

Let $I_A$ is the moment of inertia about an axis passing through A and parallel to ${\text{BD}}$. $I_A=M . I$ due to the point mass at $B+$M. $I$ due to the point mass at $D+$M. I due to the point mass at $C$. $\;I_A=2 \times m (\;{ℓ}/{{\sqrt{2}}}) ^{2+}m({\sqrt{2}} ℓ)^2$ $\;=m ℓ^{2+}2 m ℓ^2=3 m ℓ^2$

Here angular momentum is conserved.Applying conservation of angular momentum $I^{'} \omega ^{'}=I \omega$ $\; (m R^{2+}2 M R^2) \omega ^{'}=m R^2 \omega$ $\;\Rightarrow \omega ^{'}=\omega [\;\frac{m}{m+2 M}]$

We know, Torque $\vec{\tau }=\vec{r} \times \vec{F}$ $\;=({i}^{∧}-{j}^{∧}) \times (-F {k}^{∧})$ $\;=F({i}^{∧}+{j}^{∧})$

Let the mass of each particle is ${\text{m}}$.Then force experienced by each particle, $F=m \omega ^2 R$ $\; \therefore \;\frac{F_1}{F_2}=\;\frac{m \omega ^2 R_1}{m \omega ^2 R_2}$ $\;\Rightarrow \;\frac{F_1}{F_2}=\;\frac{R_1}{R_2}$

Let mass of the semi circular disc $=M$Now assume a disc which is combination of two semi circular parts. Let $I$ be the moment of inertia of the uniform semicircular disc. So $2 I$ will be the moment of inertia of the full circular disc and $2 {\text{M}}$ will be the mass.$\;\Rightarrow 2 I=\;\frac{2 M r^2}{2}$ $\;\Rightarrow I=\;\frac{M r^2}{2}$

For solid sphere the moment of inertia of $A$ about its diameter$I_A=\;\frac{2}{5} M R^2 .$The moment of inertia of a hollow sphere $B$ about its diameter$I_B=\;\frac{2}{3} M R^2$ $\therefore I_A< I_B$

Solid sphere is rotating in free space that means no external torque is operating on the sphere.Angular momentum will remain the same since external torque is zero.

We know that density $(d)=\;{{\text{mass}}(M)}/{\;\text{ volume }\;(V)}$ $\therefore$ Mass of ${\text{disc}} M=d \times V=d \times (\pi R^2 \times t)$.The moment of inertia of any disc is $I=\;\frac{1}{2} M R^2$ $\; \therefore I=\;\frac{1}{2} (d \times \pi R^2 \times t) R^2=\;\frac{\pi d}{2} t \times R^4$ $\; \therefore \;\frac{I_X}{I_Y}=\;\frac{t_X R_X^4}{t_Y R_Y^4}$ $\;=\;\frac{t \times R^4}{\frac{t}{4} \times (4 R)^4}$ $\;=\;\frac{1}{64}$

As we know $\vec{\tau }=\vec{r} \times \vec{F}$So the angle between $\vec{\tau }$ and $\vec{r}$ is $90^{\circ}$ and the angle between $\vec{t}$ and $\vec{F}$ is also $90^{\circ}$.We also know that the dot product of two vectors which have an angle of $90^{\circ}$ between them is zero.$\therefore \vec{r} \vec{\tau }=0$ and $\vec{F} \cdot \vec{\tau }=0$Therefore $(d)$ is the correct option.

We know Rotational Kinetic Energy $=\;\frac{1}{2} I \omega ^2$, Angular Momentum $L=I \omega \Rightarrow I=\;\frac{L}{\omega }$ $\therefore$ Initial $K . E .=\;\frac{1}{2} \;\frac{L}{\omega } \times \omega ^2=\;\frac{1}{2} L \omega$ Final $K . E^{'}=\;\frac{K . E}{2}=\;\frac{1}{2} L^{'} \times 2 \omega$ $\; \therefore \;\frac{K \cdot E}{K \cdot E^{'}}=\;\frac{L \times \omega }{L^{'} \times \omega ^{'}}$ $\;\Rightarrow \;\frac{K \cdot E}{\frac{K \cdot }{2}}=\;\frac{L \times \omega }{L^{'} \times 2 \omega }$ $\therefore L^{'}=\;\frac{L}{4}$

Angular momentum $(L)$ $=($ linear momentum $) \times ($ perpendicular distance of the line of action of momentum from the axis of rotation) $=m v \times r$ $=m v \times 0$ $=0$ [ Here $r=0$ because the particle is moving through the line PQ and $r$ is the perpendicular distance from line PQ of the particle ]

When two small spheres of mass $m$ are attached gently, the external torque, about the axis of rotation, is zero. So, $\;\frac{d \vec{L}}{d t}=\overline{z}=0$ $\vec{L}=$ conserved So the angular momentum about the axis of rotation is conserved. $\therefore I_1 \omega _1=I_2 \omega _2$ $\Rightarrow \omega _2=\;\frac{I_1}{I_2} \omega _1$ Here Moment of inertia of Disc $I_1=\;\frac{1}{2} M R^2$ and After adding two sphere Moment of Inertia of disc and two sphere, $I_2=\;\frac{1}{2} M R^{2+}2(\;\frac{1}{2} m R^{2+}\;\frac{1}{2} m R^2)$ $\therefore \omega _2=\;\frac{\;\frac{1}{2} M R^2}{\frac{1}{2} M R+2 m R^2} \times \omega _1=\;\frac{M}{M+4 m} \omega _1$

Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same $(g \sin \theta )$.

Moment of Inertia of a circular wire about an axis $n n^{'}$ passing through the centre of the circle and perpendicular to the plane of the circle $=M R^2$ As shown in the figure, $X$-axis and $Y$-axis lies in the plane of the ring. Then by perpendicular axis theorem $I_X+I_Y=I_Z$ $\Rightarrow 2 I_X=M R^2[.$ as $I_X-I_Y$ (by symmetry) and $.I_Z=M R^2]$ $\therefore I_X=\;\frac{1}{2} M R^2$