Rotational Motion Practice Questions

The moment of inertia ($I$) of a cylindrical rod about an axis perpendicular to its length and passing through its center is $I = \frac{ML^2}{12}$, where $M$ is the mass and $L$ is the length. The radius of gyration ($K$) is related to the moment of inertia by the formula $I = MK^2$.

By equating these two expressions, we get $MK^2 = \frac{ML^2}{12}$.
This simplifies to an expression for the radius of gyration in terms of length: $K^2 = \frac{L^2}{12}$, or $K = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.

Given the length of the rod is $L = 10\sqrt{3}$ m.
Substituting the value of $L$, we find the radius of gyration:
$K = \frac{10\sqrt{3}}{2\sqrt{3}} = \frac{10}{2} = 5$ m.

Loss in P.E. = Gain in K.E.2 m g R=\;\frac{1}{2}\[;\frac{1}{2} m R^{2+}m R^2]$ w^2$$2 m g R=;\frac{1}{2} \times ;\frac{3}{2} m R^2 w^2$$w^2=;\frac{8 g}{3 R}$$w=\sqrt{;\frac{8 g}{3 R}}=4 \sqrt{;\frac{g}{2 \times 3 R}}$$\Rightarrow x=;\frac{g}{2}=5$

From energy conservation${\text{mgh}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} {\text{I}} \omega ^2$ ${\text{mgh}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} \;{ {\text{mR}}^2}/{2} \omega ^2$ $10 {\text{h}}=\;\frac{16}{2}+\;\frac{16}{4} \Rightarrow {\text{h}}=1.2 {\text{m}}=120 {\text{cm}}$

$I_1=\;\frac{M L^2}{3} . . . . .(1)$For ring: $I_2=\;\frac{M R^2}{2}$ and $2 \pi R=L$ $\Rightarrow I_2=\;\frac{M}{2} (\;\frac{L^2}{4 \pi ^2}) . . . . .$ $\Rightarrow \;\frac{I_1}{I_2}=\;\frac{8 \pi ^2}{3}$ $\Rightarrow x=8$

$I_1 \omega _1=I_2 \omega _2$ $M R^2 \omega _1=(M R^{2+}2 m R^2) \omega _2$ $\omega _2=(\;\frac{M}{M+2 m}) \omega _1$ $\omega _2=2(\;\frac{M}{M+2 m})$

$a=\;{g \sin \theta }/{1+\;\frac{K^{2}}{R^{2}}}$ $v=\sqrt{\;{2 S g \sin \theta }/{1+\;\frac{K^{2}}{R^{2}}}}$ $\Rightarrow \;\frac{v_{c}}{v_{s s}} \sqrt{\;{1+\frac{K_{s s}^{2}}{R^{2}}}/{1+\;\frac{K_{c}^{2}}{R^{2}}}}=\sqrt{\;\frac{1+\frac{2}{5}}{1+\;\frac{1}{2}}}$ $\Rightarrow \sqrt{\;\frac{\;\frac{7}{5}}{\frac{3}{2}}}=\sqrt{\;\frac{14}{15}}$

If $\lambda$ is the mass per unit length of the scale then$0.02 \times (30) \times 10+\lambda 40 \times 20 \times 10=\lambda 60 \times 30 \times 10$ $0.006=\lambda 10$ $\;\text{ or }\; 100 \lambda =0.06 {\text{kg}}$ $=6 \times 10^{-2} {\text{kg}}$ $\Rightarrow x=6$

$|\vec{i}|=|\vec{r} \times (m \vec{v})|$ $=|(3 \hat{i}-\hat{j}) \times (3 \hat{j}+\hat{k})|$ $=|-\hat{i}-3 \hat{j}+9 \hat{k}|$ $={\sqrt{91}}$

${\text{L}}_0=$ angular momentum of shell about ${\text{O}}$.As shell is rolling$\;\text{ so }\; V_{c m}=\omega R$ $L_0=m V_{c m} R+I \omega$ $=1 \times \omega {\text{R}} \times {\text{R}}+\;\frac{2}{3} {\text{R}}^2 \omega$ $=\;\frac{5}{3} {\text{R}}^2 \omega$so $a=5$

${\text{I}}_1= {\text{I}}_3=\;{ {\text{MR}}^2}/{4}$ ${\text{I}}_2=\;{ {\text{MR}}^2}/{4}+ {\text{MR}}^2=\;\frac{5}{4} {\text{MR}}^2= {\text{I}}_4$So ${\text{I}}= {\text{I}}_1+ {\text{I}}_2+ {\text{I}}_3+ {\text{I}}_4$ $=\;{ {\text{MR}}^2}/{2}+\;\frac{5}{2} {\text{MR}}^2$ $=3 {\text{MR}}^2, \;\text{ Putting }\; {\text{R}}=\;{ {\text{a}}}/{2}$ ${\text{I}}=\;{3 {\text{Ma}}^2}/{4}, \;\text{ So }\; {\text{x}}=3$

Taking torque from B$F_w \times 5=\;\frac{3}{2} m g$ $\Rightarrow F_w=\;\frac{3}{10} \times 10 \times 10$ $=30 N$ $N=m g=100 N$and $f_r=F_w=30 {\text{N}}$so $F_f={\sqrt{N^{2+}f_r^2}}={\sqrt{10900}}=10 {\sqrt{109}} N$so $\;\frac{F_w}{F_f}=\;{3}/{{\sqrt{109}}}$

$\vec{\tau }=\vec{r} \times \vec{F}$ $=(2 \hat{i}+\hat{j}+2 \hat{k}) \times (3 \hat{i}+4 \hat{j}-2 \hat{k})$ $=-10 \hat{i}+10 \hat{j}+5 \hat{k}$

$\vec{\tau } \;\; ={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{vmatrix}}$ $\;\; =\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10)$ $\;\; =-17 \hat{i}+19 \hat{j}-4 \hat{k}$

$20-T=2 a$and $0.1 \times {\text{T}}=0.02 \alpha =\;\frac{0.02 a}{0.1}$ ${\text{T}}=2 {\text{a}}$ $\Rightarrow {\text{a}}=5 {\text{m}} / {\ sec}^2$So ${\text{T}}=10 {\text{N}}$

$I= (m_A+m_B+m_C+m_D) r^2$ ${\text{I}}_{\;\text{net }\;}=(1+2+3+4) \cdot (\;{a}/{{\sqrt{2}}}) ^2$, where ${\text{a}}=$ side of square$=10 \times \;\frac{1^2}{2}=5 {\text{kg}} {\text{m}}^2$

For rolling wheel${[12 g sin \;\alpha -3 g sin \;\alpha ] \times R= (2 \times 12 R^{2+}3 R^2) \times \;\frac{a}{R}}$ $\Rightarrow \;\frac{9 g sin \;\alpha }{27}=a$ $\Rightarrow a=\;\frac{g sin \;\alpha }{3}$ $\therefore v=\sqrt{2 \times \;\frac{g sin \;\alpha }{3} \times \;\frac{h}{ sin \;\alpha }}=\sqrt{\;\frac{2}{3} g h}$ $=\;\frac{1}{2} \times \sqrt{\;\frac{8}{3} g h}$ $\therefore x=\;\frac{8}{3}=2.67$

Let's draw the free body diagram of the solid cylinder. Using the condition of the equilibrium of the cylinder; In the direction of inclined plane, $T+f-m g \sin 60^{\circ}=0$ In the perpendicular direction of inclined plane,In the perpendicular direction of inclined plane, $N-m g \cos 60^{\circ}=0$ $\Rightarrow \;\; N=m g \cos 60^{\circ}$ The frictional force between the rough surface and cylinder is $f=\mu _{s} N$ $\Rightarrow \;\; f=0.4 {mg} \cos 60^{\circ} \Rightarrow f=0.2 {mg}$ or $f=\;\frac{m g}{5}$

The acceleration ($a$) of an object of mass $m$ and radius $R$ rolling down an inclined plane with angle $\theta$ is given by:

$a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$

1. Moment of Inertia ($I$) Comparison:

For a ring, solid cylinder, and solid sphere of the same mass and radius:

• Ring: $I = mR^2 \implies \frac{I}{mR^2} = 1$
• Solid Cylinder: $I = \frac{1}{2}mR^2 \implies \frac{I}{mR^2} = 0.5$
• Solid Sphere: $I = \frac{2}{5}mR^2 \implies \frac{I}{mR^2} = 0.4$

Therefore: $I_{\text{ring}} > I_{\text{cylinder}} > I_{\text{sphere}}$

2. Acceleration ($a$) Comparison:

Since acceleration is inversely proportional to $(1 + \frac{I}{mR^2})$, the object with the smallest Moment of Inertia will have the greatest acceleration:
$a_{\text{ring}} < a_{\text{cylinder}} < a_{\text{sphere}}$

3. Velocity ($v$) at the Bottom:

From the equation of motion $v^2 = u^2 + 2as$ (starting from rest, $u=0$):
$v = \sqrt{2as}$
Thus, the sphere (highest acceleration) will have the greatest velocity, and the ring (lowest acceleration) will have the least velocity.

$v_{\text{ring}} < v_{\text{cylinder}} < v_{\text{sphere}}$

Correct Option: (A)