Rotational Kinetic Energy Questions in English

(D) Let $m$ be the mass,$v$ be the velocity of the center of mass,$R$ be the radius,and $I$ be the moment of inertia of the body.
The translational kinetic energy is $KE_{\text{trans}} = \frac{1}{2} mv^2$.
The rotational kinetic energy is $KE_{\text{rot}} = \frac{1}{2} I \omega^2$.
Since the body is rolling without slipping,$v = \omega R$,so $\omega = \frac{v}{R}$.
Substituting this into the rotational kinetic energy formula: $KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 = \frac{1}{2} \left(\frac{I}{R^2}\right) v^2$.
Given $KE_{\text{trans}} = KE_{\text{rot}}$,we have $\frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{I}{R^2}\right) v^2$.
This implies $m = \frac{I}{R^2}$,or $I = mR^2$.
The moment of inertia $I = mR^2$ corresponds to a ring (or a thin hollow cylinder) about its central axis.

(D) The rotational kinetic energy of a body is given by $x = \frac{1}{2} I \omega^2$.
The angular momentum of a body is given by $y = I \omega$.
We can express the kinetic energy in terms of angular momentum as:
$x = \frac{(I \omega)^2}{2 I} = \frac{y^2}{2 I}$.
Rearranging this equation to solve for the moment of inertia $I$,we get:
$I = \frac{y^2}{2 x}$.

(B) The rotational kinetic energy $K$ of a body rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2} I \omega^2$.
The angular momentum $L$ is given by $L = I \omega$.
We can express the kinetic energy in terms of angular momentum as:
$K = \frac{1}{2} I \left( \frac{L}{I} \right)^2 = \frac{L^2}{2I}$.
Given $K = x$ and $L = y$,we have $x = \frac{y^2}{2I}$.
Rearranging for the moment of inertia $I$,we get $I = \frac{y^2}{2x}$.
Therefore,the correct option is $B$.

(B) The moment of inertia of a solid sphere about its diameter is $I_s = \frac{2}{5} mR^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_c = \frac{1}{2} mR^2$.
Let the angular speed of the sphere be $\omega$. Then the angular speed of the cylinder is $2\omega$.
The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Kinetic energy of the sphere: $K_s = \frac{1}{2} I_s \omega^2 = \frac{1}{2} (\frac{2}{5} mR^2) \omega^2 = \frac{1}{5} mR^2 \omega^2$.
Kinetic energy of the cylinder: $K_c = \frac{1}{2} I_c (2\omega)^2 = \frac{1}{2} (\frac{1}{2} mR^2) (4\omega^2) = mR^2 \omega^2$.
The ratio of kinetic energy of the sphere to the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{5} mR^2 \omega^2}{mR^2 \omega^2} = \frac{1}{5}$.
Thus,the ratio is $1: 5$.

(C) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
For a solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} M R^2$.
Let the angular speed of the cylinder be $\omega_c = \omega$.
So,$K_c = \frac{1}{2} (\frac{1}{2} M R^2) \omega^2 = \frac{1}{4} M R^2 \omega^2$.
For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} M R^2$.
The angular speed of the sphere is $\omega_s = \frac{\omega}{2}$.
So,$K_s = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{\omega}{2})^2 = \frac{1}{5} M R^2 (\frac{\omega^2}{4}) = \frac{1}{20} M R^2 \omega^2$.
The ratio of the kinetic energy of the sphere to that of the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{20} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} = \frac{1}{20} \times \frac{4}{1} = \frac{4}{20} = \frac{1}{5}$.

(B) The moment of inertia of a solid sphere about its diameter is $I_S = \frac{2}{5} M R^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_C = \frac{1}{2} M R^2$.
Let $\omega_C$ be the angular speed of the cylinder and $\omega_S$ be the angular speed of the sphere.
Given that $\omega_S = \frac{\omega_C}{2}$.
The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega^2$.
The ratio of the kinetic energy of the sphere to that of the cylinder is:
$\frac{K.E._S}{K.E._C} = \frac{\frac{1}{2} I_S \omega_S^2}{\frac{1}{2} I_C \omega_C^2} = \frac{I_S}{I_C} \times \left( \frac{\omega_S}{\omega_C} \right)^2$.
Substituting the values:
$\frac{K.E._S}{K.E._C} = \frac{\frac{2}{5} M R^2}{\frac{1}{2} M R^2} \times \left( \frac{\omega_C / 2}{\omega_C} \right)^2 = \frac{2/5}{1/2} \times \left( \frac{1}{2} \right)^2 = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}$.
Thus,the ratio is $1: 5$.

(D) Given that the rotational kinetic energies are equal:
$(K.E.)_A = (K.E.)_B$
$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$
$\frac{I_1}{I_2} = \frac{\omega_2^2}{\omega_1^2} \implies \frac{\omega_2}{\omega_1} = \sqrt{\frac{I_1}{I_2}} \quad ...(i)$
Also,rotational kinetic energy is given by $K.E. = \frac{L^2}{2I}$. Since kinetic energies are equal:
$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$
$\frac{I_2}{I_1} = \frac{L_2^2}{L_1^2}$
Given the ratio of angular momenta $\frac{L_1}{L_2} = \frac{1}{\sqrt{3}}$,we have $\frac{L_2}{L_1} = \sqrt{3}$.
Therefore,$\frac{I_2}{I_1} = (\sqrt{3})^2 = 3$.
Thus,$I_2 = 3 I_1$.

(D) The moment of inertia of a solid sphere about its diameter is $I_{\text{sphere}} = \frac{2}{5} M R^2$. The rotational kinetic energy is $K_{\text{sphere}} = \frac{1}{2} I_{\text{sphere}} \omega_{\text{sphere}}^2 = \frac{1}{2} \times \frac{2}{5} M R^2 \omega_{\text{sphere}}^2 = \frac{1}{5} M R^2 \omega_{\text{sphere}}^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_{\text{cylinder}} = \frac{1}{2} M R^2$. Given the angular speed $\omega_{\text{cylinder}} = 2 \omega_{\text{sphere}}$,the rotational kinetic energy is $K_{\text{cylinder}} = \frac{1}{2} I_{\text{cylinder}} \omega_{\text{cylinder}}^2 = \frac{1}{2} \times \frac{1}{2} M R^2 (2 \omega_{\text{sphere}})^2 = \frac{1}{4} M R^2 (4 \omega_{\text{sphere}}^2) = M R^2 \omega_{\text{sphere}}^2$.
The ratio of their kinetic energies is $\frac{K_{\text{sphere}}}{K_{\text{cylinder}}} = \frac{\frac{1}{5} M R^2 \omega_{\text{sphere}}^2}{M R^2 \omega_{\text{sphere}}^2} = \frac{1}{5}$.

(C) The moment of inertia $I$ of the system about the axis passing through the midpoint is given by $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = 2m(\frac{d^2}{4}) = \frac{md^2}{2}$.
Rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Substituting the value of $I$,we get $K = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2 \omega^2}{4}$.
Rearranging for $\omega^2$,we get $\omega^2 = \frac{4K}{md^2}$.
Taking the square root on both sides,we get $\omega = \sqrt{\frac{4K}{md^2}} = \frac{2}{d} \sqrt{\frac{K}{m}}$.

(A) The rotational kinetic energy $E$ of a rigid body is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular frequency.
For a molecule consisting of two atoms of mass $m$ separated by a distance $d$,the moment of inertia about an axis passing through the center of mass and perpendicular to the line joining the atoms is $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = 2m(\frac{d^2}{4}) = \frac{md^2}{2}$.
Substituting $I$ into the energy equation: $E = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2}{4} \omega^2$.
Solving for $\omega^2$: $\omega^2 = \frac{4E}{md^2}$.
Taking the square root: $\omega = \sqrt{\frac{4E}{md^2}} = \frac{2}{d} \sqrt{\frac{E}{m}}$.

(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$.
Given that the angular velocity $\omega = 1 \ rad/s$,we substitute this into the formula: $K = \frac{1}{2} I (1)^2 = \frac{1}{2} I$.
According to the problem,the moment of inertia $(I)$ is numerically equal to '$P$' times the rotational kinetic energy $(K)$: $I = P \cdot K$.
Substituting the expression for $K$: $I = P \cdot (\frac{1}{2} I)$.
Dividing both sides by $I$ (assuming $I \neq 0$): $1 = P \cdot \frac{1}{2}$.
Therefore,$P = 2$.

(A) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} MR^2$.
Let the angular speed of the sphere be $\omega_s = \omega$.
Thus,the kinetic energy of the sphere is $K_s = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2$.
For a disc rotating about an axis passing through its centre and perpendicular to its plane,the moment of inertia is $I_d = \frac{1}{2} MR^2$.
The angular speed of the disc is given as $\omega_d = 2\omega$.
Thus,the kinetic energy of the disc is $K_d = \frac{1}{2} (\frac{1}{2} MR^2) (2\omega)^2 = \frac{1}{4} MR^2 (4\omega^2) = MR^2 \omega^2$.
The ratio of the kinetic energy of the disc to that of the sphere is $\frac{K_d}{K_s} = \frac{MR^2 \omega^2}{\frac{1}{5} MR^2 \omega^2} = 5$.
Therefore,the ratio is $5: 1$.

(A) The moment of inertia $I$ of a diatomic molecule consisting of two atoms of mass $m$ separated by a distance $d$ about an axis passing through its center of mass and perpendicular to the line joining the atoms is given by:
$I = m(d/2)^2 + m(d/2)^2 = 2m(d^2/4) = \frac{md^2}{2}$.
The rotational kinetic energy $E$ is given by $E = \frac{1}{2} I \omega^2$,where $\omega$ is the angular frequency.
Substituting the value of $I$ into the energy equation:
$E = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2}{4} \omega^2$.
Solving for $\omega$:
$\omega^2 = \frac{4E}{md^2}$.
$\omega = \sqrt{\frac{4E}{md^2}} = \frac{2}{d} \sqrt{\frac{E}{m}}$.

(A) Given: Moment of inertia $I = 3 \ kg \ m^{2}$,angular velocity $\omega = 3 \ rad \ s^{-1}$,and mass $m = 27 \ kg$.
The rotational kinetic energy of the body is given by $K_{rot} = \frac{1}{2} I \omega^{2}$.
The translational kinetic energy of the second body is given by $K_{trans} = \frac{1}{2} m v^{2}$.
According to the problem,$K_{rot} = K_{trans}$.
Therefore,$\frac{1}{2} I \omega^{2} = \frac{1}{2} m v^{2}$.
$I \omega^{2} = m v^{2}$.
$v^{2} = \frac{I \omega^{2}}{m}$.
$v = \omega \sqrt{\frac{I}{m}}$.
Substituting the values: $v = 3 \times \sqrt{\frac{3}{27}} = 3 \times \sqrt{\frac{1}{9}} = 3 \times \frac{1}{3} = 1 \ m \ s^{-1}$.

(D) The moment of inertia $I$ of a circular disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
Given: Mass $M = 20 \ kg$,Radius $R = 1 \ m$,Angular velocity $\omega = 2 \ rad \ s^{-1}$.
Calculating the moment of inertia: $I = \frac{1}{2} \times 20 \times (1)^2 = 10 \ kg \ m^2$.
The rotational kinetic energy $K_{rot}$ is given by $K_{rot} = \frac{1}{2}I\omega^2$.
Substituting the values: $K_{rot} = \frac{1}{2} \times 10 \times (2)^2 = 5 \times 4 = 20 \ J$.

(C) The rotational kinetic energy of a rod is given by the formula:
$K_{rot} = \frac{1}{2} I \omega^2$
For a rod of mass $M$ and length $L$ rotating about an axis passing through its center and perpendicular to its length,the moment of inertia is:
$I = \frac{M L^2}{12}$
Substituting this into the kinetic energy formula:
$K_{rot} = \frac{1}{2} \left( \frac{M L^2}{12} \right) \omega^2 = \frac{1}{24} M L^2 \omega^2$ $(i)$
The mass $M$ of the rod can be expressed in terms of its volume and density:
$M = \text{Volume} \times \text{Density} = (A \times L) \times \rho = A L \rho$ $(ii)$
Substituting equation $(ii)$ into equation $(i)$:
$K_{rot} = \frac{1}{24} (A L \rho) L^2 \omega^2 = \frac{1}{24} A L^3 \rho \omega^2$

(D) Given,mass of a circular disc,$M = 12 \,kg$,
radius,$R = 0.5 \,m$,
angular velocity,$\omega = 100 \,rad/s$.
The moment of inertia of a thin circular disc about its central axis is $I = \frac{1}{2} MR^2$.
The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.
Substituting the value of $I$ in the formula,we get $K = \frac{1}{2} (\frac{1}{2} MR^2) \omega^2 = \frac{1}{4} MR^2 \omega^2$.
Now,substituting the given values:
$K = \frac{1}{4} \times 12 \,kg \times (0.5 \,m)^2 \times (100 \,rad/s)^2$
$K = 3 \times 0.25 \times 10000 \,J$
$K = 0.75 \times 10000 \,J = 7500 \,J$.
Since $1 \,kJ = 1000 \,J$,we have $K = 7.5 \,kJ$.
Therefore,the rotational kinetic energy of the disc is $7.5 \,kJ$.

(D) The moment of inertia $(I)$ of a sphere about its diameter is given by $I = \frac{2}{5}MR^2$,where $M$ is the mass and $R$ is the radius.
Since the spheres have equal outer radii $(R_1 = R_2 = R)$ and equal moments of inertia $(I_1 = I_2)$,it follows that $\frac{2}{5}M_1R^2 = \frac{2}{5}M_2R^2$,which implies $M_1 = M_2$.
However,if the spheres are hollow or have different internal structures,they could have the same moment of inertia despite different masses if their mass distributions differ.
Given the problem states they have the same moment of inertia,the rotational kinetic energy $(K_r)$ is defined as $K_r = \frac{1}{2}I\omega^2$.
Since $I_1 = I_2$ and both are rotated with the same angular speed $(\omega_1 = \omega_2 = \omega)$,their rotational kinetic energies must be equal $(K_{r1} = K_{r2} = \frac{1}{2}I\omega^2)$.
Thus,option $(D)$ is the correct statement.

(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^{2}$.
Since the kinetic energies are equal,we have $I_{1} \omega_{1}^{2} = I_{2} \omega_{2}^{2} = I_{3} \omega_{3}^{2}$,which implies $\omega \propto \frac{1}{\sqrt{I}}$.
For a square plate of side $a$ and mass $M$:
$1$. For axis $1$ (passing through the center and parallel to sides),$I_{1} = \frac{Ma^{2}}{12}$.
$2$. For axis $2$ (passing through the center and parallel to the diagonal),$I_{2} = \frac{Ma^{2}}{12}$.
$3$. For axis $3$ (passing through the center and perpendicular to the plane of the plate),by the perpendicular axis theorem,$I_{3} = I_{x} + I_{y} = \frac{Ma^{2}}{12} + \frac{Ma^{2}}{12} = \frac{Ma^{2}}{6}$.
Thus,the moments of inertia are in the ratio $I_{1}: I_{2}: I_{3} = \frac{1}{12}: \frac{1}{12}: \frac{1}{6} = 1: 1: 2$.
The ratio of angular speeds is $\omega_{1}: \omega_{2}: \omega_{3} = \frac{1}{\sqrt{I_{1}}}: \frac{1}{\sqrt{I_{2}}}: \frac{1}{\sqrt{I_{3}}} = \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}} = 1: 1: \frac{1}{\sqrt{2}} = \sqrt{2}: \sqrt{2}: 1$.