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Kinetic Energy of a Rotating Body | Physics 11, JEE & NEET

Here you will study work done in rotation motion and Kinetic Energy of a Rotating Body. WORK DONE IN ROTATORY MOTION When a body rotates under action of torque then work is done by torque. If $\tau$ is uniform and body rotates by an angle $\theta$ then Work done = $\tau \theta$ But if $\tau$ depends on $\theta \quad$ for example in case of twisting a wire, $\tau=c \theta \quad$ where $c$ is const. $\begin{array}{ll}{\text { Then work done }} & {\mathrm{dW}=\tau \mathrm{d} \theta \quad \text { (since variable } \tau)} \\ {} & {\quad W=\int \tau \mathrm{d} \theta}\end{array}$ $\mathrm{w}=\int \tau \mathrm{d} \theta$ So for twisting of wire $\quad \mathrm{W}=\int_{0}^{\theta} \mathrm{c} \theta \mathrm{d} \theta=\mathrm{c} \frac{\theta^{2}}{2}$ For twisting of wire for angle $\theta_{1}$ to $\theta_{2}$ W $=\int_{\theta_{1}}^{\theta_{2}} c \theta d \theta=\frac{c}{2}\left(\theta_{2}^{2}-\theta_{1}^{2}\right)$   KINETIC ENERGY OF ROTATION The energy due to rotational motion of a body is known as rotational kinetic energy. If be moment of inertia of body about axis of rotation and $\omega$ is its angular velocity.  Then kinetic energy of rotation. $\mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{I} \omega^{2} \quad$ or $\quad \mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{MK}^{2} \omega^{2}$ $E_{r}=\frac{J^{2}}{2 I}=\frac{\vec{J} \cdot \vec{J}}{2 I}$                      $(\because J=\mathrm{I} \omega)$\ If $\omega$ is constant, $\quad \mathrm{E}_{\mathrm{r}} \propto \mathrm{I}$ If $\mathrm{J}$ is constant, $\quad \mathrm{E}_{\mathrm{r}} \propto \frac{1}{\mathrm{I}}$    
Rotational Kinetic Energy
$E_{\text {rot }}=\frac{1}{2} I \omega^{2}=\frac{(I \vec{\omega})^{2}}{2 I}=\frac{(\vec{I} \vec{\omega}) \cdot(\vec{I} \vec{\omega})}{2 I}=\frac{(\vec{J}) \cdot(\vec{J})}{2 I}$
Work energy theorem: The work done by torque = change in kinetic energy of rotation
$=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)$
Rotational Power $P_{\text {rot }}=\frac{d}{d t}\left(E_{\text {rot }}\right)=\frac{d}{d t}\left(\frac{\vec{J} \cdot \vec{J}}{2 I}\right)$
$=\frac{1}{2 \mathrm{I}}\left(\overrightarrow{2 \mathrm{J}} \cdot \frac{\overrightarrow{\mathrm{d} \mathrm{J}}}{\mathrm{dt}}\right)=\left(\frac{\overrightarrow{\mathrm{J}}}{\mathrm{I}}\right) \cdot \frac{\overrightarrow{\mathrm{d}} \mathrm{J}}{\mathrm{dt}}$  $\because \vec{\omega}=\frac{\vec{J}}{\mathrm{I}} \quad$ and $\quad \vec{\tau}=\frac{\overrightarrow{\mathrm{d}} \mathrm{J}}{\mathrm{dt}}$
$P_{\text {rot }}=\vec{\tau} . \vec{\omega}$
  Ex. If the rotational rotational kinetic energy of a body is increased by $300 \%$ then find percentage increament in its angular momentum. J\alpha $\sqrt{\mathrm{E}} \quad \mathrm{E}_{1}=\mathrm{E} \quad \mathrm{E}_{2}=4 \mathrm{E}$ Percentage increament in its angular momentum $=\frac{\mathrm{J}_{2}-\mathrm{J}_{1}}{\mathrm{J}_{1}} \times 100$ $=(\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}}-1) \times 100=(\sqrt{\frac{4 \mathrm{E}}{\mathrm{E}}}-1) \times 100=100 \%$
Ex. If the angular momentum of a body in increased by $200 \%$ find percentage increament in its rotational kinetic energy. $\mathrm{J}_{1}=\mathrm{J} \quad \mathrm{J}_{2}=3 \mathrm{J} \quad \mathrm{E}=\frac{\mathrm{J}^{2}}{2 \mathrm{I}}$ percentage increament in its rot $\mathrm{K} . \mathrm{E}$. $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100=\left(\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}-1\right) \times 100$ $=\left(\frac{\mathrm{J}_{2}^{2}}{\mathrm{J}_{1}^{2}}-1\right) \times 100=\left(\frac{9 \mathrm{J}^{2}}{\mathrm{J}^{2}}-1\right) \times 100=800 \%$
    Introduction to Rotational Dynamics  Moment of Inertia Moment of Inertia: Perpendicular and Parallel axis theorem Radius of Gyration Law of Conservation of Angular Momentum Conservation of Angular Momentum Examples Kinetic Energy of a Rotating Body Work done in rotatory Motion Rotational Power Combine Translational and Rotational Motion Rolling without slipping Rolling on a plane surface Rolling on a Inclined Plane For Latest updates, Subscribe our Youtube Channel       

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