Answer any three questions. a. What is escape velocity? Derive an expression for it on the surface of the earth. [4] b. Define angle of friction and angle of repose. Establish a relation between…

a. Escape Velocity and Its Derivation

Definition

Escape velocity is the minimum velocity required by an object to escape the gravitational pull of a massive body (like Earth) without any further propulsion. In other words, it is the velocity at which the total mechanical energy (kinetic + potential) of the object becomes zero or positive, allowing it to move infinitely far away from the gravitational field.

Derivation of Escape Velocity on Earth’s Surface

To derive the expression for escape velocity ($v_e$), we use the principle of conservation of energy.

1. Total Mechanical Energy at Earth’s Surface When an object of mass $m$ is projected from the Earth’s surface with velocity $v_e$, its total mechanical energy is the sum of its kinetic energy (KE) and gravitational potential energy (PE): $$\text{Total Energy}=\text{KE}+\text{PE}=\frac{1}{2}m{v}_e^2-\frac{G{M}_{E}m}{R_E}$$ where:

   • $G$ = Universal gravitational constant ($6.67\times 10^{-11}{\text{N m}}^2/{\text{kg}}^2$),
   • $M_E$ = Mass of Earth ($5.98\times 10^{24}\text{kg}$),
   • $R_E$ = Radius of Earth ($6.37\times 10^6\text{m}$).

2. Condition for Escape For the object to escape Earth’s gravity, its total energy must be zero or positive (i.e., it must overcome Earth’s gravitational pull). Thus: $$\frac{1}{2}m{v}_e^2-\frac{G{M}_{E}m}{R_E}\ge 0$$ Simplifying: $$\frac{1}{2}m{v}_e^2\ge \frac{G{M}_{E}m}{R_E}$$ The mass $m$ cancels out: $$\frac{1}{2}{v}_e^2\ge \frac{G{M}_E}{R_E}$$ $$v_e^2\ge 2\frac{G{M}_E}{R_E}$$ $$v_e\ge \sqrt{2\frac{G{M}_E}{R_E}}$$

3. Final Expression The minimum escape velocity is: $$v_e=\sqrt{2\frac{G{M}_E}{R_E}}$$ Substituting known values: $$v_e=\sqrt{2\times \frac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{6.37\times 10^6}}$$ $$v_e\approx 11.2\text{km/s}$$

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b. Angle of Friction and Angle of Repose

Definitions

• Angle of Friction ($\varphi$): The angle between the resultant reaction force ($R$) and the normal reaction ($N$) when a body is on the verge of sliding.
• Angle of Repose ($\theta$): The angle of inclination of a plane at which a body just begins to slide down due to gravity.

Relation Between Angle of Friction and Angle of Repose

Consider a block of mass $m$ placed on an inclined plane with angle $\theta$ (angle of repose). The forces acting on the block are:

• Weight ($mg$): Acts vertically downward.
• Normal reaction ($N$): Acts perpendicular to the plane.
• Frictional force ($f$): Acts parallel to the plane, opposing motion.

At the limiting equilibrium (just before sliding): $$f=\mu N$$ where $\mu$ is the coefficient of static friction.

Resolving forces:

• Along the plane: $mg\sin \theta =f=\mu N$
• Perpendicular to the plane: $N=mg\cos \theta$

Substituting $N$: $$mg\sin \theta =\mu mg\cos \theta$$ $$\tan \theta =\mu$$ $$\theta ={\tan }^{-1}\mu$$

Now, consider the angle of friction ($\varphi$). When a force $P$ is applied horizontally to the block, the resultant reaction $R$ makes an angle $\varphi$ with the normal $N$. At limiting equilibrium: $$\tan \varphi =\frac{f}{N}=\frac{\mu N}{N}=\mu$$ Thus: $$\varphi ={\tan }^{-1}\mu$$

Conclusion: Since both $\theta$ and $\varphi$ depend on $\mu$, we have: $$\boxed{\theta =\varphi }$$ The angle of repose ($\theta$) is equal to the angle of friction ($\varphi$).

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c. Moment of Inertia of a Thin Uniform Rod

Definition

The moment of inertia (I) of a body about an axis is a measure of its resistance to rotational motion. It depends on the mass distribution relative to the axis of rotation.

Derivation for a Thin Uniform Rod About Its Center

Consider a thin uniform rod of mass $M$ and length $L$, rotating about an axis perpendicular to its length and passing through its center.

1. Mass per Unit Length ($\lambda$) $$\lambda =\frac{M}{L}$$

2. Infinitesimal Mass Element Consider a small element of length $dx$ at a distance $x$ from the center. Its mass is: $$dm=\lambda dx=\frac{M}{L}dx$$

3. Moment of Inertia Contribution The moment of inertia of this element about the axis is: $$dI=x^{2}dm=x^2\left(\frac{M}{L}dx\right)$$

4. Total Moment of Inertia Integrate from $-L/2$ to $+L/2$ (since the rod is symmetric about the center): $$I={\int }_{-L/2}^{L/2}{x}^2\left(\frac{M}{L}\right)dx=\frac{M}{L}{\int }_{-L/2}^{L/2}{x}^{2}dx$$ $$I=\frac{M}{L}{\left[\frac{x^3}{3}\right]}_{-L/2}^{L/2}=\frac{M}{L}\left(\frac{(L/2{)}^3}{3}-\frac{(-L/2{)}^3}{3}\right)$$ $$I=\frac{M}{L}\left(\frac{L^3}{24}+\frac{L^3}{24}\right)=\frac{M}{L}\left(\frac{L^3}{12}\right)$$ $$I=\frac{M{L}^2}{12}$$

Final Expression: The moment of inertia of a thin uniform rod about an axis through its center and perpendicular to its length is: $$\boxed{I=\frac{1}{12}M{L}^2}$$

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d. Simple Harmonic Motion (SHM) and Total Energy

Definition of SHM

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from the mean position and acts in the opposite direction. Mathematically: $$F=-kx$$ where:

• $F$ = Restoring force,
• $k$ = Force constant (or spring constant),
• $x$ = Displacement from equilibrium.

Total Energy in SHM

The total energy ($E$) of a particle in SHM is the sum of its kinetic energy (KE) and potential energy (PE). It remains constant throughout the motion.

1. Kinetic Energy (KE) $$KE=\frac{1}{2}m{v}^2$$ For SHM, velocity $v$ is given by: $$v=\omega \sqrt{A^2-x^2}$$ where:

   • $\omega =\sqrt{\frac{k}{m}}$ (angular frequency),
   • $A$ = Amplitude (maximum displacement).

   Thus: $$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$$

2. Potential Energy (PE) $$PE=\frac{1}{2}k{x}^2$$ Substituting $k=m{\omega }^2$: $$PE=\frac{1}{2}m{\omega }^2{x}^2$$

3. Total Energy ($E$) $$E=KE+PE=\frac{1}{2}m{\omega }^2(A^2-x^2)+\frac{1}{2}m{\omega }^2{x}^2$$ $$E=\frac{1}{2}m{\omega }^2{A}^2-\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m{\omega }^2{x}^2$$ $$E=\frac{1}{2}m{\omega }^2{A}^2$$

   Since $\omega =\sqrt{\frac{k}{m}}$, we can also write: $$E=\frac{1}{2}k{A}^2$$

Final Expression: The total energy of a particle in SHM is: $$\boxed{E=\frac{1}{2}k{A}^2\text{or}E=\frac{1}{2}m{\omega }^2{A}^2}$$