ENGINEERING PHYSICS UNIT 1 PART B

Step 1: Define Mass Density Assuming the mass is uniformly distributed, the volume density \( \rho \) is: $$ \rho = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{M}{\pi R^2 L} $$

Step 2: Choose a Differential Element Imagine the cylinder is composed of many thin, concentric cylindrical shells. Consider one such shell with:

• Radius: \( r \)
• Thickness: \( dr \)
• Length: \( L \)

The volume \( dV \) of this thin shell is circumference \( \times \) length \( \times \) thickness: $$ dV = 2\pi r \cdot L \cdot dr $$

Step 3: Determine Mass of the Element (dm) The mass of the element is density \( \times \) volume: $$ dm = \rho \cdot dV $$ Substitute \( \rho \) from Step 1: $$ dm = \left( \frac{M}{\pi R^2 L} \right) (2\pi r \cdot L \cdot dr) $$ Cancel common terms (\( \pi \) and \( L \)): $$ dm = \frac{2M}{R^2} r \, dr $$

Step 4: Integrate The moment of inertia of the small element is \( dI = dm \cdot r^2 \). To find the total \( I \), integrate from the center (\( r=0 \)) to the outer surface (\( r=R \)): $$ I = \int_{0}^{R} r^2 \cdot dm $$ $$ I = \int_{0}^{R} r^2 \left( \frac{2M}{R^2} r \, dr \right) $$ Pull constants out: $$ I = \frac{2M}{R^2} \int_{0}^{R} r^3 \, dr $$ $$ I = \frac{2M}{R^2} \left[ \frac{r^4}{4} \right]_{0}^{R} $$ $$ I = \frac{2M}{R^2} \left( \frac{R^4}{4} \right) = \frac{1}{2} M R^2 $$

Step 1: Define Mass Density The volume of the material (solid part) is \( V = \pi (R_2^2 - R_1^2) L \). $$ \rho = \frac{M}{\pi (R_2^2 - R_1^2) L} $$

Step 2: Choose a Differential Element Consider a thin shell within the material boundaries (where \( R_1 < r < R_2 \)) with thickness \( dr \). $$ dm = \rho \cdot (2\pi r \cdot L \cdot dr) $$ Substitute \( \rho \): $$ dm = \frac{2M}{R_2^2 - R_1^2} r \, dr $$

Step 3: Integrate Integrate from the Inner Radius (\( R_1 \)) to the Outer Radius (\( R_2 \)): $$ I = \int_{R_1}^{R_2} r^2 \cdot dm $$ $$ I = \frac{2M}{R_2^2 - R_1^2} \int_{R_1}^{R_2} r^3 \, dr $$ $$ I = \frac{2M}{R_2^2 - R_1^2} \left[ \frac{r^4}{4} \right]_{R_1}^{R_2} $$ $$ I = \frac{2M}{R_2^2 - R_1^2} \cdot \frac{1}{4} (R_2^4 - R_1^4) $$

Step 4: Mathematical Simplification Use the identity \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \): $$ (R_2^4 - R_1^4) = (R_2^2 - R_1^2)(R_2^2 + R_1^2) $$ Substitute this back into the equation: $$ I = \frac{M}{2(R_2^2 - R_1^2)} (R_2^2 - R_1^2)(R_2^2 + R_1^2) $$ Cancel the term \( (R_2^2 - R_1^2) \).

1. Flywheels (Solid Cylinders)

Flywheels in engines are essentially heavy solid cylinders. They utilize a high Moment of Inertia (\( I = \frac{1}{2}MR^2 \)) to store rotational kinetic energy (\( KE = \frac{1}{2}I\omega^2 \)). This resists sudden changes in speed, smoothing out the power delivery in internal combustion engines.

2. Drive Shafts (Hollow Cylinders)

Automobile drive shafts are often hollow. A hollow shaft has a high moment of inertia relative to its mass because the mass is distributed further from the center. This provides a high strength-to-weight ratio, allowing it to transmit torque efficiently without adding excess weight to the vehicle.

3. Industrial Rollers

Conveyor belt rollers are hollow cylinders designed to resist bending forces while maintaining specific rotational properties to move goods efficiently.

Step 1: Identify Parameters

• Parallel Axis through Center: The Diameter of the disc.
• Known \( I_{CM} \): \( I_{\text{diameter}} = \frac{1}{4}MR^2 \)
• Distance (\( d \)): The radius \( R \).

Step 2: Apply Theorem $$ I = I_{\text{diameter}} + Md^2 $$ $$ I = \frac{1}{4}MR^2 + M(R)^2 $$ $$ I = \frac{1}{4}MR^2 + \frac{4}{4}MR^2 $$

Step 1: Identify Parameters

• Parallel Axis through Center: The Central Axis (Polar axis).
• Known \( I_{CM} \): \( I_{\text{central}} = \frac{1}{2}MR^2 \)
• Distance (\( d \)): The radius \( R \).

Step 2: Apply Theorem $$ I = I_{\text{central}} + Md^2 $$ $$ I = \frac{1}{2}MR^2 + M(R)^2 $$ $$ I = \frac{1}{2}MR^2 + \frac{2}{2}MR^2 $$

Step 1: Geometry of Twist Consider a cylindrical shell of radius \( x \) and thickness \( dx \) within the cylinder.

• Let \( \phi \) be the angle of shear.
• Let \( \theta \) be the angle of twist.

From the arc length relation: \( L\phi = x\theta \) $$ \implies \phi = \frac{x\theta}{L} $$

Step 2: Shear Stress and Force Using Modulus of Rigidity (\( G \)): $$ \text{Shear Stress} = G \times \text{Strain} = G\phi = \frac{Gx\theta}{L} $$ The area of the shell cross-section is \( dA = 2\pi x \, dx \). The shearing force (\( dF \)) on this shell is Stress \( \times \) Area: $$ dF = \left( \frac{Gx\theta}{L} \right) (2\pi x \, dx) = \frac{2\pi G \theta}{L} x^2 \, dx $$

Step 3: Torque on the Shell The torque (moment of force) \( d\tau \) about the axis is force \( \times \) distance (\( x \)): $$ d\tau = dF \cdot x = \left( \frac{2\pi G \theta}{L} x^2 \, dx \right) \cdot x $$ $$ d\tau = \frac{2\pi G \theta}{L} x^3 \, dx $$

Step 4: Total Twisting Couple Integrate from the center (\( x=0 \)) to the surface (\( x=R \)): $$ \tau = \int_{0}^{R} \frac{2\pi G \theta}{L} x^3 \, dx $$ $$ \tau = \frac{2\pi G \theta}{L} \left[ \frac{x^4}{4} \right]_{0}^{R} $$ $$ \tau = \frac{\pi G R^4 \theta}{2L} $$

Step 5: Torsional Couple per Unit Twist (C) The couple per unit twist is \( C = \tau / \theta \):

Given Data:

• Mass (\( M \)): 2 kg
• Radius (\( R \)): 25 cm = 0.25 m
• Frequency (\( f \)): 5 rot/s
• Angular Velocity (\( \omega \)): \( 2\pi f = 2\pi(5) = 10\pi \) rad/s

Step 1: Apply Formula For a solid sphere rotating about its diameter, the Moment of Inertia is: $$ I = \frac{2}{5} MR^2 $$

Step 2: Substitution $$ I = \frac{2}{5} \times 2 \times (0.25)^2 $$ $$ I = 0.8 \times 0.0625 $$

Step 1: Apply Formula The rotational kinetic energy is given by: $$ K = \frac{1}{2} I \omega^2 $$

Step 2: Substitution $$ K = \frac{1}{2} (0.05) (10\pi)^2 $$ $$ K = 0.025 \times 100 \pi^2 $$ $$ K = 2.5 \pi^2 $$ (Using \( \pi^2 \approx 9.87 \)) $$ K \approx 2.5 \times 9.87 $$

Step 1: Bending Moment Equation Consider a section at distance \( x \) from the fixed end. The external bending moment due to load \( W \) (weight) acting at the free end distance \( (l-x) \) is: $$ M_{ext} = W(l - x) $$ The internal bending moment is related to the radius of curvature \( R \) and flexural rigidity \( EI \): $$ M_{int} = \frac{EI}{R} = EI \frac{d^2y}{dx^2} $$ Equating for equilibrium: $$ EI \frac{d^2y}{dx^2} = W(l - x) $$

Step 2: Slope (First Integration) Integrate w.r.t \( x \): $$ EI \frac{dy}{dx} = \int (Wl - Wx) dx = Wlx - \frac{Wx^2}{2} + C_1 $$ Boundary Condition: At the fixed end (\( x=0 \)), the beam is horizontal, so slope \( \frac{dy}{dx} = 0 \). $$ \therefore C_1 = 0 \implies EI \frac{dy}{dx} = Wlx - \frac{Wx^2}{2} $$

Step 3: Deflection (Second Integration) Integrate again w.r.t \( x \): $$ EI y = \int \left( Wlx - \frac{Wx^2}{2} \right) dx = \frac{Wlx^2}{2} - \frac{Wx^3}{6} + C_2 $$ Boundary Condition: At the fixed end (\( x=0 \)), there is no depression, so \( y = 0 \). $$ \therefore C_2 = 0 $$

Step 4: Maximum Depression at Free End The maximum depression \( y \) occurs at the free end where \( x = l \). Substitute \( x=l \) into the equation: $$ EI y = \frac{Wl(l)^2}{2} - \frac{W(l)^3}{6} $$ $$ EI y = \frac{Wl^3}{2} - \frac{Wl^3}{6} = \frac{3Wl^3 - Wl^3}{6} = \frac{2Wl^3}{6} $$ $$ y = \frac{Wl^3}{3EI} $$

Step 5: Young's Modulus Formula For a rectangular beam of breadth \( b \) and thickness \( d \), the geometric moment of inertia is \( I = \frac{bd^3}{12} \). Substitute \( I \) into the depression formula: $$ y = \frac{Wl^3}{3E (bd^3/12)} = \frac{4Wl^3}{Ebd^3} $$ Rearranging to find Young's Modulus (\( E \)): Where \( W = Mg \) (Mass \(\times\) Gravity).

Apparatus Cantilever beam (wooden scale or metal bar), G-clamp, Weight hanger, Slotted weights, Traveling microscope (or pin and scale), Vernier caliper, Screw gauge.

Procedure

1. Clamp one end of the beam firmly to the table edge.
2. Fix a pin vertically at the free end of the beam (or use a microscope to focus on a mark).
3. Suspend a weight hanger at the free end. Note the initial reading of the pin position using a traveling microscope.
4. Add weights in steps (e.g., 50g, 100g) and record the microscope reading for the depression (\( y \)) each time.
5. Remove weights in the same steps and record readings again (loading and unloading) to minimize error.
6. Calculate the mean depression \( y \) for a specific load \( M \).
7. Measure the length (\( l \)) of the cantilever from the fixed end to the load point.
8. Measure the breadth (\( b \)) using a Vernier caliper and thickness (\( d \)) using a Screw gauge at multiple places.

Calculation Plot a graph of Load (\( M \)) vs. Depression (\( y \)). It should be a straight line. Find the slope \( \frac{M}{y} \) from the graph. Calculate \( E \) using the derived formula: $$ E = \frac{4 g l^3}{b d^3} \times \left( \frac{M}{y} \right) $$

Step 1: Radius of Curvature The external bending moment is \( M = Wa \) (Force \(\times\) distance). The internal bending moment is \( \frac{EI}{R} \). Equating them: $$ \frac{EI}{R} = Wa \implies R = \frac{EI}{Wa} $$ (Where \( R \) is the radius of curvature, \( E \) is Young's modulus, \( I \) is geometric moment of inertia).

Step 2: Geometry of the Circle Since the beam bends into a circular arc, consider the property of intersecting chords. Let \( y \) be the elevation at the center and \( l \) be the distance between knife edges. $$ y(2R - y) = \frac{l}{2} \cdot \frac{l}{2} $$ $$ 2Ry - y^2 = \frac{l^2}{4} $$ Since the elevation \( y \) is very small compared to \( R \), we neglect \( y^2 \). $$ 2Ry = \frac{l^2}{4} \implies y = \frac{l^2}{8R} $$

Step 3: Substitution Substitute the value of \( R \) from Step 1: $$ y = \frac{l^2}{8 (EI / Wa)} = \frac{Wal^2}{8EI} $$

Step 4: Geometric Moment of Inertia For a rectangular beam of breadth \( b \) and thickness \( d \): $$ I = \frac{bd^3}{12} $$ Substitute \( I \) into the elevation equation: $$ y = \frac{Wal^2}{8E (bd^3/12)} = \frac{12Wal^2}{8Ebd^3} $$

Step 5: Final Expression $$ y = \frac{3Wal^2}{2Ebd^3} $$ Rearranging to find Young's Modulus (\( E \)):

Procedure

1. Place the beam symmetrically on two knife edges separated by a distance \( l \).
2. Suspend two weight hangers of equal mass from the ends of the beam (at distance \( a \) from knife edges).
3. Focus the traveling microscope on the tip of the pin placed at the center of the beam.
4. Note the initial reading of the microscope with the dead load (weight of hangers only).
5. Add weights \( M \) (e.g., 50g) to both hangers simultaneously and record the new position of the pin (elevation).
6. Continue adding weights in steps and noting the readings. Then decrease the weights in the same steps and note readings again (Loading and Unloading).
7. Calculate the mean elevation \( y \) for a specific load \( M \).
8. Measure the length between knife edges (\( l \)) and the length of overhang (\( a \)).
9. Measure breadth (\( b \)) and thickness (\( d \)) of the beam using calipers/screw gauge.

Calculation Plot a graph of Load (\( M \)) vs Elevation (\( y \)). It will be a straight line. Find the slope \( \frac{M}{y} \) from the graph. Using the formula (where \( W = Mg \)): $$ E = \frac{3 g a l^2}{2 b d^3} \left( \frac{M}{y} \right) $$

Step 1: Bending Moment Equation The reaction at each support (A and B) is \( W/2 \) due to symmetry. Consider a section at a distance \( x \) from support A (where \( x < l/2 \)). The external bending moment is: $$ M_{ext} = \text{Reaction} \times \text{Distance} = \frac{W}{2} \cdot x $$ The internal bending moment is related to the flexural rigidity \( EI \): $$ EI \frac{d^2y}{dx^2} = \frac{Wx}{2} $$

Step 2: First Integration (Slope) Integrate w.r.t \( x \): $$ EI \frac{dy}{dx} = \frac{Wx^2}{4} + C_1 $$ Boundary Condition: At the center of the beam (\( x = l/2 \)), the tangent is horizontal, so the slope \( \frac{dy}{dx} = 0 \). $$ 0 = \frac{W(l/2)^2}{4} + C_1 \implies C_1 = -\frac{Wl^2}{16} $$ Substituting \( C_1 \): $$ EI \frac{dy}{dx} = \frac{Wx^2}{4} - \frac{Wl^2}{16} $$

Step 3: Second Integration (Depression) Integrate again w.r.t \( x \): $$ EI y = \frac{Wx^3}{12} - \frac{Wl^2x}{16} + C_2 $$ Boundary Condition: At the support A (\( x=0 \)), there is no vertical displacement, so \( y = 0 \). $$ \therefore C_2 = 0 $$

Step 4: Maximum Depression at Center The maximum depression \( y \) occurs at the center where \( x = l/2 \). Substitute this into the equation: $$ EI y = \frac{W(l/2)^3}{12} - \frac{Wl^2(l/2)}{16} $$ $$ EI y = \frac{Wl^3}{96} - \frac{Wl^3}{32} $$ Finding a common denominator (96): $$ EI y = \frac{Wl^3 - 3Wl^3}{96} = -\frac{2Wl^3}{96} = -\frac{Wl^3}{48} $$ Considering only the magnitude (dropping the negative sign): $$ y = \frac{Wl^3}{48EI} $$

Step 5: Young's Modulus Formula For a rectangular beam, \( I = \frac{bd^3}{12} \). Substitute this into the depression formula: $$ y = \frac{Wl^3}{48E (bd^3/12)} = \frac{12Wl^3}{48Ebd^3} = \frac{Wl^3}{4Ebd^3} $$ Rearranging for \( E \):

Procedure

1. The given beam is placed symmetrically on two knife edges separated by a distance \( l \).
2. A weight hanger is suspended exactly at the center of the beam.
3. A pin is fixed vertically at the center of the beam to serve as a pointer.
4. A traveling microscope is placed in front of the setup and focused on the tip of the pin.
5. Note the initial reading of the microscope (vertical scale) with the dead load (hanger only).
6. Add weights \( M \) (e.g., 50g) to the hanger and record the new reading of the pin (depression). Increase the load in steps.
7. Decrease the load in the same steps and record the readings again (unloading) to check for elasticity.
8. Calculate the mean depression \( y \) for a given load \( M \).
9. Measure the length \( l \) (distance between knife edges), breadth \( b \) (using Vernier caliper), and thickness \( d \) (using Screw gauge).

Calculation Plot a graph of Load (\( M \)) vs Depression (\( y \)). It should be a straight line passing through the origin. Find the slope \( \frac{M}{y} \) from the graph. Calculate Young's Modulus using the formula (where \( W = Mg \)): $$ E = \frac{g l^3}{4 b d^3} \times \left( \frac{M}{y} \right) $$