Ms Word to PowerPoint Converter Tool.

Ms Word to PowerPoint Converter Tool: Convert Word (.docx) Q&A notes to PowerPoint slides instantly. One slide per question-answer pair with automatic detection and numbering. Free, browser-based tool for students and teachers.

Ms Word to PowerPoint Converter Tool

---

---

How to Use This Tool

1. Upload File: Click the file input box and select a .docx (Microsoft Word) file containing Q&A notes.Word to PowerPoint Q&A Converter Tool
2. Format Requirement: Your Word file should have questions starting with words like “Q”, “Question”, etc., and answers starting with “A”, “Ans”, “Answer”, etc. The tool automatically detects these and pairs each question with its answer.
3. Convert: Click the Convert to PPT button. The tool will process the file and create one PowerPoint slide per Q&A pair (Question on top, Answer below, with proper numbering).
4. Download: Once conversion is complete, the Download PPTX button becomes active. Click it to download the generated PowerPoint file.
5. Reset: Use the Reset button to clear the upload and start over.

The tool works entirely in your browser—no server upload, fully client-side for privacy.Word to PowerPoint Q&A Converter Tool

Who Can Use This Tool

Anyone with a web browser can use it, especially:

• Students preparing revision slides from Q&A notes
• Teachers and educators creating quick presentation slides from question banks
• Professionals making training or interview preparation materials
• No login or subscription required—just add it to a WordPress page/post.Word to PowerPoint Q&A Converter Tool

---

Ms Word To Pdf File Converter	Click here
MS Word to Excel MCQ Converter	Click here

Test Your Merit Here

---

Thought for 30s

Rotational Dynamics: Detailed Explanation

Rotational dynamics is the branch of physics that deals with the motion of rigid bodies rotating about a fixed axis. It is completely analogous to linear (translational) dynamics, but instead of force, mass, and linear acceleration, we use torque, moment of inertia, and angular acceleration. All the important equations and concepts are given below with proper mathematical expressions using KaTeX.

1. Basic Angular Kinematics

• Angular displacement: $\theta$θ (in radians)
• Angular velocity: $\omega = \frac{d\theta}{dt}$ω=dtdθ​ (rad/s)
• Angular acceleration: $\alpha = \frac{d\omega}{dt}$α=dtdω​ (rad/s²)

The three kinematic equations for constant angular acceleration are exactly parallel to the linear ones:

$$\omega = \omega_0 + \alpha t$$ω=ω0​+αt

$$\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$$θ=θ0​+ω0​t+21​αt2

$$\omega^2 = \omega_0^2 + 2\alpha (\theta – \theta_0)$$ω2=ω02​+2α(θ−θ0​)

2. Moment of Inertia (I)

Moment of inertia is the rotational analogue of mass. It measures how difficult it is to change the rotational motion of a body.

For a system of point masses:Ms Word to PowerPoint Converter Tool

$$I = \sum m_i r_i^2$$I=∑mi​ri2​

For continuous bodies, it is calculated by integration. Standard values (about axis through centre of mass unless stated):

• Thin rod (length $L$L, mass $M$M) about centre ⊥ to length: $I = \frac{1}{12}ML^2$I=121​ML2
• Thin rod about end ⊥ to length: $I = \frac{1}{3}ML^2$I=31​ML2
• Thin circular ring or hoop about central axis: $I = MR^2$I=MR2
• Thin disc or solid cylinder about central axis: $I = \frac{1}{2}MR^2$I=21​MR2
• Solid sphere about diameter: $I = \frac{2}{5}MR^2$I=52​MR2
• Hollow sphere about diameter: $I = \frac{2}{3}MR^2$I=32​MR2

Theorems for moment of inertia:Ms Word to PowerPoint Converter Tool

• Parallel-axis theorem: $I = I_{\text{cm}} + Md^2$I=Icm​+Md2 (where $d$d is distance between parallel axes)
• Perpendicular-axis theorem (for planar lamina): $I_z = I_x + I_y$Iz​=Ix​+Iy​

3. Torque ($\tau$τ)

Torque is the rotational analogue of force:Ms Word to PowerPoint Converter Tool

$$\vec{\tau} = \vec{r} \times \vec{F}$$τ=r×F

Magnitude: $\tau = r F \sin\theta$τ=rFsinθ (where $\theta$θ is angle between $\vec{r}$r and $\vec{F}$F).

Newton’s second law for rotation (about a fixed axis):Ms Word to PowerPoint Converter Tool

$$\tau = I \alpha$$τ=Iα

4. Rotational Kinetic Energy and Work–Power

Rotational KE:

$$K_{\text{rot}} = \frac{1}{2} I \omega^2$$Krot​=21​Iω2

Work done by torque:

$$W = \int \tau \, d\theta$$W=∫τdθ

(for constant torque: $W = \tau \theta$W=τθ)

Power:

$$P = \tau \omega$$P=τω

5. Angular Momentum (L)

Angular momentum is the rotational analogue of linear momentum:Ms Word to PowerPoint Converter Tool

$$\vec{L} = I \vec{\omega} \quad \text{(for rotation about fixed axis)}$$L=Iω(for rotation about fixed axis)

or in general $\vec{L} = \vec{r} \times \vec{p}$L=r×p​.

Newton’s second law in angular form:

$$\vec{\tau} = \frac{d\vec{L}}{dt}$$τ=dtdL​

Law of conservation of angular momentum: If net external torque $\tau_{\text{ext}} = 0$τext​=0, then $L$L is conserved:Ms Word to PowerPoint Converter Tool

$$I_1 \omega_1 = I_2 \omega_2$$I1​ω1​=I2​ω2​

6. Rolling Motion (without slipping)

For pure rolling: $v = \omega R$v=ωR and $a = \alpha R$a=αR.

Total kinetic energy:

$$K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 \left(1 + \frac{I}{mR^2}\right)$$Ktotal​=21​mv2+21​Iω2=21​mv2(1+mR2I​)

Acceleration of a body rolling down an incline (angle $\theta$θ) without slipping:

$$a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}}$$a=1+mR2I​gsinθ​

5 Numerical Problems with Detailed Solutions

Problem 1: Torque and Angular Acceleration A wheel of moment of inertia $I = 0.4$I=0.4 kg m² is rotating at $\omega_0 = 20$ω0​=20 rad/s. A constant torque $\tau = 8$τ=8 N m is applied for 5 s. Find (a) the angular acceleration, (b) the final angular velocity, and (c) the work done by the torque.

Solution: (a) $\alpha = \frac{\tau}{I} = \frac{8}{0.4} = 20$α=Iτ​=0.48​=20 rad/s² (b) $\omega = \omega_0 + \alpha t = 20 + 20 \times 5 = 120$ω=ω0​+αt=20+20×5=120 rad/s (c) Angular displacement $\theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 20 \times 5 + \frac{1}{2} \times 20 \times 25 = 100 + 250 = 350$θ=ω0​t+21​αt2=20×5+21​×20×25=100+250=350 rad Work done $W = \tau \theta = 8 \times 350 = 2800$W=τθ=8×350=2800 J (Alternatively, $W = \frac{1}{2} I (\omega^2 – \omega_0^2) = \frac{1}{2} \times 0.4 \times (14400 – 400) = 2800$W=21​I(ω2−ω02​)=21​×0.4×(14400−400)=2800 J)

Problem 2: Rotational Kinetic Energy & Angular Momentum A solid cylinder of mass 4 kg and radius 0.1 m rotates about its central axis at 30 rad/s. Calculate (a) its rotational kinetic energy and (b) its angular momentum.

Solution: Moment of inertia of solid cylinder: $I = \frac{1}{2} M R^2 = \frac{1}{2} \times 4 \times (0.1)^2 = 0.02$I=21​MR2=21​×4×(0.1)2=0.02 kg m² (a) $K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.02 \times (30)^2 = 9$Krot​=21​Iω2=21​×0.02×(30)2=9 J (b) $L = I \omega = 0.02 \times 30 = 0.6$L=Iω=0.02×30=0.6 kg m²/s

Problem 3: Conservation of Angular Momentum A child of mass 40 kg stands at the centre of a frictionless rotating platform of moment of inertia 200 kg m² and angular speed 2 rad/s. The child walks radially outward to a distance of 1 m from the centre. The moment of inertia of the platform + child system now becomes 280 kg m². Find the new angular speed of the system.Ms Word to PowerPoint Converter Tool

Solution: Initial angular momentum $L_i = I_i \omega_i = 200 \times 2 = 400$Li​=Ii​ωi​=200×2=400 kg m²/s (External torque = 0, so angular momentum is conserved) Final angular momentum $L_f = I_f \omega_f$Lf​=If​ωf​ $400 = 280 \times \omega_f$400=280×ωf​ $\omega_f = \frac{400}{280} = \frac{10}{7} \approx 1.43$ωf​=280400​=710​≈1.43 rad/s

Problem 4: Rolling without Slipping down Incline A solid sphere (mass 2 kg, radius 0.05 m) rolls down a frictionless incline of angle 30° without slipping. Take $g = 10$g=10 m/s². Find (a) linear acceleration of centre of mass and (b) speed after travelling 2 m down the incline (starting from rest).

Solution: For solid sphere, $I = \frac{2}{5} M R^2$I=52​MR2, so $\frac{I}{M R^2} = \frac{2}{5}$MR2I​=52​ (a) $a = \frac{g \sin\theta}{1 + \frac{I}{M R^2}} = \frac{10 \times \sin 30^\circ}{1 + 0.4} = \frac{10 \times 0.5}{1.4} = \frac{5}{1.4} \approx 3.57$a=1+MR2I​gsinθ​=1+0.410×sin30∘​=1.410×0.5​=1.45​≈3.57 m/s² (b) Using $v^2 = u^2 + 2 a s$v2=u2+2as (u = 0, s = 2 m): $v^2 = 2 \times 3.57 \times 2 \approx 14.28$v2=2×3.57×2≈14.28 $v \approx \sqrt{14.28} \approx 3.78$v≈14.28​≈3.78 m/s

Problem 5: Atwood’s Machine with Massive Pulley Two blocks of masses 5 kg and 3 kg are connected by a light string passing over a pulley of mass 4 kg and radius 0.1 m (assume pulley is a uniform disc, $I = \frac{1}{2} M R^2$I=21​MR2). The system is released from rest. Find the acceleration of the blocks (take $g = 10$g=10 m/s², neglect friction).Ms Word to PowerPoint Converter Tool

Solution: Pulley $I = \frac{1}{2} \times 4 \times (0.1)^2 = 0.02$I=21​×4×(0.1)2=0.02 kg m² Let tension on 5 kg side = $T_1$T1​, on 3 kg side = $T_2$T2​, acceleration of blocks = $a$a, angular acceleration of pulley $\alpha = a / R$α=a/R.

Equations: For 5 kg: $5g – T_1 = 5a$5g−T1​=5a → $50 – T_1 = 5a$50−T1​=5a For 3 kg: $T_2 – 3g = 3a$T2​−3g=3a → $T_2 – 30 = 3a$T2​−30=3a For pulley: Torque $\tau = (T_1 – T_2) R = I \alpha$τ=(T1​−T2​)R=Iα $(T_1 – T_2) \times 0.1 = 0.02 \times (a / 0.1)$(T1​−T2​)×0.1=0.02×(a/0.1) $T_1 – T_2 = 2a$T1​−T2​=2a

Solving the three equations: Add first two: $20 = 8a + (T_1 – T_2)$20=8a+(T1​−T2​) Substitute $T_1 – T_2 = 2a$T1​−T2​=2a: $20 = 8a + 2a = 10a$20=8a+2a=10a $a = 2$a=2 m/s²

---