Kinematics of particle and rigid bodies
- What are tangential and normal components of acceleration in curvilinear motion? In one line On a curved path, acceleration splits into a tangential part that changes speed and a normal part, towards the centre of curvatu... 1
- How do you find the radius of curvature of a particle's path from its x and y motion? In one line Split the acceleration into tangential and normal parts using the velocity direction, then use ρ = v²/a_n . Explained simply The... 1
- What is general plane motion of a rigid body and how does it differ from translation and rotation? In one line General plane motion is the combination of translation and rotation of a rigid body in a plane, like a rolling wheel or a connec... 1
- What is the instantaneous centre of rotation (ICR) and how is it located? In one line The ICR is the point, in or outside the body, that has zero velocity at a given instant, so the body seems to rotate purely abou... 1
- How is the ICR method applied to a wheel rolling without slipping? In one line For rolling without slipping, the contact point with the ground is the ICR, so every point's velocity equals ω times its distanc... 1
- How do you use the ICR method for a crank and connecting rod (two-link) mechanism? In one line Use the crank's known rotation to get the velocity of the crank pin, then locate the connecting rod's ICR from the crank-pin and... 1
- How do you analyse straight-line motion of a particle when acceleration varies with time? In one line When acceleration is a function of time, integrate it once to get velocity and again to get position, using the given initial co... 1
- How do you handle variable acceleration given as a function of position or velocity using v dv/dx? In one line When acceleration depends on position or velocity, use a = v dv/dx or a = dv/dt , separate the variables, and integrate. Explain... 1
- How are velocity and acceleration of a particle on a curved path found using rectangular components? In one line Write the position as x(t) and y(t); differentiate each to get velocity and acceleration components, then combine them. Explaine... 1
- The acceleration of a particle in rectilinear motion is given by \(a = 100 - 3x^2\ \text{m/s}^2\). Knowing that at t = 0, v = 0 and x = 0 find the velocity of the particle at x = 8 m. Since a is a function of x, use a = vdv/dx : v dv = (100 - 3x²) dx ⇒ displaystyle∫₀^v v dv = ∫₀^x (100 - 3x²) dx v²/2 = 100x - x³ At x = 8 m... 1
- The position vector of a particle which moves in the X-Y plane is given by \(\bar r = (3t^3 - 4t^2)\,i + (0.5t^4)\,j\) m. Calculate Velocity and acceleration at t = 4 Sec. Velocity: bar v = dbar r/dt = (9t² - 8t) i + 2t³ j At t = 4 s: bar v = (144 - 32) i + 128 j = 112 i + 128 j m/s |v| = √(112² + 128²) = √(289... 1
- Collar B moves up with constant velocity of \(V_B = 2\) m/s. Rod AB is pinned at B. Find the angular velocity of rod AB and velocity of A.[Figure: collar B slides on a fixed vertical rod. Rod AB of length 1.5 m runs from B down to the left to end A;… Given: v_B = 2 m/s ↑, AB = 1.5 m at 50° to horizontal, A moves along a 30° incline. Relative velocity: vec v_B = vec v_A + vecω × vec r_(B/A... 1
- The acceleration of the particle is defined by the relation \(a = 25 - 3x^2\ \text{mm/s}^2\). The particle starts with no initial velocity at the position x = 0. Determine the position where the velocity is maximum and the corresponding maximum veloc… Using vdv/dx = a : displaystyle∫₀^v v dv = ∫₀^x (25 - 3x²) dx ⇒ v²/2 = 25x - x³ Maximum velocity when dv/dx = 0 , i.e. when a = 0 : 25 - 3x²... 1
- The motion of a particle is defined by the position vector, \(\bar r = 5t\,i + 3t^2\,j\) where ‘r’ is in meters and t is in seconds. At the instant when t = 2 sec, find i) Tangential and normal components of accelerations ii) Radius of curvature. bar v = 5 i + 6t j ; at t = 2 s: bar v = 5 i + 12 j , v = √(25 + 144) = 13 m/s bar a = 6 j m/s² (constant), a = 6 m/s² i) Tangential compone... 1
- In the mechanism shown in fig, the angular velocity of link AB is 5 rad/sec anticlockwise. At the instant shown, determine the angular velocity of link BC and velocity of piston C.[Figure: link AB hangs vertically from a fixed pivot A; AB = 200 mm, r… Velocity of B: v_B = ω_AB × AB = 5 × 0.2 = 1 m/s, perpendicular to AB. With B directly below A and AB rotating anticlockwise, v_B is horizon... 1