This page contains summary of formula in college physics, step-by-step solutions to problems in statics, dynamics, vibrations, simple harmonic motion. The basic concepts of physics such as FBD (Free-Body-Diagram) has relevance throughout the engineering course as well as design solutions to industrial needs. The Newton's law of motion, energy-mass-momentum conservation, work energy theorem are the basic building blocks to address any problem in mechanics. Please note that some prior knowledge of kinematics, mechanics and statics at college level is required.
The principle of the conservation of momentum can be used only for systems not acted upon by external forces. This principle is most useful when it can be recognized that by treating several bodies together as a system certain unknown forces will occur as equal and opposite pairs and will thus cancel.
Conservation of Mechanical EnergyThe principle of the conservation of energy is applicable only to systems for which a potential energy can be defined. The work-energy equation (derived earlier) is more general than the equation of conservation of energy. This is so because work-energy equation applies to non-conservative systems (such as systems with frictional losses) as well as conservative systems.
CollisionDirect central impact refers to collision situation when "direction of rebound" is same as "direction of approach". The velocities after impact in case of perfectly elastic collision is:
Now let’s analyze situations when two projectiles are launched with a time gap. The conditions of two projectiles to collide with each other will be investigated. Now, the necessary and sufficient condition is that the two projectiles are at same coordinates (x, y) at later instant of time to collide. Thus: Equating the x-locations:
There are 5 parameters here:Δt, v01, v02, α1, α2. Thus, for any four given parameters the fifth can be calculated.
Collision is based on conservation of momentum and loss of energy defined by coefficient of restitution. This solution describes a case where one of the colliding objects is attached to a spring.
A gun whose barrel makes an angle 'θ' with the horizontal fires a bullet having muzzle velocity ‘U’ with respect to the barrel. The system is mounted on a frictional horizontal track with equal coefficients of static and dynamic coefficients. The mass of the gun-system is M and the mass of the bullet is m. Find the recoil velocity 'V' and the magnitude of the absolute velocity of the bullet as it leaves the gun.
Note that the problem changes state and there are two events in the system. Before bullet is fired and after bullet is fired. This is similar to situations encountered in problems dealing with collisions. Thus:Momentum of the system (gun + bullet) before firing = 0
To a stationary observer, the gun-system is moving left and the bullet is moving right. From equation of relative motion – for horizontal components of velocities:
Two persons A and B of equal weight hold on to the two ends of a massless inextensible rope which passes over a frictionless massless pulley. 'A' starts climbing up the rope with a velocity VA,R relative to the rope, while B hangs on without climbing. Find the absolute velocity of B.
M * (VA,G + VB,G) = 0. That is: (VA,R + VB,G) + VB,G = 0.Hence,
A cylinder of radius R is rolling without slipping on a horizontal surface. Find the relationship between the linear velocity V of the centre of the cylinder and the angular velocity ω of the cylinder.
Linear displacement = OO’ = V * Δt
A particle of mass m and radius r starts from rest and slides down the side of a hemispherical surface of radius R under the action of gravity. If there is no friction, and the particle starts from A, what force will be exerted on the surface by the particle at the instant when it is located at B? Assume R >> r.
Torque acting about centre of mass, τCG = ICG * α = R * f ... [I] where ICG is the mass moment of inertia about centre of mass (or centre of gravity) and α is the angular acceleration.Applying Newton's second law for linear motion: m * a = -f, thus a = -f / m ... [II].
Chindren are playing in rain which is falling steadily at 10 [cm/hr]. The terminal velocity of rain drops can be assumed 5 [m/s] and the average radius of the head of the children to be 7.5 [cm]. What is the average force acting on the children's head dut to vertically falling rain drops.
Note that the rain intensity is specified in length per unit time such as 10 [cm/hr]. This means over unit m2, the water collected will have height of 10 [cm] in 1 [hr].
The mass flow rate of water in rain = 10 [cm/hr] * 1000 [kg/m3] * π * 7.52 [cm2] = 4.91 * 10-7 [kg/s].Assuming all the water sticks with the head that is no splashing of the rain water in vertical direction, all the momentum is lost. Hence, force acting on the head = 4.91 * 10-7 [kg/s] * 5 [m/s] = 2.45 * 10-6 [N].
A frictionless piston of mass M is attached to a mass-less spring though a rod and disk arrangement of negligible mass. The pressure and temperature inside the gas chamber is p and T and the walls of the chamber have infinite thermal conductivity. The volume, cross-section area and diameter of the gas chamber are V, A and D respectively. Find the frequency of oscillation for small displacement of the piston.
Steps to solve SHM problem:
Step-1: Identify the mean (equilibrium) position. This is the static condition when any temporary force or support are not present and only forces continuourly acting in the system (gravity, buoyancy, hydrostatic pressure, spring force) are present. These forces are called "forces internal to the system".Look for statement like "a mass is released from free length of spring, a ball is dropped from... All these cases might be a situation where SHM motion does not start from mean position.
Step-2: Draw the Free-Body-Diagram [FBD] of mean position
Step-3: Assume a small displacement (disturbance) 'x' along the direction of force. Create new FBD under displacement condition
Step-4: Find the unbalanced (restoring) force and divide it by the (equivalent) mass of the system
Step-5: The coefficient of 'x' is the square of "angular frequency". Note that "angular frequency" and "angular velocity" both have same units but they have different meaning.For example, a rod pinned at one end and rotating about it will have both angular velocity [rate of range of angle: dθ/dt] and the angular frequency [ω] - how fast is oscillates about its mean position.
A body of mass M is falling through a viscous resisting medium subjected to a drag force proportional to the velocity squared that is FV = KD * V. If the body starts from rest, find expression for velocity at later times.
A player hits the football which is moving away from him with velocity 0.5 [m/s] with 100.0 [N] of force. If the ball has a mass of 0.25 kg and is in contact with the strings of the racket for 0.025 seconds, what will be the velocity of the ball when it leaves the player’s foot?.
We know by experience that pulling a cart is easier than pushing - if frictional force is significant. Qualitatively this is because pull-action acts against gravity causing the normal reaction to reduce and hence the redcution in frictional force occurs.
Force balance in X- and Y-directions:
If frictional is negligible:
A uniform solid rope of mass per unit length λ is coiled in a small pile on the floor. Its one end is tied with a massless inextensible cord which passes over a pulley. The other send of this cord is pulled down with a constant speed v. Find the force required when the rope has been raised to height x.
Let's assume that a time t, a length of rope x has been raised and a small segment of mass Δm is raised up in time Δt. Hence, impulse = f(t) * Δt. The change in momentum = Δm * v.
From Impulse-Momentum equation: f(t) * Δt = Δm * v, that is f(t) = v * Δm / Δt = v * (λ * v)
Weight of rope of length x, W = λ * x * g. Hence, total force require to be applied = F(t) = f(t) + λ * x * g = λ * [v2 + x * g].
A uniform solid rope of mass per unit length λ is free falling on the floor. Assume that the rope hits the same location. Find the force acting on the floor when a length x has already fallen over.This problem is opposite to the previous problem. Additionally, the rope is falling with an acceleration 'g' and not at constant velocity.
Let's assume that a time t when a length of rope x has already fallen, a small segment of mass Δm fall in time Δt. Hence, impulse = f(t) * Δt. The change in momentum = Δm * v.
From Impulse-Momentum equation: f(t) * Δt = Δm * v, that is f(t) = v * Δm / Δt = v * (λ * v).
From equation v2=u2 + 2gs, v2 = 2gxWeight of rope of length x, W = λ * x * g. Hence, total force require to be applied = F(t) = f(t) + λ * x * g
A person of mass M stands on a platform of mass m and pulls himself up by a rope that runs over a mass-less pulley. He pulls the rope with a force of magnitude F. Find the acceleration of platform.
Since the vertical displacement of both the platform and the person is same, their accelerations are also equal. Free body diagrams of the person and the platform:
A person of mass 'm' is suspended from a roof on a cord of negligible mass having a length L. Assuming cord is behaving as a spring with spring constant k, and that L is the stretched length, calculate the work which the person would have to do to climb to the roof. Compare this with the work required to climb when cord is inextensible. Treat person as a point mass.
Extension of the cord: x = m * g / k
Potential energy of the system when person is at the floor:Potential energy of spring + potential energy of the person, VA = 1/2 * k * x2 + 0 = (1/2k) * m2 * g2.
Potential energy of the system when person is at the roof:Potential energy of spring + potential energy of the person, VB = 0 + m * g * L = m * g * L.
Work done by the person, W1 = - Work done by the (gravity + spring) forces = VB - VB
W1 = m * g / (2k) * [2 * k * L - m * g]Work done when the cord is inextensible, W2 = m * g * L
Two blocks of mass MA and MB are in contact as shown. The coefficient of kinetic friction between all surfaces is μk. Find the magnitude of the horizontal force F necessary to drag block B to the left at a constant speed if A and B are connected by a massless flexible but inextensible thread passing around a fixed, frictionless pulley
Since there is no acceleration, the problem gets highly simplified if the system can be visualized as shown below. Note that this is applicable only when the blocks are moving at constant velocity. This is not applicable if block B and A are accelerating in opposite direction.
Two blocks of mass M and m are in contact as shown. The coefficient of kinetic friction between all surfaces is 0. Find the magnitude of the horizontal velocity V of the block M when mass m just reached the ground level.
v.sinβ = u.sinφ, v.cosβ = e × (u. cosφ)
Finally: cotβ = e.cotφ
The minimum velocity required to climb a curb of height 'b' by a hollow spherical ball of radius R and mass m can be calculated using conservation of angular momentum and conservation of energy.
Conservation of angular moment about top-left corner of the curb:
Minimum velocity required to climb the curb can be obtained by conservation of energy after impact and body has just reached over the top face of curb.
The rate of cooling of a block considered as lumped mass is directly proportional to the difference of instantaneous temperature and the temperature of the surroundings.
A particle is dropped with zero initial velocity at the surface of the earth through a hole drilled diametrically in the earth. What would be the velocity of the particle when it reachs the center of the earth?
In a conservative force field, change in the potential energy = negative of the work done by the force. Potential at the suface of the earth (centre of any solid sphere of mass M), V1 = -GM/R, where G = Universal Constant of Gravitation, M = Mass of the earth and R is the radius.Potential at the centre of the earth, V2 = -3GM/2R.
From conservation of total energy, (P.E. + K.E.)SURFACE = (P.E. + K.E.)CENTRE, the change in kinetic energy = 1/2 * m * v2 = m * [-GM/R - (-3GM/2R)] = GMm/2R. Also, acceleration due to gravity, 'g' = GM/R2.Thus: 1/2 * m * v2 = m * g * R/2. Hence, v = [gR]0.5.
Another approach to solve this question would be to apply Newton's law of motion where the force acting on the particle varies as F(r) = GM(r)m/r2. Here, r = instantaneous radius of the particle, M(r) = Mass of earth of radius r = M * r3 / R3.
dW = F.dr = GM/R3.rdr.
Also, work done upon the mass m by the force F is equal to the change in the kinetic energy of the mass. Hence,
ΔKE = GM/2R = 1/2 * m * v2.
Two vertical cylindrical tanks of area A1 and A2 are inter-connected. Initally water level in each tank is H above the horizontal pipe connecting the two. The fluid in tank '1' is pushed down by a distance x. Find the potential energy of the system as a function of x, the density of the fluid ρ, and the areas of the tanks. Neglect frictional losses while the liquid level was depressed.
From conservation of mass: A1 * ρ * x = A2 * ρ * x2. Thus: x2 = A1/A2 * x.
[Incorrect Method]: As per change in centre of mass of the displaced volume of the liquid in two tanks:Change (reduction) in P.E. of tank 1: ΔPE1 = - [A1 * x * ρ] * g * x/2.
ΔPE = ΔPE1 + ΔPE2 = A1 * ρ * g * x2/2 [A1/A2 - 1].
[Correct Method]: Change in potential energy = Negative of work done by force acting on left column
F(x) = p(x) * A1, p(x) = ρ * g * [x + x2] = ρ * g * x [1 + A1/A2].
Since, force and displacements are in opposite directions: dW = -F(x)* dxW = - A1 * ρ * g * x2/2 [1 + A1/A2].
Thus, ΔPE = -W = A1 * ρ * g * x2/2 [1 + A1/A2].
Bead Sliding on a Rod Rotating in Horizontal Plane about it mid-point:
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