Rotational Motion Flashcards
Condition for rigid body dynamics
Relative distance between two points should not change
Moment of inertia of discrete particle system
Σmiri^2, where mi is the total mass of the body and ri is the perpendicular distance of the body from the axis of rotation
Moment of inertia of a continuous body
∫miri^2, because we are adding small-small particles forming the body
M.O.I of rod when the axis is the passing from one end
ML^2/3, where m= mass of the body and L = length of the body from the axis
M.O.I of rod when the axis is the passing through the centre of mass
ML^2/12, where m= mass of the body and L = length of the body from the axis
M.O.I of ring when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
MR^2, where m = mass of the body and r = radius of the body
M.O.I of disc when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
MR^2/2, where m = mass of the body and r = radius of the body
M.O.I of hollow cylinder, when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
MR^2, where m = mass of the body and r = radius of the body
M.O.I of solid cylinder, when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
MR^2/2, where m = mass of the body and r = radius of the body
M.O.I of solid sphere, when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
2/5MR^2, where m = mass of the body and r = radius of the body
M.O.I of hollow sphere, when the axis is the passing through the centre of mass, perpendicular to the plane of the ring
2/3MR^2, where m = mass of the body and r = radius of the body
Parallel Axis Theorem
I = Ic +md^2, where Ic = moment of inertia from the centre of the body and d = perpendicular distance.
This theorem should only be applied on a axis parallel to Ic
Calculate the MOI of a solid sphere with axis on the tangent of the sphere.
MOI = 2/5Mr^2 + M(R)^2
Perpendicular Axis Theorem
Iz = Ix + Iy, where all the three are perpendicular to each
Iz is perpendicular to the plane of the body, whereas Ix and Iy are parallel to the body
Calculate MOI of ring when the axis is perpendicular to the surface.
Iz = MR^2+Mr^2 = 2MR^2
Radius of gyration
i = Mk^2,
where k = radius of gyration, distance a single particle has to be placed for it to have the same MOI
Torque(τ) and its three definitions
Force because of the rotational effect of a body
τ=rFsinθ
τ = rfperpendicular
τ = rperpendicular F
What does equilibrium mean in a rotating and translating body?
Translational Equilibrium = Fnet = 0
Rotational Equilibrium = τnet = 0(about any point)
Which point should we take the torque about in a fixed axis rotation?
Hinge Point
Angular Momentum
For a body: I(about axis)*ω, ω= angular velocity of the body
For a particle: Mvr, where v = velocity vector and r = radius vector
Rotational kinetic energy
Iω^2/2
Work Energy Theorem
Work done by all forces = Change in kinetic Energy
Acceleration of a pulley when two masses(m1 and m2) are attached by a string with a pulley having mass
Acceleration = (m2g-m1g)/m1+m2+I/R^2
What is the conditions for conservation of angular momemtum?
Torque net = 0
Initial Momentum = Final Momentum
Angular Impulse
l*J, where j = impulse and l = length from which it is applied
Change in angular momentum
What is the conditions for pure rolling?
The velocity of the lowest point of the body should be at rest w.r.t ground.
v = Rω
a = Rα
Acceleration of a body rolling from an inclined plane
a= gsinθ/1+I/mR^2
What can you say about the kinetic energy of various body rolling down from an inclined plane?
The work done by gravity is same, hence the kinetic energy is same even though everything is different.
Angular Momentum of a body rolling
Icω + Mrvc, Ic = MOI about COM and vc = velocity of centre of mass
Angle of repose and Angle of toppling
Angle of repose = μs(Coefficient of static friction)
Angle of topping = L/h
