Rotational Dynamics | Mechanics & General Properties of Matter - Physics PDF Download

Angular Momentum Due to Particle

The particle has momentum , then angular momentum about point O is given, where is position vector of particle with respect to origin. Consider motion in the x-y plane, first in the x - direction and then in the y - direction, as shown below in figure (a) and (b).

The most general case involves both these motions simultaneously, as shown in above figure. Hence, L_z = xp_y - yp_x
As you can verify by inspection or by evaluating the cross product as follows. Using r = (x, y, 0) and p = (p_x, P_y, 0), we have

Motion of a rigid body involving translation and rotation

Assume a sphere of mass M and radius R are rolling on a rough surface, where v_cm is velocity  of center of mass and ω is angular velocity
Pure translation: when sphere perform pure translation motion then every point has same speed of speed of center of mass
Pure Rotation: when sphere perform pure rotation about center of mass then center of mass have zero velocity and all  other  point have velocity
so upper and lower points have v = ωR but in opposite direction
Rolling - when translation and rotation can be combined then sphere will rolling

The condition of Rolling without slipping

A rigid body such as sphere having translation motion as velocity of center of mass vcm and angular velocity about center of mass is ω. The condition of rolling without slipping will achieve if v_cm = ωR where R is radius of rolling body.

Similarly, if acceleration of center of mass is a_cm and α is angular velocity of about center of mass then a_cm = ωR , where R is radius of rolling body.

Angular Momentum of Rigid body involve Rolling

The angular momentum  About any point O of rigid body of mass M and radius R

Where  is velocity vector of center of mass
 is position vector of center of mass from origin O
is  angular velocity about center of mass
I_c.m is moment of inertia about center of mass

Example 1: A sphere of mass M and radius R having velocity of center of mass v0 . As friction of surface is enough so that sphere Achieving condition of rolling without slipping.
(a) Find the angular momentum about Center of mass C
(b) Find the Angular momentum about point of contact P
(c) Find Angular momentum about Point O

(a) 

for rolling without slipping, v_c.m = ωR ω = V₀/R

|r| = R and angle between
for rolling without slipping, v_c.m = ωR ω
(c)
|r| = r and angle between

For rolling without slipping, v_c.m = ωR ω = v₀/R

Kinetic Energy of Rigid Body

The rigid body of mass M and radius R . The moment of inertia about center of mass is I_cm. The velocity of center of mass is v_cm and angular velocity about center of mass is ω, then kinetic energy is

Example 2: A sphere of mass M and radius R having velocity of center of mass v₀. As friction of surface is enough so that sphere Achieving condition of Rolling without Slipping . Find the kinetic energy of sphere.

Rolling without slipping, v_c.m = ωR ω = V₀/R 1/2

Example 3: A rod of length l is pivoted at one of its end. If rod is constrained to move in y - z plane, then
(a) find the total energy of system
(b) find angular frequency for small oscillation.
 

(a)

Potential energy V is given by (θ) = -

(b) Hence, energy is conserved
for small oscillation, sinθ = θ , so

Example 4: A rod of mass m and length l is suspended from two mass less vertical springs with a spring constants k₁ and k₂. If x₁ and x₂ be the displacements from equilibrium position of the two ends of the rod, find the kinetic energy of system.

T = 1/2
for small oscillation
(keep in mind that there is not pure rotation case)

Example 5: A stick of length l and mass M initially upright on a frictionless table, starts falling. The problem is to find the speed of the center of mass as a function of position.

The key lies in realizing that since there are no horizontal forces, the center of mass must fall straight down. Since we must find velocity as a function of position, it is natural to apply energy methods.

The sketch shows the stick after it has rotated through angle θ and the center of mass has fallen distance y. The total energy at rest is E = K0 + U0 = Mgl/2
The kinetic energy at a later time is and the corresponding potential energy is
Since, there are no dissipative forces, mechanical energy is conserved and
Hence,
We can eliminate θ by turning to the constraint equation. From the sketch we see that
 we obtain

Torque

The torque where  are always perpendicular. There can be a torque on a system with zero net force, and there can be force with zero net torque. In general, there will be both torque and force. These three cases are illustrated in the sketches as shown below. (The torques are evaluated about the centers of the disks.)

If is the component of the torque about the center of mass andis the total applied force, then torque is given by
The first term is the torque about the center of mass due to the various external forces, and the second term is the torque due to the total external force acting at the center of mass.

Conservation of Angular Momentum

Torque is important because it is intimately related to the rate of change of angular momentum:
 
But since the cross product of two parallel vectors is zero. Also, by Newton’s second law. Hence, the second terms isand we have If external torque is zerothen angular momentum,is constant and the angular momentum is conserved.

Example 6: A disk of mass M and radius b is pulled with constant force F by a thin tape wound around its circumference. The disk slides on ice without friction.
(a) What will be torque about Center of mass
(b) What will be torque about Point A
(c) If Icm is moment of inertia about center of mass. Find the acceleration of center of mass from result obtain by a)
(d) Find the acceleration of center of mass from result (a)

where and because 

as center of mass is origin
(b) Choose a coordinate system whose origin A is along the line of F . The torque about center of mass  C is,

(c) τcm = I_cmα ⇒ bF = I_cmα ⇒ α = bF/I_cm 
(d) The torque about A is zero. As we expect that angular momentum about the origin is conserved.
The angular momentum about A is Lz = I_cmω + ( R × MV)_z = I_cmω -bMV.
Since, dL_z/dt = 0, we have 0= I_cmα - bMα or α = as before.

Example 7: A uniform rod of mass and length a lies on smooth horizontal plane. A particle of mass m moving at speed v0 perpendicular to length of rod strikes it at a distance 4 a from the center and stops after the collision.

(a) Find the velocity of center of mass of rod
(b) Find the angular velocity of rod about its center of mass just after collision.

From conservation of linear momentum mv₀ + M × 0 = m ×0 + Mv_cm

System before collision

System after collisionIn this problem Angular momentum is conserve about any point because external torque is zero so from conservation of angular momentum about center of mass

about origin

Example 8: A sphere of mass M and radius R having velocity of center of mass v0 and angular velocity V₀/2R. The friction of surface are enough that after some time sphere achieve the case of rolling without slipping. Find the velocity of center of mass.

In this problem, the angular momentum is conserved about point of contact P . Initial angular momentum is

The angular momentum about point P when sphere achieve the condition of rolling without slipping



From conservation of angular momentum

Newton’s Law of Motion for Rigid Body

A rigid body having mass M and moment of inertia about center of mass is I_cm
External Force  is acting on center of mass and is linear acceleration of center of mass.
From Newton’s law of motion,
Torque about center of mass is  whereis angular acceleration about center of mass
For rolling without slipping, α = α_cm/R

Direction of frictional force in case of rotational dynamics

Assume a sphere of mass M and radius R are rolling on a rough surface, where α is acceleration of center of mass and α is angular acceleration.
Case 1: If acceleration of center of mass α is more than Rα , then friction will opposite to direction of  acceleration of center of mass F
Case 2: If acceleration of center of mass α is less than Rα, then friction will same  to direction of  acceleration of center of mass

Case 3: In case of rolling without slipping point of contact A will remain at rest, then the direction of frictional force depends on applied torque and possible relative velocity between surface and sphere.

For example if applied force F is at topmost point of sphere then due to torque lowest point will move clock wise direction then frictional force is in same direction of F.
In another example, if sphere is rolling up on inclined plane and by force F and sphere is rolling upward so point of contact A is tend to move clockwise direction to keep this point rest frictional force will in upward direction.

Example 9: A uniform drum of radius R and mass M rolls without slipping down a plane inclined at angle θ. Find its acceleration along the plane. The moment of inertia of the drum about its axis is I0 = MR²/2.

The forces acting on the drum are shown in the diagram. f is the force of friction. The translation of the center of mass along the plane is given by
W sin θ - f = Mα
and the rotation about the center of mass by torque equation Rf = I₀α.

For rolling without slipping, we also have
α = Rα.
If we eliminate f , we obtain W sin θ - I₀ α/b = Mα
Using we obtain
Mg sin θ -  Mα/2 = Mα or α = 2/3 g sin θ

Example 10:  A sphere of mass M and radius R rolling down the wedge with angle θ. Find the minimum value of coefficient of static friction to support pure rolling.

Force equation Mg sin θ - f = Mα ....(i)

The torque equation is given by τ = I_cmα

For limiting case μ_sN ≥ f ⇒ μ_s.Mg cos θ = 2/7 mg sin θ ⇒  μ_s ≥ 2/7 tan θ

Example 11: A wheel of radius R and weight W is to be raised over an obstacle of height h by a horizontal force F applied to the centre as shown in figure. Find the minimum value of F.

Taking torque about D, the corner of the obstacle,

F × CD = W × BD

Example 12: A force F acts tangentially at the highest point of a sphere of mass M and radius R kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of center of sphere.

f is frictional force on the lowest point P The force equation is equivalent to

F + f = Ma ....(i)

The torque equation is equivalent to FR - fR = I_cmα

For case of rolling without slipping
So torque equation can be reduce to F - f = 2/5 Mα .....(ii)
So solving equation (i) and (ii) we will get 2F = 7/5 Mα ⇒ α = 10F/7M

Example 13: A thin mass less rod of length 2 l has equal point masses m attached at its ends (see figure). The rod is rotating about an axis passing through its centre and making angle θ with it.
(a) Find the angular momentum of system
(b) Find the magnitude of the rate of change of its angular momentum i.e., 

All component of product of inertia is zero,

x L = 2ml²ω²sinθcosθ ml²ω² sin2 θ

Example 14: The linear mass density of a rod of length L varies from one end to the other as , where x is the distance from one end with tensions T₁ and T₂ in them (see figure), and λ₀ is a constant. The rod is suspended from a ceiling by two massless strings.
(a) Find mass of Rod
(b) Find center of mass of Rod
(c) Find tension T₁ and T₂

(a) The mass of rod is
(b) The centre of gravity of the rod is located at

(c) Force equation T₁ + T₂ = 4λ₀Lg/3
System is in equilibrium about center of mass so
From torque equation
By putting value of T₂ = 9/7 T₁ in equation, we get T₁ + T₂ = 4λ₀Lg/3
and