Torque and Angular Momentum

The fact that the North Pole star is a good indicator of the direction of true (geographic) north, shows that the Earth’s axis of rotation maintains an essentially fixed orientation in space, despire the Earth’s orbital motion about the Sun. However, detailed observations show that the Earth’s rotation axis is very gradually sweeping out a cone, causing the north celestial pole to move in a circle on the sky, with an angular diameter of47° and a period of 25,800 years.  The additional element of the Earth’s rotational motion that causes the gradual change in axial orientation is an example of precession, a phenomenon that affects a wide range of rotating bodies.

The motion of a rigid body is, in general, a combination of translational motion and rotational motion. To achieve any kind of turning effect about an axis of rotation it is essential that the line of action of the force does not go through the axis of rotation. In this case, if r is a horizontal displacement vector that stretches from the axis of rotation to the point of application of the force, then the orientation of F in the horizontal plane is determined by the angle θ between r and F. The torque magnitude would be given by,

Γ = rF sinθ

Where r is the magnitude of the displacement vector r.

Given a point O, and a force F that acts at some other point whose displacement vector from O is r, then if the angle between r and F is θ, the torque about O due to F is a vector Γ whose magnitude is rF sinθ, and which points in the direction perpendicular to the plane containing r and F, in the sense specified by the right-hand rule.

The torque about a point O due to a force F is,

Γ = r x F

Where r is the displacement vector from O to the point of application of F.

When several different torques act at a point, the total torque at that point is given by the vector sum of the individual torques.

The instantaneous velocity of any point in a body, relative to O is,

v = ω x r

and the angular acceleration of the body is a = dω/dt

N.B. there are situations where a body can exhibit angular acceleration even when there is no external torque to cause it.

Given any two vectors, a and b, separated by an angle θ, it is always possible to form their vector product a X b (also known as the cross product). The vector product a X b is defined as a vector, with magnitude ab sin θ, that points in the direction perpendicular to both a and b, as specified by the right-hand rule.

If the components of a and b are known,

a X b = (aybzazby, azbxaxbz, axbyaybx)

The scalar product of a and b is represented by a ∙ b, and is defined by the relation a ∙ b = ab cos θ. In terms of components the scalar product of two vectors may be written,

a ∙ b = axbx + ayby + azbz

N.B. The cross product is non-commutative, i.e.

a x b = – (a x b)

A body that is not undergoing translational acceleration is said to be in a state of translational equilibrium, and Newton’s first law implies,

Σi Fi = 0, where the Fi are the external forces.

We can write down a corresponding condition for a rotating rigid body by demanding that the vector sum of the external torques about the body’s centre of mass should be zero. Thus we have the condition for rotational equilibrium,

Σi Γi = 0, where the Γi are the external torques.

When both conditions are true we have mechanical equilibrium. A static system is a special case of a system in equilibrium, so it must necessarily satisfy this condition. However, the mechanical equilibrium condition is not sufficient to guarantee that a system will be static.

The special case of mechanical equilibrium that occurs when a body or a system of bodies is completely motionless, i.e. when vcm = 0 and ω = 0, is known as static equilibrium. Under these circumstances the absence of torques around any axis will ensure that the angular acceleration about the axis is zero and that it therefore continues to be true that ω = 0. Under the conditions of static equilibrium the requirement that the resultant torques about the centre of mass should be zero may be shown to be equivalent to saying: For a system in static equilibrium the resultant torque about any point must be zero.

There are three possible types of static equilibrium: Stable equilibrium, unstable equilibrium and neutral equilibrium.  (Think of a particle in a well, a particle on top of a hill, a particle on a plane).

A particle of mass m, moving uniformly around a circle of radius r centred on a point O, will have kinetic energy by virtue of its motion. If the particle has speed v, its kinetic energy will be ½ mv2. However, in this case, v = rω where ω is the angular speed of the particle around O, we can regard the kinetic energy as an example of rotational kinetic energy,

Erot = ½ mr2ω2

For all the particles in a body,

Erot = ½ ω2 Σi miri2

And this sum within this equation is said to represent the moment of inertia of the system of particles, about the rotation axis, and is usually given the symbol Ι,

Erot = ½ Ι ω2, where Ι = Σi miri2

The moment of inertia of a system depends on the axis about which it is determined.

The moment of inertia of a system about a given axis characterizes the way in which the mass of the system is distributed about that axis, and thus plays an important role in determining the rotational energy of the system about the axis.

For a cylindrical shell or ring of mass M and radius R, rotating about its axis of rotational symmetry,

Ics = MR2

For a hollow cylinder of mass M and internal radius R2 and external radius R1, about its axis of rotational symmetry,

Ihc = ½ M(R12 + R22)

For a solid cylinder or disc of mass M and radius R, rotating about its axis of rotational symmetry,

Isc = ½ MR2

For a rectangular plate of mass M and sides a and b, rotating about its axis of rotational symmetry,

Irp = 1/12 M(a2 + b2)

For a long thin rod of mass M and length L, rotating about an axis perpendicular to the rod and through its centre,

Itr = 1/12 ML2

For a long thin rod of mass M and length L, about an axis perpendicular to the rod and through one of its ends,

Itr = 1/3 ML2

For a solid sphere of mass M and radius R, rotating about an axis through its centre,

Iss = 2/5 MR2

For a thin spherical shell of mass M and radius R, about an axis through its centre,

Itss = 2/3 MR2

If forces are applied to a body in such a way taht they apply a net torque Γ that is parallel to ω the resulting angular acceleration of the body will be,

Γ = Ι dω/dt

The angular momentum l of a particle about a point O is defined by the vector product of the displacement vector r of the particle from point O and the particle’s linear momentum p relative to O,

l = r x p

For a rotating rigid body, the angular momentum L about a given point depends on the way the body’s mass is distributed and on the component of its angular velocity ω. It is not generally true that L is parallel to ω nor that L = Ιω.

The fundamental equation is,

Γ = dL/dt where Γ represents the sum of the external torques about a point and L is the angular momentum of the body about that same point.

For any system the total angular momentum about any point remains constant as long as no net external torque acts on the system. This is the law of conservation of angular momentum.

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~ by jamesdow2013 on April 9, 2013.

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