Define angular momentum. Derive an expression for it in vector form.

Angular momentum of a rotating body is measure of the quantity of angular motion contained in the body. It is also called moment of linear momentum of the body.

Due to rotational inertia possessed by the body about the axis of rotation, the rotating body cannot change its angular momentum itself. The torque produced by applied force on the rotating body changes its angular momentum and rate of change of angular momentum is equal to torque on the body, i.e.


Angular momentum of a rotating body is measure of the quantity of ang

Let us consider a body moving in XY plane. The torque on the body revolving in XY plane is,

τz =xFy-yFx ...(1)

Let px and vx be the X-component and py and vy be the Y component of linear momentum and velocity of body in XY plane respectively.


Angular momentum of a rotating body is measure of the quantity of ang

Angular momentum of a rotating body is measure of the quantity of ang

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Define couple. Derive an expression for the moment of couple.

It is the set of two equal and opposite force having different line of action. A couple produces the torque and tends to cause the rotational motion.

Let us consider two equal and opposite forces acting at points A and B having position vectors rightwards arrow for straight r subscript 1 of and space rightwards arrow for straight r subscript 2 of as shown in figure.

It is the set of two equal and opposite force having different line o

It is the set of two equal and opposite force having different line o
perpendicular distance between the lines of action of two forces of couple. Thus the magnitude of moment of couple is equal to magnitude of one of the force of couple and perpendicular distance between the lines of action of two forces of couple.
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Show that torque produced by force on the body is equal to product of the force and perpendicular distance of line of action of force from the axis of rotation or it is equal to the product of radial distance of point of action of force from the axis of rotation and transverse component of force.

Let us consider a particle moving along the curve PQ under the influence of a force rightwards arrow for F of let a any instant r, the particle be at A and its position vector is rightwards arrow for r of


Let us consider a particle moving along the curve PQ under the influe

where Fϕ is transverse component of force. Therefore torque is equal to product of radial distance and transverse component of force.

Also x = τ Fsin ϕ = F(rsin ϕ)

= Fd

Therefore, torque is equal to product magnitude of force and perpendicular distance of line of action of force from the axis of rotation.

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In which condition do the centre of mass and centre of gravity of body coincide?

When the size of a body is small, the centre of mass and centre of gravity coincide. 

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Under what condition torque acting on body by a given force is zero?


W e space k n o w space t h a t comma space

space space space space space space space tau space equals space r with rightwards arrow on top x space F with rightwards harpoon with barb upwards on top space

rightwards double arrow space space space straight tau space equals space r space Fsinθ space

If sin θ = 0 or r = 0

then, θ= 0°, 180°

That is, Torque = 0.

That is, when line of action of forces passes through the axis of rotation, torque is zero. 
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