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Rotational motion: Angular momentum and Torque

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Explain the concepts of angular momentum and torque with reference to the rotational motion of a rigid body. Derive the relevant expressions and illustrate with solved examples.

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This solution contains concepts of angular momentum and torque with reference to rotational motion of a rigid body have been explained in details. Relevant expressions have been derived and solved examples have been included. Multiple diagrams are present to illustrate concepts.

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Angular momentum: Angular moment to rotational motion is as linear momentum is to translational motion. Hence, before we consider the concept of angular momentum with reference to rotational motion, let's review the concept of linear momentum with reference to translational motion.

For a body of mass m moving with a velocity v at a given instant, we define its linear momentum at that instant as the product of mass and velocity i.e. mv.

For a body undergoing rotational motion about an axis with a rotational (or angular) speed ω (defined as the angle swept by a straight line drawn from the axis of rotation to any point on the object expressed in radians/sec), we can define its angular momentum (L) as the product of its moment of inertia (I) about the given axis of rotation and the angular velocity ω. (For an understanding of the concept of moment of inertia the student may refer to Solutions Library posting no. 563269).

L = Iω ...(1)

Using this definition we derive a general expression for angular momentum of a body of arbitrary shape, this figure can be seen in the attachment.

Let a body of any arbitrary shape be rotating about an axis passing through point O as shown in the fig..

Let us consider the given body is notionally replaced by a point mass m placed at point P, such that the moment of inertia of the point mass placed at P about the given axis is same as the moment of inertia of the body about the axis. Position vector of point P is shown as vector r.

Let us assume, at an instant the velocity vector v of the mass m be as shown in the fig.. Let velocity vector v be resolved into components vcosθ and vsinθ. Component vcosθ must be zero as the body can't move in that direction, the body having only rotational motion about the given axis. Hence, the imaginary mass m can only have the translational velocity vsinθ.

Angular speed ω = vsinθ/r

Further, moment of inertia of the imaginary mass m about the given axis of rotation is given by:

I = mr2

Substituting for I and ω in (1) we get: L = Iω = mr2 x vsinθ/r = mrvsinθ

As mv represents the linear momentum p of the mass, L = rpsinθ

Like linear momentum, the angular momentum is also a vector quantity. Above equation can be rewritten in vector form as: L = r X p (cross or vector product)
where θ is the angle between the position vector r of point P and the velocit vector v.

Magnitude of the angular momentum is equal to rpsinθ. In the fig., the velocity vector v has been extended backwards and a perpendicular OQ has been drawn on this line. Length of perpendicular OQ is rsinθ. Hence,

Magnitude of angular momentum = rpsinθ = p(rsinθ) = Magnitude of the linear momentum x Perpendicular distance of the axis of rotation from the velocity vector

Direction of the angular velocity vector is given by the right hand screw rule of vector multiplication which can be seen in figure in attachment.

If vector C = A X B then i) vector C is perpendicular to the plane in which vectors A and B lie and ii) points in the direction in which a right handed screw placed at the point where the tails of the two vectors meet (point O) advances when turned from the first vector (A) towards the second vector (B). For the ...

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