Question

Class 11th !

Derive the relation between torque and angular momentum of particle about an axis .


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Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the Torque be "τ",
• And, Angular Momentum be "L".

$\huge{\bold{\underline{Explanation:-}}}$

Before Deriving the relation between Torque and Angular momentum we need to know some more equations,

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Let a particle "P" of mass "m" is moving in a circular path of radius "r" about the axis of rotation.

Now, Its Tangential acceleration is

$\large{\tt a_T = \dfrac{dv}{dt}}$

After rearranging we get,

$\large{\boxed{\tt a_T = r \alpha}}$

Now, Applying Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Substituting the values,

$\large{\tt F = m \times a_T}$

$\large{\tt F = m \times r \alpha}$

$\large{\boxed{\tt F = mr \alpha}}$

Now, Torque Due to this Force is,

$\large{\boxed{\tt \tau = r \times F}}$

Substituting the values,

$\large{\tt \tau = r \times mr \alpha}$

$\large{\tt \tau = mr^2 \alpha}$

This Torque is For only Particle "P",

Now, Total Torque on the body can be obtained by summation of these terms,

$\large{\tt \tau = ( \sum mr^2) \times \alpha}$

(Σmr² = I {Moment of Inertia})

This comes as,

$\large{\tt \tau = I \times \alpha \: -----(1)}$

$\large{\boxed{\tt \tau = I \times \alpha}}$

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Angular momentum (L) of a rigid body rotating about a fixed axis :-

Taking here the same particle "p"

Therefore, The momentum will be Perpendicular to the The axis of rotation (Say AB as the axis of rotation)

$\large{\therefore \tt mv \perp AB}$

So, The Angular momentum will be,

$\large{\tt L = mvr}$

As we know v = rω, Substituting it,

$\large{\tt L = mr \times (r \omega)}$

$\large{\tt L = mr^2 \omega}$

This Angular momentum is For only Particle "P",

Now, Total Angular momentum on the body can be obtained by summation of these terms,

$\large{\tt L = ( \sum mr^2) \times \omega}$

(Σmr² = I {Moment of Inertia})

Therefore,

$\large{\tt L = I \times \omega \: -----(2)}$

$\large{\boxed{\tt \tau = I \times \omega}}$

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Relation b/w Torque (τ) and Angular momentum(L) :-

From the above equation (2),

∵$\large{\tt L = I \omega}$

Differentiating w.r.t time

$\large{\tt \dfrac{dL}{dt} = \dfrac{d(I \omega)}{dt}}$

Here Moment of Inertia is constant and cannot be differentiated,

$\large{\tt \dfrac{dL}{dt} = I \times \dfrac{d( \omega)}{dt}}$

It becomes,

$\large{\tt \dfrac{dL}{dt} = I \times \alpha}$

$\large{\tt \dfrac{dL}{dt} = I \alpha}$

From equation (1)

${\underline{\tt \tau = I \alpha}}$

Substituting it,

$\large{\tt \dfrac{dL}{dt} = \tau}$

$\huge{\boxed{\boxed{\tt \tau = \dfrac{dL}{dt}}}}$

Hence derived !

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Answer 2

Answer:

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