Derive the relation between torque and angular momentum

Asked by admin @ in Physics viewed by 309 People

Class 11th !

Derive the relation between torque and angular momentum of particle about an axis .No Spam! ​

Answered by admin @


\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,

\rule{300}{1.5}

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}}

\rule{300}{1.5}

\rule{300}{1.5}

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}}

\rule{300}{1.5}

\rule{300}{1.5}

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 !

\rule{300}{1.5}


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