Learn kinetic energy of a rotating body, Find the angular velocity of the rod when it becomes vertical and the work done by gravity when the rod turned by an angle ?.
Rotation about a fixed axis :
In rotational motion, the particles forming the rigid body, move in parallel planes along circles centered on the same fixed axis, called the axis of rotation.
Axis of rotation :
Figure shows a rigid body of arbitrary shape in rotation about a fixed axis, called the axis of rotation or the rotation axis.
\( K E= \int \dfrac{1}{2} (dm) v^2 \)
Since, v = r \(\omega\)
\( K E= \int \dfrac{1}{2} (dm) \;r^2_{{p}/{o}} \;\omega^2 \)
\( K E= \dfrac{\omega^2}{2} \int(dm) \;r^2_{{p}{/o}} \)
Note :- \(\omega\) of all the particles on a rigid body is same.
\(I_0 = \int dm (r_ {{p}/{o}})^2\)
A \( \sqrt {\dfrac{2mg X_{cm} sin\,\theta} {I}}\)
B \( \sqrt {\dfrac{mg X_{cm} sin\,\theta}{I}}\)
C \( \sqrt {\dfrac{2mg X_{cm}}{ I}}\)
D \( \sqrt {\dfrac{2mg X_{cm} sin\theta}{3 I}}\)
A \( \sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}\)
B \( \sqrt {2(m_1 + m_2){\dfrac{g}{\ell}}}\)
C \( \sqrt {\dfrac{3}{2} \dfrac{g}{\ell}}\)
D \( \sqrt {(m_1-2m_2) \over \ell g}\)
\(W_ g = mg\dfrac{\ell}{2} sin\theta\)
A \(mg\ell sin\theta\)
B \(mg\ell cos\theta\)
C \(mg \dfrac{\ell}{2} cos\theta\)
D \(mg \dfrac{\ell}{2} sin\theta\)
Change in kinetic energy = Total work done
\( K E_f \;– \;KE _i =W_g\)
\( K E_f =W_g\) [ Since, \(K E _i =0\) ]
\( K E =mg \dfrac{\ell}{2}\) ...........(1)
Also, \( K E = \dfrac{1}{2} I \omega^2\) ..............( 2)
where \(I\) is the moment of inertia of the body about the given axis of rotation.
From (1) and (2), \(\omega\) can be calculated.
A \(\sqrt {\dfrac{2g} {\ell}}\)
B \(\sqrt {\dfrac{g} {\ell}}\)
C \(\sqrt {\dfrac{3g}{\ell}}\)
D \(\sqrt {2g{\ell}}\)