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.
A
B
C
D
As the center moves perpendicular to the plane of the disk, its center lies at O and the chord AB, passing through O.
Therefore, the line perpendicular to the planes of circular motion i.e., in the plane of disk is the axis of rotation. Hence, option (C) is correct.
In option (A), distance between center of mass and axis of rotation is \(\dfrac{R}{3}\) . So, it can't rotate in a circle of radius \(\dfrac{R}{2}\) .
Hence, option (A) is incorrect.
In option (B), the distance between axis of rotation and center of mass is R. So, it can't rotate in a circle of radius \(\dfrac{R}{2}\) .
Hence, option (B) is incorrect.
In option (D), the distance between axis of rotation and center mass is zero. So, it can't rotate in a vertical circle of radius \(\dfrac{R}{2}\) .
Hence, option (D) is incorrect.
Option C is Correct
\( K E= \int \dfrac{1}{2} (dm) v^2 \)
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 14 J
B 20 J
C 16 J
D 18 J
Moment of inertia of rod about A
\(I_ {{rod}/{A}} = \dfrac{ML^2}{3}\)
Kinetic energy , \(K E = \dfrac {1}{2} \; \dfrac{ML^2}{3} \omega^2\)
\( = \dfrac {1}{2}\times\left( \dfrac{6(2)^2\times(2)^2}{3}\right)\)
= 16 Joules
Option C is Correct
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 \( \sqrt {\dfrac{2mg X_{cm} sin\theta}{3 I}}\)
Work done by gravity,
\(W_g = mg h _{cm}\)
where \(h _{cm}\) is displacement of center of mass in vertical direction.
So, \(h _{cm} = X_{cm}sin\,\theta\)
\(\because\) \(W_g = mg \;X_{cm}sin\,\theta\)
\(K.E=mg X_{cm} sin\,\theta \)............(1)
Also, \(K.E=\dfrac{1}{2} I\omega^2\)................(2)
\(I\) is moment of inertia of the body about the given axis of rotation.
From (1) and (2)
\( mg X_{cm} sin\,\theta = \dfrac{1}{2} I \omega^2 \)
\( \omega =\sqrt {\dfrac{2mg X_{cm} sin\,\theta}{I}}\)
Option A is Correct
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 \( \sqrt {(m_1-2m_2) \over \ell g}\)
Center of mass, C of the rod is at a distance \(\dfrac{\ell}{6}\) from O when it turns by an angle \(\dfrac{\pi}{6}\), where \(h_{cm}\) is the vertical displacement of center of mass.
Vertical displacement of center of mass of rod
\({(h_{cm})_{rod}} = \dfrac{\ell}{6} sin \dfrac{\pi}{6} \)
\(= \dfrac{\ell}{12}\) (downwards)
Vertical displacement of center of mass of particle
\({(h_{cm})_{particle}} = \dfrac{\ell}{3} sin \dfrac{\pi}{6} \)
\(= -\dfrac{\ell}{6}\) (upwards)
Total work done under gravity
\(W_{g} = m_{1} g{(h_{cm})}_{rod} + {m_2}g {(h_{cm})_{particle}}\)
\(W_{g}= m_{1}g\dfrac{\ell}{12} - m_{2}g\dfrac{\ell}{6}\)
Then, by work - energy theorem
\(\Delta KE = W_g\)
\(K E_{f} - KE_{i}= (m_{1}g\dfrac{\ell}{12} - m_{2}g\dfrac{\ell}{6})\)
\(KE_{f} = (\dfrac{m_{1}}{2}-m_{2})g\dfrac{\ell}{6}\)
\(KE_{f} = ({m_{1}}- 2m_{2})g\dfrac{\ell}{12}\)..................(1)
Moment of inertia of a system containing two or more bodies about an axis, is sum of \(I\) of the bodies about the same given axis.
Thus, \(I = I_{rod} + I_{particle}\)
\(I = \big[ \dfrac {m_{1}\ell^2}{12}+ m_1(\dfrac{\ell}{2})^2\big]+ m_{2} (\dfrac{\ell}{3})^2\)
\(I = \dfrac{m_1\ell^2}{9} +\dfrac{m_2\ell^2}{9} = (m_{1} + m_{2}) \dfrac{\ell^2}{9}\)
Therefore, \(KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} (m_{1}+m_{2}) \dfrac{\ell^{2}}{9}\omega^2\) .......................(2)
From (1) and (2),
\( = (m_{1}-2m_{2})g\dfrac{\ell}{12} = \dfrac{1}{2}(m_{1}+m_{2})\dfrac{\ell}{9} \omega^2\)
\(\omega= \sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}\)
Option A is Correct
\(W_ g = mg\dfrac{\ell}{2} sin\theta\)
A \(mg\ell sin\theta\)
B \(mg\ell cos\theta\)
C \(mg \dfrac{\ell}{2} cos\theta\)
D \(mg \dfrac{\ell}{2} sin\theta\)
Center of mass (C) of a uniform rod lies at a distance \(\dfrac{L}{2}\) from one end.
Vertical displacement of center of mass, \(h_ {cm} = \dfrac{\ell}{2} sin\theta\)
Work done by gravity, \(W_g = m g h_{cm}\)
\( = mg\dfrac{\ell}{2} sin\theta\)
Option D is Correct
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 \(\sqrt {2g{\ell}}\)
Work done by gravity,
\(W_g = mg h _{cm}\)
where \(h _{cm}\) is displacement of center of mass in vertical direction.
By work–Energy theorem,
change in kinetic energy = Total work done
\( K E_f \;– \;KE _i =W_g\)
Since, \(KE _i =0\)
Therefore, \( K E_f =W_g\)
Kinetic Energy \( K E =mg \dfrac{\ell}{2}\)............(1)
Also, \( K E = \dfrac{1}{2} I\omega^2\)................(2)
\(I\) is moment of inertia of the body about the given axis of rotation.
From (1) and (2)
\(mg \dfrac{\ell}{2} = \dfrac{1}{2} I\omega^2\)
\(mg \dfrac{\ell}{2} = \dfrac{1}{2}\dfrac{m\ell^2}{3} \omega^2\) \([\because\,\,I = \dfrac{m\ell^2}{3}]\)
\(\omega = \sqrt {\dfrac{3g}{\ell}}\)
Option C is Correct
