Informative line

Kinetic Energy And Work Energy Theorem

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 ?.

Fixed Axis

  • 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.

Illustration Questions

A disk of radius R rotates in such a manner that its center moves in a vertical circle of radius \(\dfrac{R}{2} \)  and its velocity is always perpendicular to the plane of the disk. Then, the axis of rotation is 

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.

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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.  

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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.

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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.

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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.

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A disk of radius R rotates in such a manner that its center moves in a vertical circle of radius \(\dfrac{R}{2} \)  and its velocity is always perpendicular to the plane of the disk. Then, the axis of rotation is 

A image
B image
C image
D image

Option C is Correct

Kinetic Energy of a Rotating Body 

  • The kinetic energy of a particle of mass m, moving with velocity v, is given by \(K = \dfrac{1}{2} m v^2\).
  • In linear motion, the linear variables i.e., displacement (s), velocity (v) and acceleration (a) are same for all particles of a body.

 

  • In a pure rotational motion, these linear variables have different values for different particles of the body. Thus, these variables cannot be associated with the rotating body.

 

  • For a body which is in pure rotational motion the angular variables i.e., angular displacement, angular velocity and angular acceleration have same value for all particles of a body.
  • Using this, the kinetic energy of a rigid body can be calculated by adding kinetic energy of all the particles

                    \( K E= \int \dfrac{1}{2} (dm) v^2 \)  

                   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.

  • Moment of inertia of a body about an axis passing through O is given by : 

                  \(I_0 = \int dm (r_ {{p}/{o}})^2\)

  • Therefore, Kinetic energy, \(K = \dfrac{1}{2} I_0 \;\omega^2\)  
  • Here, \(r_ {{p}/{o}}\) is the perpendicular distance of the points from an axis passing through O.
  • \(I_0 \) is the moment of inertia of the body about the axis passing through O i.e., axis of rotation.

Illustration Questions

A rod AB of mass (M) 6kg and length (L) 2m is rotating about A with an angular velocity (\(\omega\)) 2 rad/sec. Find its kinetic energy.

A 14 J

B 20 J

C 16 J

D 18 J

×

Moment of inertia of rod about A

                          \(I_ {{rod}/{A}} = \dfrac{ML^2}{3}\)

 

  

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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

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A rod AB of mass (M) 6kg and length (L) 2m is rotating about A with an angular velocity (\(\omega\)) 2 rad/sec. Find its kinetic energy.

A

14 J

.

B

20 J

C

16 J

D

18 J

Option C is Correct

Work Done by Gravity, when a Rod Turns by an Angle \(\theta\)

  • Consider a rod of mass m and length \(\ell\), which is hinged at one end and kept in horizontal position initially, as shown in figure.

  • When the rod turned by an angle \(\theta\), then its position can be shown as in figure. 
  • The center of mass (C) of a uniform rod lies at a distance \(\dfrac{L}{2}\)  from one end.
  • The vertical displacement of center of mass is, \(h_ {cm} = \dfrac{\ell}{2} sin\theta\)
  • The work done by gravity is \(W_g = m g h_{cm}\)

                                                        \(W_ g = mg\dfrac{\ell}{2} sin\theta\)      

Illustration Questions

A uniform rod of mass m, length \(\ell\) is hinged at one end and kept in horizontal  position initially, as shown. Find the work done by gravity when the rod turned by an angle \(\theta\).

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.

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Vertical displacement of center of mass, \(h_ {cm} = \dfrac{\ell}{2} sin\theta\)

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Work done by gravity, \(W_g = m g h_{cm}\)

                                  \( = mg\dfrac{\ell}{2} sin\theta\)

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A uniform rod of mass m, length \(\ell\) is hinged at one end and kept in horizontal  position initially, as shown. Find the work done by gravity when the rod turned by an angle \(\theta\).

image
A

\(mg\ell sin\theta\)

.

B

\(mg\ell cos\theta\)

C

\(mg \dfrac{\ell}{2} cos\theta\)

D

\(mg \dfrac{\ell}{2} sin\theta\)

Option D is Correct

Angular Velocity and Kinetic Energy of Rigid Body when it  Rotates under Gravity

  • Consider a uniform rod of mass m and length \(\ell\), which is hinged at one end and kept in horizontal position initially as shown in figure.  

  • When it becomes vertical under the action of gravity. then the position of rod will be as shown in figure.
  • The work done by gravity is \(W_ g = mgh_{cm}\), where \(h_{cm}\) is displacement of center of mass in vertical direction. 
  • By work energy-theorem,

Change in kinetic energy = Total work done 

                          \( K E_f \;– \;KE _i =W_g\)

                           \( K E_f =W_g\)                  [ 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. 

 

Illustration Questions

A uniform rod of mass m and length \(\ell\) is hinged at one end and kept in horizontal position initially, as shown. Find the angular velocity of the rod when it becomes vertical. Consider downward motion to be positive.

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. 

 

image

By work–Energy theorem,

change in kinetic energy = Total work done 

\( K E_f \;– \;KE _i =W_g\)
Since, \(KE _i =0\)

Therefore, \( K E_f =W_g\) 

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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. 

image

From (1) and (2)

     \(mg \dfrac{\ell}{2} = \dfrac{1}{2} I\omega^2\)

     \(mg \dfrac{\ell}{2} = \dfrac{1}{2}\dfrac{m\ell^2}{3} \omega^2\)                  \([\because\,\,I = \dfrac{m\ell^2}{3}]\) 

         \(\omega = \sqrt {\dfrac{3g}{\ell}}\)   

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A uniform rod of mass m and length \(\ell\) is hinged at one end and kept in horizontal position initially, as shown. Find the angular velocity of the rod when it becomes vertical. Consider downward motion to be positive.

image
A

\(\sqrt {\dfrac{2g} {\ell}}\)

.

B

\(\sqrt {\dfrac{g} {\ell}}\)

C

\(\sqrt {\dfrac{3g}{\ell}}\)

D

\(\sqrt {2g{\ell}}\)

Option C is Correct

Illustration Questions

A rigid body having moment of inertia \(I\), about the given axis of mass m, hinged at O is kept in a position such that the line joining center of mass to O is horizontal. Given position of center of mass from O is \(X_{cm}\). Find the angular velocity under gravity when CO turns by an angle \(\theta\).

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. 

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So, \(h _{cm} = X_{cm}sin\,\theta\)

\(\because\) \(W_g = mg \;X_{cm}sin\,\theta\)

 

 

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\(K.E=mg X_{cm} sin\,\theta \)............(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. 

image

From (1) and (2) 

     \( mg X_{cm} sin\,\theta = \dfrac{1}{2} I \omega^2 \)

\( \omega =\sqrt {\dfrac{2mg X_{cm} sin\,\theta}{I}}\)

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A rigid body having moment of inertia \(I\), about the given axis of mass m, hinged at O is kept in a position such that the line joining center of mass to O is horizontal. Given position of center of mass from O is \(X_{cm}\). Find the angular velocity under gravity when CO turns by an angle \(\theta\).

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

\( \sqrt {\dfrac{2mg X_{cm} sin\theta}{3 I}}\)

Option A is Correct

Illustration Questions

A uniform rod AB of mass \(m_1\), and length \(\ell \) is hinged at a point O such that AO is \(\dfrac{\ell}{3}\). If a point mass m2 is attached at point A. Then, find the angular velocity of the system, when it is released from horizontal position and turned by an angle \(\dfrac {\pi}{6}\). Take downward direction to be positive.

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.

image

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)       

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Total work done under gravity

\(W_{g} = m_{1} g{(h_{cm})}_{rod} + {m_2}g {(h_{cm})_{particle}}\)

\(W_{g}= m_{1}g\dfrac{\ell}{12} - m_{2}g\dfrac{\ell}{6}\)

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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)

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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, \(I = I_{rod} + I_{particle}\)

\(I = \big[ \dfrac {m_{1}\ell^2}{12}+ m_1(\dfrac{\ell}{2})^2\big]+ m_{2} (\dfrac{\ell}{3})^2\)

\(I = \dfrac{m_1\ell^2}{9} +\dfrac{m_2\ell^2}{9} = (m_{1} + m_{2}) \dfrac{\ell^2}{9}\)

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Therefore, \(KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} (m_{1}+m_{2}) \dfrac{\ell^{2}}{9}\omega^2\) .......................(2)

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From (1) and (2),

 \( = (m_{1}-2m_{2})g\dfrac{\ell}{12} = \dfrac{1}{2}(m_{1}+m_{2})\dfrac{\ell}{9} \omega^2\)

\(\omega= \sqrt {3(m_1-2m_2)g \over 2(m_1 + m_2)\ell}\)   

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A uniform rod AB of mass \(m_1\), and length \(\ell \) is hinged at a point O such that AO is \(\dfrac{\ell}{3}\). If a point mass m2 is attached at point A. Then, find the angular velocity of the system, when it is released from horizontal position and turned by an angle \(\dfrac {\pi}{6}\). Take downward direction to be positive.

image
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

\( \sqrt {(m_1-2m_2) \over \ell g}\)

Option A is Correct

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