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Kinetic Energy In Combined Motion

Learn rotational & translational kinetic energy about center of mass, practice gravitational potential & work energy in combined motion.

Translational and Kinetic Energy of Center of Mass

  • Consider a body of mass \(M\) and radius \(R\) rolling on a horizontal rough surface.
  • The body is rotating with angular velocity \(\omega\).
  • The center of mass of the body is moving horizontally with velocity \(v\).

  • Since, the rolling motion is a combination of rotational motion and translational motion.
  • Thus, the kinetic energy of rolling body is a combination of rotational kinetic energy and translational kinetic energy.

Rotational Kinetic Energy about Center of Mass 

  • Since, body is in pure rotation about center of mass.
  • Thus, the rotational kinetic energy of body is

\(KE_{\text{Rotational}}=\dfrac{1}{2}I_{cm}\,\omega^2\) ...(1)

where \(I_{cm}\) is the moment of inertia of body about center of mass.

Translational Kinetic Energy of Center of Mass 

\((KE)_{\text{Translational}}=\dfrac{1}{2}Mv^2_{cm}\)  ...(2)

  • Thus, the kinetic energy of a rolling body is

\((KE)_{\text{Rolling}}=(KE)_{\text{Rotational}}+(KE)_{\text{Translational}}\)

from (1) & (2) we get

\((KE)_{\text{Rolling}}=\dfrac{1}{2}I_{cm}\,(\omega)^2+\dfrac{1}{2}Mv^2_{cm}\)

 

  • Since, the point \(A\) is in contact with surface, thus the velocity of both contact point and surface will be same.

\(v_{G}=v_{A}\)

This is also known as the condition of rolling.

  • \(v-\omega R=\) Net velocity of lowest point due to combined motion.

Illustration Questions

A disk of mass \(M=4\,kg\) and radius \(R=2\,m\) is rolling on a horizontal rough surface. The body is moving with velocity \(v=2\,m/s\). What will be the rotational and translational kinetic energy of the disk?

A \((KE)_{\text{Rot}}=2\,J,\;\;(KE)_{\text{Tra}}=6\,J\)

B \((KE)_{\text{Rot}}=4\,J,\;\;(KE)_{\text{Tra}}=8\,J\)

C \((KE)_{\text{Rot}}=1\,J,\;\;(KE)_{\text{Tra}}=2\,J\)

D \((KE)_{\text{Rot}}=6\,J,\;\;(KE)_{\text{Tra}}=2\,J\)

×

Rotational kinetic energy \((KE)_{\text{Rot}}\) of disk about center of mass is given as,

\((KE)_{\text{Rot}}=\dfrac{1}{2}I_{cm}\,\omega^2\)

\((KE)_{\text{Rot}}=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\,\omega^2\)

\(\because\;v=\omega R\)

Given : \(v=2\,m/s,\;R=2\,m,\;M=4\,kg\)

\(\therefore\;\omega =\dfrac{v}{R}\)

\(\omega =\dfrac{2}{2}\)

\(=1\,rad/s\)

\((KE)_{\text{Rot}}=\dfrac{1}{2}\dfrac{(4)(2)^2}{2}\,(1)^2\)

\(=4\,J\)

Translational kinetic energy \((KE)_{\text{Tra}}\) of center of mass is given as,

\((KE)_{\text{Tra}}=\dfrac{1}{2}Mv^2_{cm}\)

\(=\dfrac{1}{2}(4)(2)^2\)

\(=8\,J\)

A disk of mass \(M=4\,kg\) and radius \(R=2\,m\) is rolling on a horizontal rough surface. The body is moving with velocity \(v=2\,m/s\). What will be the rotational and translational kinetic energy of the disk?

image
A

\((KE)_{\text{Rot}}=2\,J,\;\;(KE)_{\text{Tra}}=6\,J\)

.

B

\((KE)_{\text{Rot}}=4\,J,\;\;(KE)_{\text{Tra}}=8\,J\)

C

\((KE)_{\text{Rot}}=1\,J,\;\;(KE)_{\text{Tra}}=2\,J\)

D

\((KE)_{\text{Rot}}=6\,J,\;\;(KE)_{\text{Tra}}=2\,J\)

Option B is Correct

Kinetic Energy of a Body Performing Combined Motion

  • Consider a body of mass \(M\) and radius \(R\) rolling on a horizontal rough surface.
  • The body is rotating with angular velocity \(\omega\).
  • The center of mass of the body is moving horizontally with velocity \(v\).

  • Since, the rolling motion is a combination of rotational motion and translational motion.
  • Thus, the kinetic energy of a rolling body is a combination of rotational kinetic energy and translational kinetic energy.

Rotational Kinetic Energy about Center of Mass 

  • Since, body is in pure rotation about center of mass.
  • Thus, the rotational kinetic energy of body is

\(KE_{\text{Rotational}}=\dfrac{1}{2}I_{cm}\,\omega^2\)   ....(1)

Where \(I_{COM}\) is the moment of inertia of body about center of mass.

Translational Kinetic Energy of Center of Mass 

\((KE)_{\text{Translational}}=\dfrac{1}{2}Mv^2_{cm}\)    .....(2)

  • Thus, the kinetic energy of a rolling body is

\((KE)_{\text{Rolling}}=(KE)_{\text{Rotational}}+(KE)_{\text{Translational}}\)

from (1) & (2) we get

\((KE)_{\text{Rolling}}=\dfrac{1}{2}I_{cm}\,(\omega)^2+\dfrac{1}{2}Mv^2_{cm}\)

 

Illustration Questions

A disk of mass \(M=2\,kg\) and radius \(R=1\,m\) is rolling on a horizontal rough surface with angular velocity \(\omega=2\,rad\,s^{-1}\), as shown in figure. The body is moving with velocity \(v=3\,m/s\). What will be the rotational and translational kinetic energy of the disk?

A \((KE)_{\text{Rotational}}=3\,J,\;\;(KE)_{\text{Translational}}=5\,J\)

B \((KE)_{\text{Rotational}}=2\,J,\;\;(KE)_{\text{Translational}}=9\,J\)

C \((KE)_{\text{Rotational}}=1\,J,\;\;(KE)_{\text{Translational}}=3\,J\)

D \((KE)_{\text{Rotational}}=3\,J,\;\;(KE)_{\text{Translational}}=8\,J\)

×

Rotational kinetic energy \((KE)_{\text{Rot}}\) of disk about center of mass is given as, 

\((KE)_{\text{Rot}}=\dfrac{1}{2}I_{cm}\,\omega^2\)

\((KE)_{\text{Rot}}=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\,\omega^2\)

Given : \(M=2\,kg,\;R=1\,m,\;\omega=2\,rad\,s^{-1}\)

\((KE)_{\text{Rot}}=\dfrac{1}{2}\left(\dfrac{(2)(1)^2}{2}\right)\,(2)^2\)

\(=2\,J\)

 

Translational kinetic energy \((KE)_{\text{Tra}}\) of center of mass is given as,

\((KE)_{\text{Tra}}=\dfrac{1}{2}Mv^2_{cm}\)

 

Given : \(M=2\,kg,\;v=3\,m/s\)

\((KE)_{Tra}=\dfrac{1}{2}(2)(3)^2\)

\(=9\,J\)

A disk of mass \(M=2\,kg\) and radius \(R=1\,m\) is rolling on a horizontal rough surface with angular velocity \(\omega=2\,rad\,s^{-1}\), as shown in figure. The body is moving with velocity \(v=3\,m/s\). What will be the rotational and translational kinetic energy of the disk?

image
A

\((KE)_{\text{Rotational}}=3\,J,\;\;(KE)_{\text{Translational}}=5\,J\)

.

B

\((KE)_{\text{Rotational}}=2\,J,\;\;(KE)_{\text{Translational}}=9\,J\)

C

\((KE)_{\text{Rotational}}=1\,J,\;\;(KE)_{\text{Translational}}=3\,J\)

D

\((KE)_{\text{Rotational}}=3\,J,\;\;(KE)_{\text{Translational}}=8\,J\)

Option B is Correct

Ratio of Rotational and Translational Kinetic Energy

  • Total kinetic energy of a rolling body is given by

\((KE)=\) Kinetic energy of pure rotation about center of mass + kinetic energy of center of mass

\(KE=\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\)

where \(I_{cm}\) is moment of inertia of body about center of mass 

\(v_{cm}\) is velocity of center of mass

Ratio of Rotational and Translational Kinetic Energy 

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{\dfrac{1}{2}I_{cm}\,\omega^2}{\dfrac{1}{2}Mv^2_{cm}}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{I_{cm}\,\omega^2}{M\,\omega^2R^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{I_{cm}}{MR^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{Mk^2}{MR^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{k^2}{R^2}\)

  • Values of \(k\) for different bodies 

For disk : \(k=\dfrac{R}{\sqrt2}\)

For ring : \(k=R\)

For solid sphere : \(k=\sqrt{\dfrac{2}{5}}R\)

For hollow sphere : \(k=\sqrt{\dfrac{2}{3}}R\)

For solid cylinder : \(k=\dfrac{R}{\sqrt2}\)

Illustration Questions

A disk of radius \(R=2\,m\) and mass \(M=2\,kg\) is rolling on a horizontal rough surface. What will be the ratio of rotational kinetic energy and translational kinetic energy?

A \(\dfrac{4}{3}\)

B \(\dfrac{1}{2}\)

C \(\dfrac{3}{4}\)

D \(2\)

×

Total kinetic energy of a rolling body is given by,

\((KE)=\) Kinetic energy of pure rotation about center of mass + kinetic energy of center of mass

\(KE=\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\)

where \(I_{cm}\) is moment of inertia of body about center of mass 

\(v_{cm}\) is velocity of center of mass

Ratio of Rotational and Translational kinetic energy 

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{\dfrac{1}{2}I_{cm}\,\omega^2}{\dfrac{1}{2}Mv^2_{cm}}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{I_{cm}\,\omega^2}{M\,\omega^2R^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{I_{cm}}{M\,R^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{Mk^2}{MR^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{k^2}{R^2}\)

For disk, \(k=\dfrac{R}{\sqrt2}\)

Thus,

\(\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{\Bigg(\dfrac{R}{\sqrt2}\Bigg)^2}{R^2}\)

\(\Rightarrow\;\dfrac{(KE)_{Rot}}{(KE)_{Tra}}=\dfrac{1}{2}\)

A disk of radius \(R=2\,m\) and mass \(M=2\,kg\) is rolling on a horizontal rough surface. What will be the ratio of rotational kinetic energy and translational kinetic energy?

A

\(\dfrac{4}{3}\)

.

B

\(\dfrac{1}{2}\)

C

\(\dfrac{3}{4}\)

D

\(2\)

Option B is Correct

Calculation of Speed of Body, when it Reaches at the Bottom of the Inclined Plane

  • Consider a disk of mass \(M\) and radius \(R\), placed at height \(h\) on a slope of angle \(\theta\), as shown in figure.

  • Initially, the angular velocity \((\omega)\) of disk and velocity of center of mass is zero.

  • Disk is released from rest and rolls down the ramp without slipping.
  • As the disk rolls down, its initial gravitational potential energy is transformed into kinetic energy.

  • By work energy theorem :

Work done = change in kinetic energy

\(\Rightarrow\;W_g=\Delta KE\)

\(\Rightarrow\;W_g=(KE)_\text{final}-(KE)_\text{initial}\)

\(\Rightarrow\;Mgh=\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_\text{final}-\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_\text{initial}\)

\(\Rightarrow\;Mgh=\left[\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_\text{final}-\left[\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_\text{initial}\)

 

\(\because\) initial \(v_{cm}=0\) & \(\omega=0\)

 

\(\therefore\,\;Mgh=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\left(\dfrac{v_{cm}}{R}\right)^2+\dfrac{1}{2}Mv^2_{cm}\)

\(\left[\because\;v_{cm}=\omega R\;\&\;(I_{cm})_{Disk}=\dfrac{MR^2}{2}\right]\)

\(\Rightarrow\;Mv^2_{cm}\left[\dfrac{1}{4}+\dfrac{1}{2}\right]=Mgh\)

\(\Rightarrow\;v^2_{cm}={\dfrac{4\,gh}{3}}\)

\(\Rightarrow\;v_{cm}=\sqrt{\dfrac{4\,gh}{3}}\)

Illustration Questions

A disk of mass \(M\) and radius \(R\) is placed at a height \(h=9\,m\) on a slope of angle \(\theta=30°\), as shown in figure. The disk is released from rest and rolls down the ramp without slipping. What will be the speed of the disk, when it reaches at the bottom of the inclined plane?  \([given: \,\,g=10\,m/s^2]\)

A \(\sqrt{30}\,m/s\)

B \(2\,m/s\)

C \(6\,m/s\)

D \(2\sqrt{30}\,m/s\)

×

As the disk rolls down, its initial gravitational potential energy is transformed into kinetic energy.

By work energy theorem :

Work done = change in kinetic energy

\(\Rightarrow\;W_g=\Delta KE\)

\(\Rightarrow\;W_g=(KE)_\text{final}-(KE)_\text{initial}\)

\(\Rightarrow\;Mgh=\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_\text{final}-\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_\text{initial}\)

\(\Rightarrow\;Mgh=\left[\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_\text{final}-\left[\dfrac{1}{2}I_{cm}\,\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_\text{initial}\)

 

\(\because\) initial \(v_{cm}=0\) & \(\omega=0\)

\(\therefore\,\;Mgh=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\left(\dfrac{v_{cm}}{R}\right)^2+\dfrac{1}{2}Mv^2_{cm}\)

\(\left[\because\;v_{cm}=\omega R\;\&\;(I_{cm})_{Disk}=\dfrac{MR^2}{2}\right]\)

\(\Rightarrow\;Mv^2_{cm}\left[\dfrac{1}{4}+\dfrac{1}{2}\right]=Mgh\)

\(\Rightarrow\;v^2_{cm}={\dfrac{4\,gh}{3}}\)

\(\Rightarrow\;v_{cm}=\sqrt{\dfrac{4\,gh}{3}}\)

Given : \(h=9\,m,\;g=10\,m/s^2\)

\(v_{cm}=\sqrt{\dfrac{4}{3}×10×9}\)

\(v_{cm}=\sqrt{120}\)

\(v_{cm}=2\sqrt{30}\,m/s\)

A disk of mass \(M\) and radius \(R\) is placed at a height \(h=9\,m\) on a slope of angle \(\theta=30°\), as shown in figure. The disk is released from rest and rolls down the ramp without slipping. What will be the speed of the disk, when it reaches at the bottom of the inclined plane?  \([given: \,\,g=10\,m/s^2]\)

image
A

\(\sqrt{30}\,m/s\)

.

B

\(2\,m/s\)

C

\(6\,m/s\)

D

\(2\sqrt{30}\,m/s\)

Option D is Correct

Illustration Questions

A rod of length \(L=6\,m\) and mass \(M=3\,kg\) is rotating about its center of mass with an angular velocity \(\omega=2\,rad/s\). The rod is moving with velocity \(v=4\,m/s\) horizontally. What will be the rotational and translational kinetic energy of the rod? 

A \((KE)_\text{Rot}=20\,J,\;\;(KE)_\text{Tra}=20\,J\)

B \((KE)_\text{Rot}=3\,J,\;\;(KE)_\text{Tra}=13\,J\)

C \((KE)_\text{Rot}=18\,J,\;\;(KE)_\text{Tra}=24\,J\)

D \((KE)_\text{Rot}=6\,J,\;\;(KE)_\text{Tra}=12\,J\)

×

Rotational kinetic energy \((KE)_\text{Rot}\) of the rod about center if mass is given by,

\((KE)_\text{Rot}=\dfrac{1}{2}\left(I_{cm}\right)\,\omega^2\)

\((KE)_\text{Rot}=\dfrac{1}{2}\left(\dfrac{ML^2}{12}\right)\,\omega^2\)

Given : \(M=3\,kg,\;\;L=6\,m,\;\;\omega=2\,rad\,s^{-1}\)

\((KE)_\text{Rot}=\dfrac{1}{2}\left(\dfrac{(3)(6)^2}{12}\right)\,(2)^2\)

\(=\dfrac{1}{2}\left(\dfrac{3×36}{12}\right)\,(4)\)

\(=18\,J\)

Translational kinetic energy \((KE)_\text{Tra}\) of center of mass is given by,

\((KE)_\text{Tra}=\dfrac{1}{2}Mv^2_{\text{cm}}\)

Given : \(M=3\,kg,\;v=4\,m/s\)

\((KE)_\text{Tra}=\dfrac{1}{2}(3)(4)^2\)

\(=24\,J\)

A rod of length \(L=6\,m\) and mass \(M=3\,kg\) is rotating about its center of mass with an angular velocity \(\omega=2\,rad/s\). The rod is moving with velocity \(v=4\,m/s\) horizontally. What will be the rotational and translational kinetic energy of the rod? 

image
A

\((KE)_\text{Rot}=20\,J,\;\;(KE)_\text{Tra}=20\,J\)

.

B

\((KE)_\text{Rot}=3\,J,\;\;(KE)_\text{Tra}=13\,J\)

C

\((KE)_\text{Rot}=18\,J,\;\;(KE)_\text{Tra}=24\,J\)

D

\((KE)_\text{Rot}=6\,J,\;\;(KE)_\text{Tra}=12\,J\)

Option C is Correct

Calculation of Height up to which a Body gets Stopped while Climbing onto a Rough Surface

  • Consider a disk of mass \(M\) and radius \(R\), moving on a rough surface.
  • The angular velocity of disk is \(\omega\) and the velocity of center of mass is \(v\).
  • The disk starts climbing onto a slope at an angle \(\theta\) with horizontal.
  • After attaining some height, it will stop.
  • Let the height at which the disk stops is \(h\).
  • The final angular velocity and linear velocity is zero because of rolling.
  • As \(v=\omega\,R\), thus, when \(v=0\)\(\omega\) must be zero to satisfy the condition of rolling.

By Work–Energy theorem

\(-Mgh=(KE)_f-(KE)_i\)

\(-Mgh=\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_f-\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_i\)

\(\Rightarrow\;-Mgh=\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_f-\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_i\)

\(\because\) For final kinetic energy, \(v_{cm}=0\) & \(\omega=0\)

\(\Rightarrow\;-Mgh=-\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]\)

\(\because\;(I_{cm})_\text{disk}=\dfrac{MR^2}{2}\)

\(\therefore\;Mgh=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\omega^2+\dfrac{1}{2}Mv^2_{cm}\)

\(\Rightarrow\;gh=\dfrac{1}{4}R^2\left(\dfrac{v_{cm}}{R}\right)^2+\dfrac{1}{2}v^2_{cm}\)

\(\Rightarrow\;gh=v^2_{cm}\left[\dfrac{1}{4}+\dfrac{1}{2}\right]\)

\(\Rightarrow\;gh=v^2_{cm}\left[\dfrac{3}{4}\right]\)

\(\Rightarrow\;h=\dfrac{3\,v^2_{cm}}{4\,g}\)

  • In case of smooth surface, the final angular velocity will not be zero while velocity of center of mass becomes zero.

As there is no torque about center of mass, so angular velocity remains constant.

Illustration Questions

A disk of mass \(M=1\,kg\) and radius \(R=1\,m\) is climbing on to a slope of angle \(\theta=30°\) with velocity \(v=10\,m/s\). At what height will it get stopped, if the slope is of sufficient rough surface for rolling?  \([given: \,\,g=10\,m/s^2]\)

A \(7.5\,m\)

B \(8\,m\)

C \(7\,m\)

D \(8.5\,m\)

×

By work–Energy theorem 

\(-Mgh=(KE)_f-(KE)_i\)

\(-Mgh=\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_f-\left[(KE)_\text{Rot}+(KE)_\text{Tra}\right]_i\)

\(\Rightarrow\;-Mgh=\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_f-\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]_i\)

\(\because\) For final kinetic energy, \(v_{cm}=0\) & \(\omega=0\)

\(\Rightarrow\;-Mgh=-\left[\dfrac{1}{2}(I_{cm})\omega^2+\dfrac{1}{2}Mv^2_{cm}\right]\)

image

\(\because\;(I_{cm})_\text{disk}=\dfrac{MR^2}{2}\)

\(\therefore\;Mgh=\dfrac{1}{2}\left(\dfrac{MR^2}{2}\right)\omega^2+\dfrac{1}{2}Mv^2_{cm}\)

\(\Rightarrow\;gh=\dfrac{1}{4}R^2\left(\dfrac{v_{cm}}{R}\right)^2+\dfrac{1}{2}v^2_{cm}\)

\(\Rightarrow\;gh=v^2_{cm}\left[\dfrac{1}{4}+\dfrac{1}{2}\right]\)

\(\Rightarrow\;gh=v^2_{cm}\left[\dfrac{3}{4}\right]\)

\(\Rightarrow\;h=\dfrac{3\,v^2_{cm}}{4\,g}\)

image

Given : \(g=10\,m/s^2,\;\;v=10\,m/s\)

\(h=\dfrac{3×(10)^2}{4×10}\)

\(h=7.5\,m\)

image

A disk of mass \(M=1\,kg\) and radius \(R=1\,m\) is climbing on to a slope of angle \(\theta=30°\) with velocity \(v=10\,m/s\). At what height will it get stopped, if the slope is of sufficient rough surface for rolling?  \([given: \,\,g=10\,m/s^2]\)

A

\(7.5\,m\)

.

B

\(8\,m\)

C

\(7\,m\)

D

\(8.5\,m\)

Option A is Correct

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