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enter image description here

A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure).

If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ represents angular velocity.

If I calculate the force using the radius of curvature of cycloid, I end up with the value of $F=mR\omega^2$, which is indeed true.

My concern is that why doesn't ICOR provide me with the correct answer.

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  • $\begingroup$ Centripetal force is what keeps the particle in circular motion. The topmost particle is moving at a radius of $R$, why have you taken $2R$? Even in the frame of reference of the lowermost point, the particle is in uniform circular motion in a circle of radius $R$. $\endgroup$ – Yashas Feb 21 '17 at 14:08
  • $\begingroup$ @Yashas, in ground frame I would perceive the particle to move in a circle of radius 4 R(as 4R is the length of radius of curvature of the cycloid at top most point) and yes since it is force so the value won't change even if I see it from another inertial frame that is the centre of the circle which is moving along a straight line. But my concern here is that ICOR is also a static point thus it will make me feel as if the whole wheel is rotating about it(momentarily) then why doesn't it yield the correct answer. $\endgroup$ – Vasu Goyal Feb 21 '17 at 14:37
  • $\begingroup$ "I would perceive the particle to move in a circle of radius $4R$ " - that is the flaw in your reasoning. Consider a particle moving in projectile motion, at the highest point, try to calculate the radius. The radius is not the maximum height of the projectile. Rather, it is $v^2/g$. $\endgroup$ – Yashas Feb 21 '17 at 14:47
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    $\begingroup$ @Yashas, I appreciate your efforts but I think that I am not able to make myself clear here. What I mean(in a general sense) here is why can't I consider centripetal force on a body about its ICOR. And yes of course I am not getting confused in considering the bottom most point as point of rotation as you said in projectile (parabolic trajectory) example. $\endgroup$ – Vasu Goyal Feb 21 '17 at 14:55
  • $\begingroup$ The topmost particle is NOT rotating about the point P. The radius is not $2R$. It doesn't matter which frame of reference you are in, the topmost point is rotating in a circle of radius $R$. Distances don't change over frames of reference. $\endgroup$ – Yashas Feb 21 '17 at 14:58
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The "true" acceleration of the rim point A is $a_A = 2 R \omega^2$. How did you decide the true value is $R \omega^2$?

You can look at the ICOR and arrive at $2 R \omega^2$, or you can add the acceleration of A due to the rotation $R \omega^2$ plus the acceleration of the center of mass $R \omega^2$ to arrive at the same value.

BTW Taking the acceleration of A and multiplying by the entire mass $m$ of the body is meaningless as a force quantity. Only the acceleration of the center of mass is important in deriving inertial forces. The is because linear momentum is defined as the mass of a body $m$ times the velocity of the center of mass $v_C = \omega R$ and force is the time derivative of momentum.

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  • $\begingroup$ firstly $m$ is the mass of a particle at top of the wheel, not the entire wheel. Secondly, if we take the same approach to find the acceleration of centre of the wheel, the acceleration comes to be $Rω²$, and is false as the trajectory of the centre is a straight line, thus the approach through ICOR method is incorrect without considering pseudo force. $\endgroup$ – Vasu Goyal Feb 24 '17 at 16:30
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$ P$ is a non-inertial frame of reference which is why it does not provides us with correct answer. In order to obtain the correct result we need to write the relation between accelerations as $a$$real$ = $a$$0$ +$a'$. Where $a$$real$ denotes acceleration of the particle(whose motion is considered) in inertial frame, $a$$0$ denotes acceleration of non-inertial frame and $a'$ denotes acceleration of particle in non-inertial frame. $a$$0$ is equal to $Rω²$ directed towards centre of the wheel whereas $a'$ is the acceleration of the top most point equal to $2Rω²$ directed radially towards the center of the wheel.

*Note: Since both points, ICOR and top most particle are diametrically opposite hence addition of acceleration vectors pointing towards centre produce a cancelation effect.

Adding both vectors results in real acceleration equal to $Rω²$ pointing towards the centre of the wheel, which is the correct result. enter image description here

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