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I am a high school student and my teacher taught me that a cyclist bend while taking a turn on a horizontal road because if he doesn't bend it then it will rotate due to net torque of friction about centre of mass,so he bend the cycle at an angle such that the net torque about COM comes out to be 0,but I want to ask that if he had bend the cycle now then why wouldn't the cycle rotate about the point where friction and normal reaction are acting now? I know that we cannot apply torque=Moment of inertia× angular acceleration anywhere, we can only apply this equation to COM and instantaneous axis of rotation, this is not the case of angular acceleration, this is the case of toppling and in toppling net torque about any point must be 0 so why it is not 0 here? I am misunderstanding something? please clear the whole picture, what is actually happening here?

this image is showing my problem

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    – imposter
    Aug 17, 2020 at 15:25

4 Answers 4

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I believe you have drawn the centrifugal force incorrectly. The centrifugal force acts through the center of gravity in the opposite direction to the centripetal force, i.e., the friction force. See red force in figure below. Now the sum of the moments about the tire contact point can be zero.

Hope this helps.

enter image description here

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  • $\begingroup$ it is not the centrifugal force, it is the rsultant of normal reaction and frictional force ,and I ant to solve it from ground frame i,e without using centrifugal force then how do I solve it , that is my question $\endgroup$ Mar 18, 2021 at 14:50
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One of your forces is marked wrong see the reference picture

why wouldn't the cycle rotate about the point where friction and normal reaction are acting now


If you calculate torque about that point you see that torque due to friction and normal reaction is 0 as distance from this point of both the forces is 0. On the other hand torque due to centrifugal force and mg is equal and opposite so net torque becomes 0 , so why would there be toppling ?

this is the case of toppling and in toppling net torque about any point must be 0 so why it is not 0 here

No , the condition (here ) for toppling will be that torque due to mg should be greater than that due to centrifugal force . The case when both are equal and opposite I.e net torque is 0 is the limiting case enter image description here

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  • $\begingroup$ I think you meant centrifugal force $\endgroup$ Aug 17, 2020 at 15:43
  • $\begingroup$ No,,bro ,what I have drawn in the figure is the net contact force ,,not centrifugal force,..as there is no such thing as Centrifugal force in reality,,,,I have to do it without considering centrifugal force,,in an inertial frame,,so please try to solve my query without considering centrifugal force $\endgroup$
    – user272523
    Aug 18, 2020 at 6:08
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No need to draw centripetal force. It is not a separate force. Rather, the friction force does the job of centripetal force in this case.

The forces acting on the cyclist + the cycle are

  1. Weight vertically downward

  2. Normal force vertically upward

  3. Friction horizontally (this provides the centripetal force)

Let us find the total torque about an axis through the center of mass. When the cyclist does not bend, then normal force passes through the COM. The only torque is due to friction, so the total torque is non-zero.

When the cyclist bends, then the normal force does not pass through the COM. The torque by the normal force opposes the torque by the friction. Therefore, the net torque can be made zero.

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I know that we cannot apply torque=Moment of inertia× angular acceleration anywhere

You can. But angular acceleration does not have to be in the form of a rotation. Linear motion can also have non-zero angular momentum.

in toppling net torque about any point must be 0

That's a simplification. It's not true if the object is accelerating. In your problem the bicycle is not static, so the unbalanced torque can cause it to rotate (topple), or it can cause it to accelerate.

If you calculate the angular momentum about that point, you'll find that both a toppling bike and an accelerating bike have the same change in angular momentum.

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