# Explanation: motorcycle leaning without centrifugal force?

bahtiyar
Hi the question is about bicycle and motorcycle dynamics
How can we explain the motorcycle leaning in turn without using the centrifugal force term?
You know the correct term for rotational motion is the centripetal force and it is toward the centre. I have a problem with rotational equilibrium it is not provided. Here the free body diagram

#### Attachments

• Untitled.png
60 KB · Views: 466

How can we explain the motorcycle leaning in turn without using the centrifugal force term?
So which frame of reference did you choose to avoid having a centrifugal force?

I have a problem with rotational equilibrium it is not provided.
Is the motorcycle in equilibrium in the frame you are using?

bahtiyar
The net force on the bike equal to centripetal force so there is no translational equilibrium for any reference frame. But there must be a rotational equilibrium for every reference frame because the bike doesn't tip over. I think it is not about the reference frame. When I use the friction force as a centripetal force and calculate the torque it is not zero.

If it's in rotational equilibrium net torque must be zero regardless the axis point that you choose.
Is that also true for non-static or non-inertial reference points?

Homework Helper
Using the attached image, and considering only the forces that act on the bike ... The downwards force of gravity acting through the center of mass could be assumed to generate no torque. The upwards normal force exerted by the pavement to the right of the center of mass results in a counter-clockwise torque. The inwards (leftwards centripetal) force exerted by the pavement below the center of mass results in a clockwise torque. In this case the sum of the torques is zero, and the bike's lean angle is not changing.

Then there are the other part of the Newton third law pairs. The upwards force gravitational force exerted on the earth. The downwards force exerted onto the pavement at the contact patch of the tire. The outwards force exerted onto the pavement at the contact patch of the tire.

bahtiyar
sorry guys I've been busy these days I couldn't look the topic. Thank you all but we are calculating the torque from the center of mass and we found it as zero. For the rotational equilibrium, net external torque must be zero regardless the axis that you choose. So two possible explanation;
1. bike not in rotational equilibrium and there is something that we didn't count
2. bike in rotational equilibrium because the gravitational force that we assume act from the center of mass not accepted as an external force and it doesn't produce external torque

I'm stuck on this question I can't find a solution

I'm stuck on this question I can't find a solution
Start by defining a reference frame.

bahtiyar
let's analyse for both reference frame
(I'm not good about the reference frame)

I'm not good about the reference frame
Then start by learning about them. It's the reference frame that decides whether you have a centrifugal force or not, and whether you can expect equilibrium or not.

bahtiyar
I'm a physics teacher and we are not teaching the centrifugal force. So we are always using inertial ref. frame. For the inertial ref. frame it seems in rotational equilibrium.

For the inertial ref. frame it seems in rotational equilibrium.
What is your reference point for rotational equilibrium? Is that point static in the inertial frame?

Homework Helper
Gold Member
2022 Award
I'm a physics teacher and we are not teaching the centrifugal force. So we are always using inertial ref. frame. For the inertial ref. frame it seems in rotational equilibrium.

If you don't want to teach the centrifugal force, then you'll have to teach angular momentum, in all its vectorial glory!

Homework Helper
let's analyze for both reference frames
Post #6 covers the case of considering torque about the center of mass, which avoids the issue that the center of mass and the frame of reference are accelerating (the related centrifugal force produces no torque about the center of mass). If considering torque about the point of contact, then the centripetal acceleration of the center of mass can no longer be ignored, and you're stuck having to deal with centrifugal force because the frame of reference is accelerating. This may be easier to conceptualize by considering the road to be accelerating horizontally (to the left using the first post's attached image) instead of the bike turning, producing the same net zero torque.

• bahtiyar
bahtiyar
If you don't want to teach the centrifugal force, then you'll have to teach angular momentum, in all its vectorial glory!
We are teaching it without deep down, we treat it as a scalar.
Are the rotational motion of bike around the curve and angular momentum of bike cause the conflict. How we can explain the case with exact scientific consistency. I'm looking an explanation for me not for my students. I don't have advanced students.

bahtiyar
Post #6 This may be easier to conceptualize by considering the road to be accelerating horizontally (to the left using the first post's attached image) instead of the bike turning, producing the same net zero torque.
we consider this friction force turn to other direction thus we add it as a centrifugal force.

• 