prof had been explaining that longitudinal force will be generated only when there is slip ( difference between peripheral velocity and ground velocity). In this condition, at 35.55, how can he draw longitudinal force variation graph for free rolling condition (where peripheral velocity and ground velocity are same)??
correct me if i am wrong. You mentioned that Rolling resistance or force is in the direction of omega i.e velocity. Then how come the velocity reduces at the bump if both are in the same direction? Should the Rolling resistance direction needed to be reversed there? or where i am going in this concept?
When he says rolling resistance, he meant reaction force. Generally we represent only the reaction forces in FBD. I think he explained it in an earlier video as a part of doubt clarification. But again, correct me if I'm wrong.
I think that the reduction of the velocity doesn't depend on the acting of the rolling resistance force. When the professor explained the variation of the radius of the tyre, on the final part you have R2< Re so the velocity of the tyre on that point lesser than the velocity of the veichle (or if you change point of view is lesser than the velocity of the ground), hence the tyre will have the tendency to be compressed. Now if you combine this tendency to get compressed to the action of the rolling resistance force you will get an "extra compression", hence there is the bump, after the bump the radius goes back to Ro. This is my interpretation of the answer, i'm not 100% sure about it, it's been a year since you asked the question so if you found your answer it would be great to hear it in order for me to understand better!
I know for sure it is formidable to delve into all these details in a lecture. To me, it looks like thinking loudly without having the tiniest fear of people's [mis]judgement.
i want to ask one question... friction does not depend on contact area, but the bicycle having less air( contact area increses) in tire required more force(more friction) to drive ... how ?
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Best explanation of this subject matter I have EVER heard.. by a WIDE margin! 👏🏾👏🏾👏🏾
Wonderfully explained the concept of vertical loading of tire.
these are excellent lectures
prof had been explaining that longitudinal force will be generated only when there is slip ( difference between peripheral velocity and ground velocity). In this condition, at 35.55, how can he draw longitudinal force variation graph for free rolling condition (where peripheral velocity and ground velocity are same)??
correct me if i am wrong. You mentioned that Rolling resistance or force is in the direction of omega i.e velocity. Then how come the velocity reduces at the bump if both are in the same direction? Should the Rolling resistance direction needed to be reversed there? or where i am going in this concept?
When he says rolling resistance, he meant reaction force. Generally we represent only the reaction forces in FBD. I think he explained it in an earlier video as a part of doubt clarification. But again, correct me if I'm wrong.
I think that the reduction of the velocity doesn't depend on the acting of the rolling resistance force. When the professor explained the variation of the radius of the tyre, on the final part you have R2< Re so the velocity of the tyre on that point lesser than the velocity of the veichle (or if you change point of view is lesser than the velocity of the ground), hence the tyre will have the tendency to be compressed. Now if you combine this tendency to get compressed to the action of the rolling resistance force you will get an "extra compression", hence there is the bump, after the bump the radius goes back to Ro. This is my interpretation of the answer, i'm not 100% sure about it, it's been a year since you asked the question so if you found your answer it would be great to hear it in order for me to understand better!
I know for sure it is formidable to delve into all these details in a lecture. To me, it looks like thinking loudly without having the tiniest fear of people's [mis]judgement.
If the tire pressure doesn't vary more then 1% then how does the tire pressure reduce, say after a month or two.???
please refer Good book for studying Tyre mechanics. We are replacing spokes of a bicycle with springs. Springs will act as shock absorber.
i want to ask one question... friction does not depend on contact area, but the bicycle having less air( contact area increses) in tire required more force(more friction) to drive ... how ?
Vertical deformation increases, this increases more hysteresis loss, that results into more effort required.
thank u sir