in the previous lectures u said if we increase velocity it will decrease friction till the point where it cant be reduced further and then the centripetal force starts to increase to adjust. But now ur saying that when velocity will decrease the centripetal force will decrease aswell???
No They both are opposite to each other but, they are not UP and DOWN, meaning that to stay in Equilbrium, the NORTH FORCE ( Rsin90-0) should be equal to the SOUTH FORCE W
Sir please reply why are we resolving the contact force because in As we studied that in inclined planes contact force is not resolved and weight is resolved instead and also that the normal contact force in 90° to the surface ao there is no point to resolve Please clarify this to me sir
That's because this time we dont want to solve parallel and perpendicular to plane, but instead solve horizontally and vertically as the circle is in the horizontal plane
It shouldn't be wrong. But you have resolved parallel and perpendicular to plane maybe that is why your forces for Fc are coming out wrong and hence the radius. The Fc is the horizontal force so we will need to resolve in the x and y plane for solving Fc (cannot do parallel and perpendicular)
Thank u so much Sir!! U give Great explanations
in the previous lectures u said if we increase velocity it will decrease friction till the point where it cant be reduced further and then the centripetal force starts to increase to adjust. But now ur saying that when velocity will decrease the centripetal force will decrease aswell???
did u get an explanation for this as im confused aswell
@@Zain65109 nope
15:26 for getting R can we use R=Wcos10
No
They both are opposite to each other but, they are not UP and DOWN, meaning that to stay in Equilbrium, the NORTH FORCE ( Rsin90-0) should be equal to the SOUTH FORCE W
@@nehalhassan8736 thanks
Sir please reply why are we resolving the contact force because in As we studied that in inclined planes contact force is not resolved and weight is resolved instead and also that the normal contact force in 90° to the surface ao there is no point to resolve
Please clarify this to me sir
That's because this time we dont want to solve parallel and perpendicular to plane, but instead solve horizontally and vertically as the circle is in the horizontal plane
I think in case of incline normal force participate in pushing car towards center
Sir are all these in the category of gravitation
I didn't get this
there are overlap questions
sir for the question at 11:00, i did R=mgcosx. but got a different answer, radius=73.6m. is my way wrong?
It shouldn't be wrong. But you have resolved parallel and perpendicular to plane maybe that is why your forces for Fc are coming out wrong and hence the radius. The Fc is the horizontal force so we will need to resolve in the x and y plane for solving Fc (cannot do parallel and perpendicular)
how is RcosQ parallel to W in the inclined plane @ 20:07