(Uniform circular motion) A car of mass m undergoes a...

(Uniform circular motion) A car of mass m undergoes a circular turn of radius R, which is banked at an angle ? with respect to the ground. Assume that there is friction between the wheels and the road, which ?? is the coefficient of static friction and g the magnitude of the gravitational acceleration. Derive an expression for the range of possible speeds v??? ? v ? v??? necessary to keep the car moving in a circle without slipping up or down the embanked turn. Express your answer in terms of some or all of the following: ??, ?, m, R and g.

Transcript

00:01 We are given here car of mass so we can write the car of mass that is small m is going to around the circular turn of the radius r so the radius we are given here which is equal to capital r which is banked at angle theta so the angle theta which is equal to b now here in this case we understand the first thing that coefficient of the static friction which is mu and and in this question we want to find let's find out the expression of the range range of the speed so the speed range is v minimum less or equal to v maximum speed limit we are given here is derive the car is moving on the circle without slipping so in this question we need to find out the velocity minimum velocity and the maximum velocity so let's find out this two velocities here and we can write here that f sine theta here in this case that n coss theta n cos beta plus we can write f s multiple by sine beta which is equal to here we can write m g now again we just simplify and we can write because we know that f s which is equal to mu multiple which is simplify and we can write that is and which is equal to n multiple by cost beta plus we can write mu and multiple by you can write sine beta which is equal to m g so this is our equation number one now here we can write n sine beta and sine beta minus we can write mu and cos beta equal to here in this question we can write that is m v minimum square divided by we can write capital r again we again simplify so here this is our equation number two this is our equation number two this is our again and this is our equation over one and this is our equation number two now we just simplify so we here we can write equation two divided by equation one so then we get that and sine beta minus we can write mu cos beta divided by we can write n cos beta plus we can write mu sine beta which is equal to we can write m v minimum square divided by we can write r multiple we can write m g and here we can write apprais simplification we can write here here we just simplify and we can write the v minimum which is equal to we can square root of r multiple by g and multiple we can write 10 beta minus we can write me divided by we can write 1 plus mu multiple by 10 beta this is the value minimum velocity now here in this case let's find out to maximum velocity so here we know that friction force already we know that now in the horizontal direction so we can write n sine beta plus we can write f s multiple by plus beta is equal to we can write here maximum velocity divided by we can write r now again we just simplify so we can write n into the vector multiple where we can write sine beta plus we can write mu cos beta which is equal to we can write here m vmex square divided by we can write capital r so this is our equation number one now again we just simplify and we can write at n cost beta and cost beta minus we can write mu and sine beta which is equal to we can write m g again we just simplify and we can write n multiple of a cost beta minus we can write mu whatever sine beta which is equal to m g which is and here this is our equation of already we have equation and equation two so here we can write equation one one prime and second here two prime because this all calculation we do for the maximum here we want to find for the finding the maximum velocity.

04:42 Now here we just simplify and we can write because equation 1 prime divided equation 2 prime which is equal to we can write that 10 beta...

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