1 Answer | Add Yours
The rider wants to ride the bike on the wall of a vertical tunnel. To stay on the wall, a minimum requirement is that the frictional force between the wheels of the bike and the wall should be equal to the gravitational force of attraction of the bike in the downward direction.
Assume the radius of the tunnel is r. If a rider moves at v m/s. the centrifugal force on the bike that acts in an outward direction from the center of the tunnel is equal to `250*v^2/r` . This is the normal force between the bike and the rider. As the coefficient of friction is 0.8, the resulting frictional force is `250*v^2/r*0.8` . This should be equal to the force in the downward direction.
`200*v^2/r = 250*9.8`
=> `v = sqrt((49*r)/4)`
=> `v = 3.5*sqrt r`
The minimum speed at which the rider has to ride the bike to not fall off the wall is `3.5*sqrt r` where r is the radius of the tunnel.
We’ve answered 318,915 questions. We can answer yours, too.Ask a question