IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

You have asked for part B alone, so I have solved that part:

At B and D, frictional force will obviously be zero as it is the topmost point, and at that point, the bike does not slide.

At C:

The forces acting on the bike are:

1) Normal reaction, N
2) mg, down

Now, mg makes an angle of 45^o with the radius vector of the first semicircle. So, the component of mg along the track = mgsin45.

Now, as the biker moves with constant velocity:

f = mgsin45

or f = 1000/1.414 = 707N.

Hence the result

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