IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

As for part D

We have to find the minimum frictional force.

Follow this argument carefully:

Frictional force =

N

So, f is dependant on N.

For f to be minimum, N should be minimum (Remember, dont think

is inversely proportional to N, as f is NOT a constant)

Now, let us consider the two normal reactions N at either of the 2 semicircular tracks:

At the first track (the one to the right)

We have mv²/R = mgcos@-N

which gives N = mgcos@-mv²/R

Consider the second part of the track:

mv²/R = N-mgcos@

or N = mgcos@+mv²/R

Obviously, N for the first track is smaller,

ie, N = mgcos@-mv²/R is smaller.

This has the value 1000/1.414 - 25 = 682N approx.

Now, as the cyclist moves with constant velocity:

mgsin@ = f (mgsin@ is the component of gravitational force along the track, @ being 45 degrees)

or 1000/1.414 =

N

N = 682N

So,

= 707.1/682 = 1.0369.

Therefore the minimum value of

is 1.0369 or 1.037 approx.

Nudge me if you have any doubts.

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