IIT JEE , AIEEE Forum - Physics , Mechanics

kishan12

HC verma pg115
Q:22
got a part but not others.
plzzzzzzzzz answer friends.
Q:25

coool_shetty

regarding the 25th one ...
when they talk abt circular motion , it is possible only if a centripetal acceleration is present ...
so at highest point a is 'g' acting downward ,. there is a tangential velocity too which is nothing but the horizontal velocity .......thats is ....
s o
(v-horz)2 /R = g....
do rate me!!!!

as an exercise ...try this problem considerintg at any other angle too ..this will improve ur confidence .....

kishan12

Oh shit stupid me!!!
thanq shetty
plzz try the other one too

kishan12

A track consists of two circular parts ABC and CDE of equal radius 100m and joined smoothly as shown in figure.Each part subtends a right angle at its centre.A cycle weighing 100kg togather with the rider travels at a constnat speed of 18km/hr on the track.
Find the force of friction exerted by the track on the tyres when cycle is at B,C,D

See cosider diagram as horizontal S
B is the highest point and D is the lowest point.
C is the in between point.
Now AnS friction at B C D are 0,707,0N!!!!!!!!!

kishan12

Someone ans

karthik2007

for which part do u want the answers?

karthik2007

do u want me to solve the whole problem?

kishan12

just tell me the B part.i will try solving the rest

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

karthik2007

any doubts?

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.

karthik2007

Sorry for the delay yaar... I had to go out for a while :)

Ask & Discuss Questions with Community & Experts