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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jun 2007 20:05:30 IST
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1.) A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt = a. The friction coefficient between the road and the tyre is u. find the speed at which the car will skid.
Ans. [(u2g2 - a2)R2]1/4
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its about how hard you can get hit and keep moving forward
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5 Jun 2007 20:20:52 IST
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dv/dt = a
Now, at every instant before slipping the frictional force f equals the resultant of centripetal and tangential forces
f2 = (mv2/R)2 + (m(dv/dt))2
For no slipping,
f<  N
(mv2/R)2 + (m(dv/dt))2 <( mg)2
(v2/R)2 + a2 <2g2
v < [(2g2 - a2 )R2]1/4
Hence the speed at which the car slips
v = [(2g2 - a2 )R2]1/4
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5 Jun 2007 20:28:20 IST
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thank you very much sir,could you please tell me exactly, or give a few hints about when to apply integration. i am really bad at sums incorporating integration.
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its about how hard you can get hit and keep moving forward
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6 Jun 2007 04:23:17 IST
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Actually, the integration wasn't necessary in this case. I have removed it.
As far as application of integration goes,
Suppose you know acceleration a as function of time. Integrate it to get velocity. Integrate it to get position.
Suppose you know acceleration as a function of position. Then use the following
a = dv/dt = (dv/dx)(dx/dt) = (dv/dx)v [As dx/dt = v]
a = v(dv/dx)
And then integrate.
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