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Question: A car goes on a horizontal circular road of radius \(R\), the speed increasing at a constant rate\(\...

A car goes on a horizontal circular road of radius RR, the speed increasing at a constant ratedvdt=a\dfrac{{dv}}{{dt}} = a. The friction coefficient between the road and the type isμ\mu . Find the speed at which car will skid.

Explanation

Solution

To solve the above question, firstly we will find net acceleration with the help of tangential acceleration which is given and centripetal acceleration, we know the relation of centripetal acceleration with velocity and radius for circular roads. We use that relation in place of centripetal acceleration.
Given:
Radius of circular road =R = R
Constant rate of increasing speed =dvdt = \dfrac{{dv}}{{dt}}
Friction coefficient =μ= \mu

Complete Step by step solution:
In circular motion, if speed is increasing then magnitude of velocity will also change and the tangential acceleration and centripetal acceleration will also produce.
Tangential acceleration a=dvdt{a_{}} = \dfrac{{dv}}{{dt}}
Centripetal acceleration ac=v2r{a_c} = \dfrac{{{v^2}}}{r}
Net acceleration =an = {a_n}
As we know that net acceleration is the sum of centripetal acceleration and tangential acceleration
an2=ac2+a2{a_n}^2 = {a_c}^2 + {a_{}}^2
⇒an=ac2+a2\Rightarrow {a_n} = \sqrt {a_c^2 + a_{}^2}……1
Putting values of ac{a_c} and at{a_t} in above equation 1.
=(v2r)2+a2= \sqrt {{{(\dfrac{{{v^2}}}{r})}^2} + {a^2}}
We are using relation of force with acceleration
F=manF = m{a_n} ……2
Putting the value of an{a_n} in equation 2.
F=m(v2r)2+a2F = m\sqrt {{{(\dfrac{{{v^2}}}{r})}^2} + {a^2}}
Friction force =μN = \mu N (here μ\mu is friction coefficient and N is normal force)
As we know that friction force will produce in motion, which will balance force produced due to tangential acceleration.
So, we can write μN=m(v2r)2+a2\mu N = m\sqrt {{{(\dfrac{{{v^2}}}{r})}^2} + {a^2}} …… 3
Here, NN is a normal force, so we can put its value mgmg
⇒μmg=m(v2r)2+a2\Rightarrow \mu mg = m\sqrt {{{(\dfrac{{{v^2}}}{r})}^2} + {a^2}}
In the above equation, mmwill be canceled out and then we will do squaring both sides.
(μg)2=(v2r)2+a2{\left( {\mu g} \right)^2} = {(\dfrac{{{v^2}}}{r})^2} + {a^2}
We need velocity, so keep velocity on left side, remaining on right side
v4r2=μ2g2−a2\dfrac{{{v^4}}}{{{r^2}}} = {\mu ^2}{g^2} - {a^2}
v4=r2(μ2g2−a2){v^4} = {r^2}({\mu ^2}{g^2} - {a^2})
Here, we got value of speed at which car will skid

v=(r2(μ2g2−a2))14v = {({r^2}({\mu ^2}{g^2} - {a^2}))^{\dfrac{1}{4}}}

Note: Points to be noted in above solution, we should keep in mind that in circular motion centripetal acceleration and tangent acceleration both applied and relationship of centripetal acceleration with velocity and radius of circular road.