A car of mass m is moving on a level circular track of radius R. If μ represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by ?
The maximum speed of the car in the circular motion .
Given that , the car of mass m , is moving on a circular track of radius R . We know that , the centripetal force provides the force for the motion . The centripetal force is given by ,
Also , the centripetal force here is provided by the frictional force. The frictional force is given by ,
The maximum speed of the car in the circular motion .
\begin{gathered}\red{\frak{Given}}\begin{cases} \textsf{ A car of mass m is moving on a level circular track of < /p > < p > radius R.} \\\\\sf \mu \ \textsf{ represents the static friction between the road and tyres of the car. }\end{cases} \end{gathered}
Given
⎩
⎪
⎪
⎨
⎪
⎪
⎧
A car of mass m is moving on a level circular track of < /p > < p > radius R.
μ represents the static friction between the road and tyres of the car.
Given that , the car of mass m , is moving on a circular track of radius R . We know that , the centripetal force provides the force for the motion . The centripetal force is given by ,
\longrightarrow\small \sf \red{ Force_{(Centripetal)}= \dfrac{ mass * ( velocity)^2}{Radius }}⟶Force
(Centripetal)
=
Radius
mass∗(velocity)
2
Also , the centripetal force here is provided by the frictional force. The frictional force is given by ,
\longrightarrow\small \sf Force_{(Friction)}= \mu * Normal \ force⟶Force
Answers & Comments
Need to FinD :-
Given that , the car of mass m , is moving on a circular track of radius R . We know that , the centripetal force provides the force for the motion . The centripetal force is given by ,
Also , the centripetal force here is provided by the frictional force. The frictional force is given by ,
And N = mg , therefore ,
Therefore ,
Answer:
Need to FinD :-
The maximum speed of the car in the circular motion .
\begin{gathered}\red{\frak{Given}}\begin{cases} \textsf{ A car of mass m is moving on a level circular track of < /p > < p > radius R.} \\\\\sf \mu \ \textsf{ represents the static friction between the road and tyres of the car. }\end{cases} \end{gathered}
Given
⎩
⎪
⎪
⎨
⎪
⎪
⎧
A car of mass m is moving on a level circular track of < /p > < p > radius R.
μ represents the static friction between the road and tyres of the car.
Given that , the car of mass m , is moving on a circular track of radius R . We know that , the centripetal force provides the force for the motion . The centripetal force is given by ,
\longrightarrow\small \sf \red{ Force_{(Centripetal)}= \dfrac{ mass * ( velocity)^2}{Radius }}⟶Force
(Centripetal)
=
Radius
mass∗(velocity)
2
Also , the centripetal force here is provided by the frictional force. The frictional force is given by ,
\longrightarrow\small \sf Force_{(Friction)}= \mu * Normal \ force⟶Force
(Friction)
=μ∗Normal force
And N = mg , therefore ,
\longrightarrow\small \sf \red{ Force_{(Friction)}= \mu \ mg}⟶Force
(Friction)
=μ mg
Therefore ,
\begin{gathered}\longrightarrow\small \sf Force_{(Friction)}= Force_{( Centripetal)} \\\\\\\sf\longrightarrow \small \mu mg = mv^2/r \\\\\\\sf\longrightarrow \small v^2 = \dfrac{ \mu mg R }{m } \\\\\\\sf\longrightarrow \small v^2 = \mu g R \\\\\\\sf\longrightarrow \small \underline{\underline{\red{\sf Velocity_{(max)}= \sqrt{ \mu R g} }}} \end{gathered}
⟶Force
(Friction)
=Force
(Centripetal)
⟶μmg=mv
2
/r
⟶v
2
=
m
μmgR
⟶v
2
=μgR
⟶
Velocity
(max)
=
μRg
\rule{200}2