
A car goes around a uniform circular track of radius $R$ at a uniform speed v once in every $T$ seconds. The magnitude of the centripetal acceleration is ac. If the car now goes uniformly around a larger circular track of radius $2R$ and experiences a centripetal acceleration of magnitude $8\,a_c$ then its time period is
A. $2T$
B. $3T$
C. $\dfrac{T}{2}$
D. $\dfrac{3T}{2}$
Answer
502.5k+ views
Hint:For solving questions on uniform circular motion, we need to know all the formulas related to angular velocity, tangential acceleration, centripetal acceleration and time period. We are already given both the value of $1^{st}$ and $2^{nd}$ in the case of centripetal acceleration, we just need to equate them.
Complete step by step answer:
We know the formula of time period (T) related to angular velocity ( ) is :
$T = \dfrac{{2\pi R}}{V}$
where $R$ is the radius of the circular path and $V$ is the velocity of the car.
Now V = $R\omega $ .Thus, T = \[\dfrac{{2\pi }}{\omega }\] .
The relation between centripetal acceleration and angular velocity is:
${a_c} = \dfrac{{{V^2}}}{R} - - - - (1)$
Thus, ${a_c} = {\omega ^2}R$ .
Now from the second condition, we see that the value of centripetal acceleration is given in terms of the $1^{st}$ case :
\[{a_c}' = \dfrac{{V{'^2}}}{{R'}} - - - - (2)\]
Now it is given that \[{a_c}' = 8{a_c}\] and \[R’ = 2R\]
Thus $1$ becomes:
\[8{a_c} = \dfrac{{V{'^2}}}{{2R}}\]
\[(8)\dfrac{{{V^2}}}{R} = \] \[\dfrac{{V{'^2}}}{{2R}}\] $ - - - (3)$
Solving equation $3$ we get :
\[V’ = 4V\]
Thus the time period will now be:
$T' = \dfrac{{2\pi R'}}{{V'}}$
$\Rightarrow T' = \dfrac{{2\pi 2R}}{{4V}}$
$\therefore T' =\dfrac{T}{2}$
Thus, we get the value of the second time period as half of the initial time period.
Hence, the correct answer is option C.
Note:To solve these questions, we have to remember the formulas mentioned above and other formulas of circular motion. Remember that there are two types of acceleration in circular motion where one is constant and the other variable. This type of problem may also be asked by giving the values of resistance and then asking the radius of the circular road so that it does not slip. The problems related to banking of roads which are related to friction and circular motion should be practised.
Complete step by step answer:
We know the formula of time period (T) related to angular velocity ( ) is :
$T = \dfrac{{2\pi R}}{V}$
where $R$ is the radius of the circular path and $V$ is the velocity of the car.
Now V = $R\omega $ .Thus, T = \[\dfrac{{2\pi }}{\omega }\] .
The relation between centripetal acceleration and angular velocity is:
${a_c} = \dfrac{{{V^2}}}{R} - - - - (1)$
Thus, ${a_c} = {\omega ^2}R$ .
Now from the second condition, we see that the value of centripetal acceleration is given in terms of the $1^{st}$ case :
\[{a_c}' = \dfrac{{V{'^2}}}{{R'}} - - - - (2)\]
Now it is given that \[{a_c}' = 8{a_c}\] and \[R’ = 2R\]
Thus $1$ becomes:
\[8{a_c} = \dfrac{{V{'^2}}}{{2R}}\]
\[(8)\dfrac{{{V^2}}}{R} = \] \[\dfrac{{V{'^2}}}{{2R}}\] $ - - - (3)$
Solving equation $3$ we get :
\[V’ = 4V\]
Thus the time period will now be:
$T' = \dfrac{{2\pi R'}}{{V'}}$
$\Rightarrow T' = \dfrac{{2\pi 2R}}{{4V}}$
$\therefore T' =\dfrac{T}{2}$
Thus, we get the value of the second time period as half of the initial time period.
Hence, the correct answer is option C.
Note:To solve these questions, we have to remember the formulas mentioned above and other formulas of circular motion. Remember that there are two types of acceleration in circular motion where one is constant and the other variable. This type of problem may also be asked by giving the values of resistance and then asking the radius of the circular road so that it does not slip. The problems related to banking of roads which are related to friction and circular motion should be practised.
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