Question

A horizontal curve on a racing track is banked at a ${45^ \circ }$ angle. When a vehicle goes around this curve at the curve’s safe speed (no friction needed to stay on the track), what is its centripetal acceleration?
A. $g$
B. $2g$
C. $0.5g$
D. None

Answer 1

Hint: To solve this question, we need to consider the free body diagram of the vehicle moving on the banked road. Then applying the condition of equilibrium of the moving vehicle, we will get the required centripetal acceleration of the vehicle.

Complete step by step answer:
Let the mass of the given vehicle be $m$ and the curve’s safe speed be $v$. If the radius of the curve s equal to $R$, then the centripetal force acting on the vehicle is given by
${F_C} = \dfrac{{m{v^2}}}{R}$______(1)
Let us consider the vehicle moving on the banked horizontal curve as represented in the figure below

Since the vehicle is moving on the horizontal curve with the curve’s safe speed, so it must be in equilibrium. So equating the force downward the slope with the force upward the slope, we get
$mg\sin {45^ \circ } = {F_C}\cos {45^ \circ }$
$ \Rightarrow \dfrac{{mg}}{{\sqrt 2 }} = \dfrac{{{F_C}}}{{\sqrt 2 }}$
Multiplying $\sqrt 2 $ both the sides, we get
${F_C} = mg$
Substituting (1) in the above equation, we get
$\dfrac{{m{v^2}}}{R} = mg$
Cancelling $m$ from both the sides, we get
$\dfrac{{{v^2}}}{R} = g$_______(2)
Now, we know from the Newton’s second law of motion that the centripetal force is equal to the mass times the centripetal acceleration, that is,
${F_C} = m{a_C}$
\[ \Rightarrow {a_C} = \dfrac{{{F_C}}}{m}\]
From (1)
${a_C} = \dfrac{{m{v^2}}}{{mR}}$
$ \Rightarrow {a_C} = \dfrac{{{v^2}}}{R}$________(3)
From (2) and (3) we get the centripetal acceleration of the vehicle as
${a_C} = g$
Thus, the centripetal acceleration of the given vehicle is equal to $g$.
Hence, the correct answer is option A.

Note: In the free body diagram of the moving vehicle, the direction of the centripetal force must be such that its component along the slope must oppose the component of weight along the slope. Then only a vehicle can move on the banked road without sliding.