Question
Question: A horizontal curve on a racing track is banked at a \({45^ \circ }\) angle. When a vehicle goes arou...
A horizontal curve on a racing track is banked at a 45∘ 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
Solution
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
FC=Rmv2______(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
mgsin45∘=FCcos45∘
⇒2mg=2FC
Multiplying 2 both the sides, we get
FC=mg
Substituting (1) in the above equation, we get
Rmv2=mg
Cancelling m from both the sides, we get
Rv2=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,
FC=maC
⇒aC=mFC
From (1)
aC=mRmv2
⇒aC=Rv2________(3)
From (2) and (3) we get the centripetal acceleration of the vehicle as
aC=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.