Question:
A car is moving with velocity V at the top of a semi-circular hill of radius 40 m such that the normal force on it is zero. Find the velocity (V) of the car. [use $g = 10\ ms^{-2}$]
A car is moving with velocity V at the top of a semi-circular hill of radius 40 m such that the normal force on it is zero. Find the velocity (V) of the car. [use $g = 10\ ms^{-2}$]
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For $N=0$ at the top of a hill, Critical Velocity $v = \sqrt{gR}$.
Updated On: May 12, 2026
- $15\ ms^{-1}$
- $20\ ms^{-1}$
- $30\ ms^{-1}$
- $40\ ms^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
At the top of a circular path, the forces acting are gravity ($mg$) downwards and normal force ($N$) upwards. The net force provides centripetal acceleration: $mg - N = mv^{2}/R$.
Step 2: Meaning
The condition "normal force is zero" ($N=0$) means the car is just about to lose contact with the road.
Step 3: Analysis
Setting $N=0$: $mg = mv^{2}/R \Rightarrow g = v^{2}/R \Rightarrow v = \sqrt{gR}$. Given $R = 40$ m and $g = 10\ ms^{-2}$, $v = \sqrt{10 \times 40} = \sqrt{400}$.
Step 4: Conclusion
$v = 20\ ms^{-1}$. Final Answer: (B)
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