A car of mass $1000 ext{ kg}$ negotiates a banked curve of radius $90 ext{ m}$ on a frictionless road. If

Question

A car of mass $1000 	ext{ kg}$ negotiates a banked curve of radius $90 	ext{ m}$ on a frictionless road. If the banking angle is $45^\circ$, the speed of the car is:

Accepted Answer

1. **Concept:** On a frictionless banked road, the horizontal component of the normal force provides the necessary centripetal force required for circular motion. The optimum speed $v_0$ (where no friction is required) depends only on the radius $R$, gravity $g$, and banking angle $	heta$ [NCERT Physics Class 11, Laws of Motion, Section 4.10, Eq 4.22].
2. **Formula:**
   $$v = \sqrt{Rg 	an	heta}$$
3. **Given Data:**
   - Radius, $R = 90 	ext{ m}$
   - Banking angle, $	heta = 45^\circ$
   - Acceleration due to gravity, $g \approx 10 	ext{ m/s}^2$ (standard approximation for integer answers).
   - Mass $m = 1000 	ext{ kg}$ (Note: Mass cancels out in the derivation and is not needed for the calculation).
4. **Calculation:**
   Substituting the values into the formula:
   $$v = \sqrt{90 	imes 10 	imes 	an(45^\circ)}$$
   Since $	an(45^\circ) = 1$:
   $$v = \sqrt{900 	imes 1}$$
   $$v = 30 	ext{ m/s}$$

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