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#banked curve

3 public questions tagged with this topic.

A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slip

Given: A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slipping? (Take g = 10 m/s², tan 15° approx 0.268 ) These values define the system as per NCERT data. Formula: Maximum speed: v_{max = sqrtrg μ_s + tanθ/1 - μ_s tanθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Numerator: μ_s + tanθ = 0.3 + 0.268 = 0.568 . Denominator: 1 - 0.3 × 0.268 = 1 - 0.0804 = 0.9196 . v_{max² = 45 × 10 × 0.568/0.9196 approx 450 × 0.6175 approx 277.875 . v_{max = sqrt277.875 approx 16.6

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

A 1200 kg car turns on a banked road ( θ = 25°, μ_s = 0.25 ) with radius 60 m . What is the optimum speed to avoid fr

Given: A 1200 kg car turns on a banked road ( θ = 25°, μ_s = 0.25 ) with radius 60 m . What is the optimum speed to avoid friction? (Take g = 10 m/s², tan 25° approx 0.466 ) These values define the system as per NCERT data. Formula: Optimum speed occurs when banking alone provides ntripetal force: v_0 = sqrtrg tanθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: r = 60 m, g = 10 m/s², tan 25° = 0.466 . v_0² = 60 × 10 × 0.466 = 600 × 0.466 = 279.6 . v_0 = sqrt279.6 approx 16.72 m/s . Result: The computed value matches

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

A 1000kg car on a banked road (θ\=15∘, μs\=0.3) turns at radius 40m. What is the maximum speed without slipping? (Take g

Max speed vmax=rgμs+tan⁡θ1−μstan⁡θ. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 14 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0