A car of mass 'm' moves on a banked road having radius 'r' and banking angle θ$\theta$ To avoid slipping from banked road, the maximum permissible speed of the car is v₀.$v_{0}.$ The coefficient of friction μ$\mu$ between the wheels of the car and the banked road is :-

Related Formula

Centripetal force configuration for simplified horizontal rotation layout:

T = mω² l$T = m\omega^2 l$

Core Logic

Given values: m = 0.5~kg$m = 0.5\mathrm{~kg}$, l = 50~cm = 0.5~m$l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, Tmax = 400~N$T_{\text{max}} = 400\mathrm{~N}$.

Equating max tension to centripetal requirement:

Step 1: Compute Angular Velocity

Pattern Recognition

Ensure units are metric (50~cm → 0.5~m$50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics.

Chapter Mix

Class 11 Physics: Laws of Motion

Related Formula

At the lowest point of a vertical circle, the equation of motion for a mass m$m$ is:

Rearranging to solve for tension T$T$:

Core Logic

Given data:

• Mass, m = 900 g = 0.9 kg$m = 900\text{ g} = 0.9\text{ kg}$
• Radius, r = 1 m$r = 1\text{ m}$
• Frequency, N = 10 rpm = (10)/(60) rps = (1)/(6) rps$N = 10\text{ rpm} = \frac{10}{60}\text{ rps} = \frac{1}{6}\text{ rps}$
• Angular velocity, ω = 2π N = 2π ((1)/(6)) = (π)/(3) rad/s$\omega = 2\pi N = 2\pi \left(\frac{1}{6}\right) = \frac{\pi}{3}\text{ rad/s}$

Step 1: Calculate Force Values

Now we substitute our parameters into the tension equation:

Since π² = 9.8$\pi^2 = 9.8$:

Pattern Recognition

Always convert Mass to kg$\text{kg}$ and rotational speed to rad/s$\text{rad/s}$ first. Using the prompt constraint π² = 9.8$\pi^2 = 9.8$ yields a perfect decimal addition match.

Chapter Mix

Class 11 Physics: Laws of Motion

Related Formula

For a particle in circular motion:

• Normal (centripetal) acceleration: ac = (v²)/(r)$a_c = \frac{v^2}{r}$
• Tangential acceleration: aₜ = (dv)/(dt)$a_t = \frac{dv}{dt}$

Core Logic

Given ac = aₜ$a_c = a_t$:

Step 1: Relate Velocity to Position and Integrate

Substitute the parameters v₀ = 4 m/s$v_0 = 4\text{ m/s}$ and r = 50 cm = 0.5 m$r = 50\text{ cm} = 0.5\text{ m}$:

Integrating this to find the position s(t)$s(t)$:

Step 2: Solve for Time of First Revolution

To complete the first revolution, the distance covered is:

Equating the distance:

Comparing this to (1)/(α)[ 1 - e-2π ] s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, we get:

α = 8$\alpha = 8$

Pattern Recognition

The condition aₜ = ac v (dv)/(ds) = (v²)/(r) (dv)/(v) = (ds)/(r)$a_t = a_c \implies v \frac{dv}{ds} = \frac{v^2}{r} \implies \frac{dv}{v} = \frac{ds}{r}$. Integrating directly gives v = v₀ es/r$v = v_0 e^{s/r}$. Substituting this back into v = ds/dt$v = ds/dt$ makes the final time integral much more intuitive.

Chapter Mix

Class 11 Physics: Laws of Motion

Related Formula

For small angles, θ ≈ θ = (h)/(d)$\tan\theta \approx \sin\theta = \frac{h}{d}$, where h$h$ is the raised height and d$d$ is the separation between the tracks.

Core Logic

Equating the two relationships:

Step 1: Compute height value

Converting to centimeters:

Pattern Recognition

Whenever θ$\theta$ is small, geometry permits approximating θ$\tan\theta$ with hwidth$\frac{h}{\text{width}}$, vastly reducing computational transcendental overhead.

Chapter Mix

Class 11 Physics: Laws of Motion

Related Formula

Core Logic

Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.

Initial mass m₁ = 1000 kg$m_1 = 1000\text{ kg}$, initial velocity v₁ = 6 m/s$v_1 = 6\text{ m/s}$. Final mass m₂ = 1000 + 200 = 1200 kg$m_2 = 1000 + 200 = 1200\text{ kg}$.

Step 1: Balance conservation equation

Pattern Recognition

Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles.

Chapter Mix

Class 11 Physics: Laws of Motion