A car of mass $m$ moves at $10$ m/s along a circular road of radius $50$ m. The angle of banking required so that the car does not slip (ignoring friction) is:

Uniform circular motion: constant speed $v$, changing direction. Angular velocity: $\omega = \Delta\theta/\Delta t = v/r$. Centripetal acceleration: $a_c = v^2/r = \omega^2 r$ (directed toward centre). Centripetal force: $F_c = mv^2/r$. Period: $T = 2\pi r/v = 2\pi/\omega$. Frequency: $f = 1/T = \omega/2\pi$.

For a particle in a vertical circle of radius $r$: At top: centripetal condition: $mg + N = mv^2/r$; minimum speed (N=0): $v_{top,min} = \sqrt{gr}$. At bottom: $N - mg = mv^2/r$; $N = mg + mv^2/r$ (larger than weight). By energy conservation from bottom to top: $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2r)$. Minimum speed at bottom to complete loop: $v_{bottom} = \sqrt{5gr}$.

Mass on string, moving in horizontal circle: $T\cos\theta = mg$, $T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta$. So $T = m\omega^2 L$ and $mg = m\omega^2 L\cos\theta$ → $\omega = \sqrt{g/L\cos\theta}$. $T = 2\pi/\omega = 2\pi\sqrt{L\cos\theta/g}$. Note: unlike simple pendulum, period depends on angle.

Torque: $\tau = rF\sin\phi = I\alpha$ (rotational analog of $F = ma$). Moment of inertia: $I = \sum m_i r_i^2$. Angular momentum: $L = I\omega$. Conservation: $L = $ const if $\tau_{net} = 0$. Rotational KE: $\frac{1}{2}I\omega^2$. Rolling without slipping: $v_{cm} = R\omega$; total KE = $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + I/mR^2)$.

On a banked curve the horizontal component of the normal reaction supplies part of the centripetal force, so a vehicle can turn faster without relying on friction. For the ideal case with no friction, $\tan\theta = \frac{v^2}{rg}$, giving the design speed for which the banking was built. With friction included the safe range widens to a maximum of $v_{max} = \sqrt{rg\frac{\mu + \tan\theta}{1 - \mu\tan\theta}}$. Note that mass cancels throughout, so the safe speed is the same for a car and a loaded truck.

For a body on a string moving in a vertical circle, tension and weight combine differently at each point. At the top, gravity alone can supply the centripetal force, so the minimum speed for the string to stay taut is $v_{top} = \sqrt{gr}$. Applying energy conservation between top and bottom then gives $v_{bottom} = \sqrt{5gr}$ as the minimum launch speed to complete the loop. A rigid rod behaves differently from a string, because it can push as well as pull, so the rod case only requires the body to reach the top with $v \geq 0$.

Centripetal force is a real inward force with a physical source: tension, friction, gravity or a normal reaction. Centrifugal force is not a real force at all but a pseudo-force that appears only when you describe the motion from inside the rotating frame, which is non-inertial. From the ground, a passenger in a turning car is pushed inward by the seat; from inside the car, the passenger appears to be flung outward. Both descriptions predict the same motion, but only the ground-frame one uses real forces — which is why free-body diagrams in the exam should be drawn in the inertial frame unless told otherwise.