HSC Physics Revision Hub

1.2

Characteristics of Circular Motion & Kinematics

In uniform circular motion a particle moves at constant speed $v$ around a circle of radius $r$ , with angular speed $\omega = v/r$ . The speed is constant but the velocity continuously changes direction, so the motion is accelerated.

Centripetal vs centrifugal force

Centripetal - the real, net force directed toward the centre that keeps the particle moving in a circle. In an inertial frame it is supplied by the actual forces (tension, friction, gravity). Its magnitude is

F_c = \frac{mv^2}{r} = m\omega^2 r

Centrifugal - the outward pseudo (fictitious) force experienced by an observer in a rotating, non-inertial frame. It has the same magnitude as the centripetal force but points radially outward, balancing the observed inward tendency.

F_{cf} = \frac{mv^2}{r} = m\omega^2 r \qquad (\text{outward})

$v$ is the linear speed, $\omega$ the angular speed, $m$ the mass and $r$ the radius of the circular path.

1.3

Applications of Uniform Circular Motion

Vehicle on a horizontal circular track - maximum speed

Static friction supplies the centripetal force. The speed is maximal when friction is at its limiting value $f_s = \mu_s N = \mu_s mg$ .

\mu_s m g = \frac{m v^2}{r} \qquad\Rightarrow\qquad v_{\max} = \sqrt{\mu_s \, r \, g}

$\mu_s$ is the coefficient of static friction, $r$ the radius of the track and $g$ the acceleration due to gravity. Above this speed the vehicle skids outward.

Well (wall) of death - minimum speed

A vehicle rides on a vertical cylindrical wall. The normal reaction supplies the centripetal force while friction balances the weight. At the minimum speed the limiting friction just balances the weight:

\mu_s \, \frac{m v^2}{r} = m g \qquad\Rightarrow\qquad v_{\min} = \sqrt{\frac{r g}{\mu_s}}

Below this speed the friction can no longer hold the vehicle and it slips down the wall.

Vehicle on a banked road

For a road banked at angle $\theta$ with no friction, the horizontal component of the normal reaction provides the centripetal force. Resolving gives the safe speed

\tan\theta = \frac{v^2}{r g} \qquad\Rightarrow\qquad v = \sqrt{r g \tan\theta}

With friction of coefficient $\mu_s$ the safe speeds lie between a lower and an upper limit:

v_1 = \sqrt{\,r g\;\frac{\tan\theta - \mu_s}{1 + \mu_s \tan\theta}\,} \qquad (\text{lower limit})

v_2 = \sqrt{\,r g\;\frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta}\,} \qquad (\text{upper limit})

$\theta$ is the banking angle and $\mu_s$ the coefficient of static friction. The radicands must stay positive: $1 - \mu_s\tan\theta > 0$ and $\tan\theta \ge \mu_s$ for the lower limit. When $\mu_s = 0$ both limits reduce to the frictionless result $\sqrt{r g \tan\theta}$ .

Conical pendulum - tension and time period

A mass on a string of length $L$ revolves in a horizontal circle while the string makes angle $\theta$ with the vertical. Balancing the vertical forces gives the tension $T_0$ , and the horizontal component supplies the centripetal force, giving the period:

T_0 = \frac{mg}{\cos\theta} \qquad\qquad T = 2\pi\sqrt{\frac{L\cos\theta}{g}}

$\theta$ is the angle of the string from the vertical, $L$ the string length, $T_0$ the tension in the string and $T$ the time period.

1.4

Vertical Circular Motion (Non-Uniform)

In a vertical circle the speed is not constant - energy is exchanged between kinetic and gravitational potential energy. For a particle on a string that just completes the whole circle (the limiting critical case, where the tension drops to zero at the top), the speeds and tensions at the key points are

Limiting complete-circle values (critical case)

\text{Top }A:\quad v_A = \sqrt{rg},\quad T_A = 0

\text{Bottom }B:\quad v_B = \sqrt{5rg},\quad T_B = 6mg

\text{Horizontal midpoint }C:\quad v_C = \sqrt{3rg},\quad T_C = 3mg

1

Radial equation at the top $A$ : both the weight and (in the critical limit) the tension act toward the centre, so gravity alone supplies the centripetal force,

mg = \frac{m v_A^2}{r} \qquad\Rightarrow\qquad v_A = \sqrt{rg},\quad T_A = 0

2

Conservation of energy from $A$ to $B$ (the bottom), height change $2r$ : $\tfrac12 m v_A^2 + 2mgr = \tfrac12 m v_B^2$ , giving

v_B^2 = v_A^2 + 4rg = 5rg \qquad\Rightarrow\qquad v_B = \sqrt{5rg}

The radial equation at $B$ (tension upward, weight downward, both toward the centre) then gives the tension,

T_B - mg = \frac{m v_B^2}{r} \qquad\Rightarrow\qquad T_B = mg + 5mg = 6mg

3

Energy from $A$ to the horizontal midpoint $C$ , height change $r$ : $\tfrac12 m v_A^2 + mgr = \tfrac12 m v_C^2$ , so

v_C^2 = v_A^2 + 2rg = 3rg \qquad\Rightarrow\qquad v_C = \sqrt{3rg}

At $C$ the weight is perpendicular to the radius, so the centripetal force comes entirely from the tension:

T_C = \frac{m v_C^2}{r} = 3mg

The critical difference across the circle is $T_B - T_A = 6mg - 0 = 6mg$ . All the values above correspond to the limiting case in which the particle just manages to complete the whole circle.

1.7

Moment of Inertia & Theorems

Definition and radius of gyration

The moment of inertia of a body about an axis measures its rotational inertia - the sum of each mass element times the square of its perpendicular distance from the axis,

I = \sum m_i r_i^2 = \int r^2\, dm

The radius of gyration $K$ is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia,

I = M K^2 \qquad\Rightarrow\qquad K = \sqrt{\frac{I}{M}}

Parallel axis theorem

The moment of inertia $I_o$ about any axis equals the moment of inertia $I_c$ about a parallel axis through the centre of mass, plus $M h^2$ , where $h$ is the perpendicular distance between the two axes:

I_o = I_c + M h^2

Derivation with the centre of mass as the origin:

1

Take coordinates about the centre of mass, so the perpendicular distances of the particles are $x_i$ and the cross term vanishes because $\sum m_i x_i = 0$ .

2

Moving to a parallel axis a distance $h$ away, each distance becomes $x_i + h$ , so expanding gives the theorem:

I_o = \sum m_i (x_i + h)^2 = \sum m_i x_i^2 + 2h\sum m_i x_i + h^2\sum m_i = I_c + 0 + M h^2

Perpendicular axis theorem (planar lamina)

For a plane lamina lying in the $xy$ plane, the moment of inertia about a perpendicular $z$ -axis through a point equals the sum of the moments of inertia about the two perpendicular $x$ and $y$ axes through that point:

I_z = I_x + I_y

It follows because $r_i^2 = x_i^2 + y_i^2$ for each particle in the lamina.

1.8

Angular Momentum & Torque

Torque and rotational dynamics for a rigid body

About a fixed principal axis, the net external torque equals the moment of inertia times the angular acceleration,

\tau = I \alpha

and the angular momentum of the body is the product of its moment of inertia and angular velocity,

L = I \omega

Here $\tau$ is the net torque, $I$ the moment of inertia about the axis, $\alpha = d\omega/dt$ the angular acceleration, $L$ the angular momentum and $\omega$ the angular velocity.

Conservation of angular momentum

If the net external torque on the system is zero ( $\sum \tau_{ext} = 0$ ), then angular momentum is conserved: $L = I\omega = \text{constant}$ . A skater pulls in their arms (reducing $I$ ) so their spin $\omega$ increases; a diver or a dancer tucks to spin faster. The total angular momentum stays the same as long as no external torque acts.

1.9

Rolling Motion (Pure Rolling, No Slipping)

Total kinetic energy of a rolling body

In pure rolling the point of contact is instantaneously at rest and $v = \omega R$ . The total kinetic energy is the sum of translational and rotational parts,

E = \frac12 M v^2 + \frac12 I \omega^2 = \frac12 M v^2\left(1 + \frac{K^2}{R^2}\right)

using $\omega = v/R$ , $I = M K^2$ , where $K^2 = I/M$ and $R$ is the rolling radius.

Motion down an inclined plane (angle theta)

Rolling without slipping down an incline of angle $\theta$ , the acceleration is smaller than pure sliding because some of the energy goes into rotation,

a = \frac{g\,\sin\theta}{1 + \dfrac{K^2}{R^2}}

Speed at the bottom from a height h

Starting from rest at height $h$ , conserving energy between the top and bottom,

Mgh = \frac12 M v^2\left(1 + \frac{K^2}{R^2}\right) \qquad\Rightarrow\qquad v = \sqrt{\frac{2gh}{1 + \dfrac{K^2}{R^2}}}

$K^2 = I/M$ is the squared radius of gyration squared of the body, $R$ its rolling (outer) radius, $g$ gravity, $h$ the vertical height and $\theta$ the angle of the incline. These hold for pure rolling without slipping.