HSC Physics

Module 5: Advanced Mechanics5.1 Projectile Motion

5.2 Circular Motion

5.3 Motion in Gravitational Fields2 Topics

Module 6: Electromagnetism6.1 Charged Particles, Conductors and Electric and Magnetic Fields

6.2 The Motor Effect1 Topic

6.3 Electromagnetic Induction

6.4 Applications of the Motor Effect1 Topic

Module 7: The Nature of Light7.1 Electromagnetic Spectrum3 Topics

7.2 Light: Wave Model

7.3 Light: Quantum Model2 Topics

7.4 Light and Special Relativity

Module 8: From the Universe to the Atom8.1 Origins of the Elements5 Topics

8.2 Structure of the Atom3 Topics

8.3 Quantum Mechanical Nature of the Atom2 Topics

8.4 Properties of the Nucleus2 Topics

8.5 Deep Inside the Atom4 Topics
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 EduKits Education
 Michael
Forces Behind Uniform Circular Motion
Cars Around Bends*
Horizontal, circular bends, that is.
 Friction between the road and tyres of a car acts towards the centre of bend curvature, so in this case it is a centripetal force.
 Remember, friction is calculated by [katex]F_f = \mu R[/katex]
 μ is the coefficient of friction, which changes by material. An ice road will have a lower value than a gravel road, for example.
 R is the reaction (or normal) force. It is equal in magnitude to the weight force (mg) but opposite in direction.
 When solving questions, resolve vectors as seen in the image.
Navigating Corners Safely
A common physics problem is to determine the maximuxm speed at which a car can safely navigate a corner. Here are some important points to consider.
 The magnitude of friction determines how strongly the car is pulled towards the centre of bend curvature.
 If the car enters the bend at too high a velocity, it will not reach circular motion but instead move out in a tangent.
 The equation [katex]F_c = \frac{mv^2}{r}[/katex] can be used to determine the centripetal force required to navigate a specific bend in a particular car. If the magnitude of friction if below this, the car will move off the circular path.
 This relationship can be expressed (and analysed) using the following inequality:
F_f\geq F_c \text{, \space \space i.e. \space \space} \mu R \geq \frac{mv^2}{r}
A Mass on a String
For a spinning mass attached to a spring, the centripetal force driving the circular motion is due to the tension in the string.
 When a mass is moving in a circle parallel to the ground, there is zero net vertical force acting on it. The mass remains in the same vertical position.
 However, there is a nonzero net horizontal force as the tension is angled towards the centre of motion while the weight force has no horizontal component to counteract this.
 Referring to the diagrams above, the forces can be broken down into horizontal and vertical components:
\begin{matrix} F_{y} = W + T\sin\theta = 0 & F_{x} = T\cos\theta \not= 0 \\ \footnotesize{\text{Vertical}} & \footnotesize{\text{Horizontal}} \end{matrix}
Objects on Banked Tracks
A banked curve has the benefit of allowing objects to maintain higher speeds when cornering without moving out on a tangent. This is because the normal reaction is tilted slightly from the vertical and can thus contribute to the centripetal force as seen below.
F_{net}=R\sin\theta+F_f\cos\theta
As with nonbanked circular motion involving friction, there is a minimum centripetal force required for any velocity to prevent a car from skidding off the curve. The centripetal force must be greater than this, seen as follows:
F_{x}\geq F_c \text{, \space \space i.e. \space \space} R\sin\theta+F_f\cos\theta \geq \frac{mv^2}{r}