Lesson 17, Topic 1
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Forces Behind Uniform Circular Motion

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Cars Around Bends*

Horizontal, circular bends, that is.

Resolution of vectors for a turning car.
  • 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 non-zero 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 non-banked 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}