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Chapter 6: Problem 22

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is \(\mu_{s}=0.800\) , how fast can you drive on a horizontal roadway around a right turn of radius 30.0 \(\mathrm{m}\) before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?

Short Answer

Expert verified
The maximum speed around the turn before the cup starts to slide is calculated as \( \sqrt{0.800 \times 9.8 \times 30.0} \) m/s. The cup will slide outward, away from the center of the turn, relative to the dashboard if the car goes too fast.

Step by step solution

01

Understand the Forces in Circular Motion

A body moving in a circle at constant speed is in a state of uniform circular motion, and is subject to an inward force called the centripetal force. The static friction between the coffee cup and the dashboard provides this centripetal force. If this static frictional force is exceeded, the object will begin to slide. The maximum frictional force can be calculated using the equation: \( F_{\text{max}} = \mu_s \times N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force, which equals the gravitational force on the cup in this case.
02

Apply Newton's Second Law for Circular Motion

For an object moving in a circle at a constant speed, Newton's second law tells us that the centripetal force \( F_c \) required to keep the object on its path is \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circular path. Since static friction provides the centripetal force up to its maximum value, \( F_{\text{max}} = F_c \).
03

Calculate the Maximum Velocity

Using the equality \( F_{\text{max}} = F_c \), and substituting the expression for centripetal force, we have \( \mu_s \times mg = \frac{mv^2}{r} \). We can cancel out the mass \( m \) from both sides, resulting in \( \mu_s \times g = \frac{v^2}{r} \). Solving for \( v \), we get \( v = \sqrt{\mu_s \times g \times r} \). Substituting the given values \( \mu_s = 0.800 \), \( g = 9.8 \text{ m/s}^2 \), and \( r = 30.0 \text{ m} \), we can calculate \( v \).
04

Solve for the Maximum Velocity

Plugging in the given values, the maximum velocity \( v \) can be calculated using the equation \( v = \sqrt{0.800 \times 9.8 \text{ m/s}^2 \times 30.0 \text{ m}} \). After computing the right-hand side, we find the value of the maximum velocity before the cup starts to slide.
05

Direction of Sliding

If the car goes too fast, the static friction will not be sufficient to provide the necessary centripetal force, and the cup will slide due to inertia. According to Newton's first law of motion, the cup will continue moving in the direction it was moving before it began to slide, which tangentially to the curve of the turn, outward relative to the curve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Imagine spinning a ball on a string; you can actually feel your hand pulling the ball inward. In circular motion physics, this inward force is known as the centripetal force. It's crucial for objects to stay on a circular path and must continuously act perpendicular to the motion.

When you're turning a corner in your car, centripetal force is needed to keep you on the curve. Without it, you'd go straight off the path due to inertia 鈥� an object in motion tends to stay in its motion's path unless acted upon by another force, according to Newton's first law. In your coffee cup scenario, the maximum static frictional force between the cup and dashboard is what's providing this necessary centripetal force.
Static Friction
Static friction is the resistance to motion that occurs when two objects are not moving relative to each other. It's what keeps that coffee cup from sliding off your dashboard as you start to turn. Static friction is fascinating because it adjusts to match the applied force, up to a maximum value.

This maximum value, Fmax, can be calculated as the product of the coefficient of static friction, 渭s, and the normal force, N. In the case of your cup on the dashboard, the normal force equals the gravitational pull on the cup, and the static frictional force equals the centripetal force needed to keep the cup making the turn with the car.
Uniform Circular Motion
The elegance of uniform circular motion lies in its constancy. An object that travels in a circular path at a constant speed, like the moon around Earth, is akin to a marathon runner maintaining a steady pace. It doesn't speed up or slow down, the speed is uniform, but the velocity is not - because velocity is a vector, and even if the magnitude (the speed) is the same, the direction changes at every instant.

In the context of your coffee cup, if it moves with the car at a constant speed and follows a path that is a segment of a circle, it's undergoing uniform circular motion, and thus constantly changing direction. From a physics standpoint, this means there must be a net force acting on it, directed toward the center of the circle.
Newton's Second Law for Circular Motion
When an object moves in circular motion, it's Newton's second law for circular motion that can explain the behavior. The law is often stated as Fc = ma, but for circular motion, acceleration a is replaced by the centripetal acceleration which is caused by the change in direction of the velocity. This centripetal acceleration (v2/r) leads to the equation Fc = mv2/r, where Fc is the centripetal force, m is mass, v is velocity, and r is the radius of the circle.

The maximum velocity before your coffee cup starts to slide can be calculated with the help of Newton's second law for circular motion. This law is at the core of solving the given problem by equating the maximum static friction force with the required centripetal force for the motion.

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Most popular questions from this chapter

Tarzan \((m=85.0 \mathrm{kg})\) tries to cross a river by swinging from a vine. The vine is 10.0 \(\mathrm{m}\) long, and his speed at the bottom of the swing (as he just clears the water) will be 8.00 \(\mathrm{m} / \mathrm{s}\) . Tarzan doesn't know that the vine has a breaking strength of 1000 \(\mathrm{N}\) . Does he make it safely across the river?

A hawk flies in a horizontal arc of radius 12.0 \(\mathrm{m}\) at a con- stant speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc but increases its speed at the rate of 1.20 \(\mathrm{m} / \mathrm{s}^{2}\) . Find the acceleration (magnitude and direction) under these conditions.

An object of mass 5.00 kg, attached to a spring scale, rests on a frictionless, horizontal surface as in Figure P6.21. The spring scale, attached to the front end of a boxcar, has a constant reading of 18.0 N when the car is in motion. (a) If the spring scale reads zero when the car is at rest, determine the acceleration of the car. (b) What constant reading will the spring scale show if the car moves with constant velocity? (c) Describe the forces on the object as observed by someone in the car and by someone at rest outside the car.

One popular design of a household juice machine is a conical, perforated stainless steel basket 3.30 cm high with a closed bottom of diameter 8.00 cm and open top of diameter 13.70 cm that spins at 20 000 revolutions per minute about a vertical axis (Figure P6.30). Solid pieces of fruit are chopped into granules by cutters at the bottom of the spinning cone. Then the fruit granules rapidly make their way to the sloping surface where the juice is extracted to the outside of the cone through the mesh perforations. The dry pulp spirals upward along the slope to be ejected from the top of the cone. The juice is collected in an enclosure immediately surrounding the sloped surface of the cone. (a) What centripetal acceleration does a bit of fruit experience when it is spinning with the basket at a point midway between the top and bottom? Express the answer as a multiple of g. (b) Observe that the weight of the fruit is a negligible force. What is the normal force on 2.00 g of fruit at that point? (c) If the effective coefficient of kinetic friction between the fruit and the cone is 0.600, with what acceleration relative to the cone will the bit of fruit start to slide up the wall of the cone at that point, after being temporarily stuck?

A hailstone of mass \(4.80 \times 10^{-4} \mathrm{kg}\) falls through the air and experiences a net force given by $$ F=-m g+C v^{2} $$ where \(C=2.50 \times 10^{-5} \mathrm{kg} / \mathrm{m} .\) (a) Calculate the terminal speed of the hailstone. (b) Use Euler's method of numerical analysis to find the speed and position of the hailstone at \(0.2-\mathrm{s}\) intervals, taking the initial speed to be zero. Continue the calculation until the hailstone reaches 99\(\%\) of terminal speed.
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