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Chapter 6: Problem 81

ssM A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg}\). Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

Short Answer

Expert verified
(a) 275.4 N; (b) 275.4 N.

Step by step solution

01

Understanding Circular Motion

When an object moves in a circle at constant speed, a centripetal force acts towards the center, keeping it in circular motion. For a bicyclist moving in a circle, this force is the friction between the bicycle tires and the road.
02

Identify Given Values

We are given the radius of the circle as \( r = 25.0 \; \text{m} \), speed of the bicyclist \( v = 9.00 \; \text{m/s} \), and the mass of the bicycle-rider system \( m = 85.0 \; \text{kg} \).
03

Calculate Centripetal Force

The centripetal force \( F_c \) needed to keep the bicyclist moving in a circle is given by the formula \( F_c = \frac{mv^2}{r} \). Substituting the values, we get:\[ F_c = \frac{85.0 \times (9.00)^2}{25.0} \]
04

Solve for Centripetal Force

Calculate using the values:\[ F_c = \frac{85.0 \times 81.0}{25.0} = \frac{6885.0}{25.0} = 275.4 \; \text{N} \]Thus, the magnitude of the force of friction, which acts as the centripetal force, is \( 275.4 \; \text{N} \).
05

Calculate Net Force

Since the only horizontal force acting on the bike in circular motion is the frictional force (which is the centripetal force), the net force on the bicycle from the road is the same as the frictional force: \( 275.4 \; \text{N} \).

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Key Concepts

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

Centripetal Force
When a bicyclist moves in a circular path, there is an invisible force that keeps it turning around the circle rather than moving in a straight line. This is called **centripetal force**. It always points towards the center of the circle, ensuring the object stays on its curved path.To calculate the centripetal force, we use the formula:\[ F_c = \frac{mv^2}{r} \]Here,
  • \( m \) is the mass of the bicyclist (and the bicycle),

  • \( v \) is the velocity or speed,

  • \( r \) is the radius of the circle.
In our problem, the mass \( m \) is 85 kg, the speed \( v \) is 9.00 m/s, and the radius \( r \) is 25 m. Plugging these into the formula, we find that the centripetal force is 275.4 N. This is the force required to keep the bike moving in a circular path.
Friction
Friction is what prevents the bike from sliding out of its path as it travels in a circle. In this context, friction acts as the centripetal force that keeps the bicycle on its circular track. **Key Points about Friction in Circular Motion:**
  • It occurs between the tires of the bicycle and the road surface.

  • For the bicycle to turn without slipping, friction must be equal to or greater than the centripetal force required.
This means that the same 275.4 N of force calculated as the centripetal force is essentially provided by friction. Therefore, the force of friction here acts to prevent the bicycle from moving out of its circular path.
Bicycle Dynamics
When considering how bicycles move, particularly around curves, we enter the world of **bicycle dynamics**. This encompasses the forces and motions experienced by the bicycle and rider. **Understanding Bicycle Dynamics in Circular Motion:**
  • As the bicycle turns, it leans inward, helping maintain balance and maximize the frictional force.

  • The center of mass (where the bicycle and rider balance) is crucial in maintaining stability.

  • Both friction and centripetal force play significant roles. Without enough friction, the bicycle could slip, causing a potential fall.
An interesting aspect of bicycle dynamics is how the rider adjusts their position and speed to ensure that all forces are in balance. This delicate balance helps ensure smooth and safe rides even at high speeds.

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Most popular questions from this chapter

The mysterious sliding stones. Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-18). For years curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard- baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about \(0.80 .\) What horizontal force must act on a \(20 \mathrm{~kg}\) stone (a typical mass) to maintain the stone's motion once a gust has started it moving? (Story continues with Problem 37.)

A baseball player with mass \(m=79 \mathrm{~kg}\), sliding into second base, is retarded by a frictional force of magnitude \(470 \mathrm{~N}\). What is the coefficient of kinetic friction \(\mu_{k}\) between the player and the ground?

A A circular-motion addict of mass \(80 \mathrm{~kg}\) rides a Ferris wheel around in a vertical circle of radius \(10 \mathrm{~m}\) at a constant speed of \(6.1 \mathrm{~m} / \mathrm{s}\). (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

Calculate the ratio of the drag force on a jet flying at 1000 \(\mathrm{km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop- driven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km}\). Assume that the air- planes have the same effective cross-sectional area and drag coefficient \(C\).

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)
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