/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A cat dozes on a stationary merr... [FREE SOLUTION] | 91Ó°ÊÓ

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A cat dozes on a stationary merry-go-round, at a radius of 6.0 m from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.0 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?

Short Answer

Expert verified
The least coefficient of static friction is approximately 0.67.

Step by step solution

01

Identify Given Values

First, let's identify the given values from the problem. The radius of the merry-go-round is 6.0 m, and the time for one complete rotation is 6.0 seconds.
02

Understand the Concept of Centripetal Force

The cat will experience a centripetal force that keeps it moving in a circular path. This force must be equal to the frictional force to prevent the cat from sliding off.
03

Calculate Angular Velocity

Angular velocity (\( \omega \)) is given by the formula:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period of one rotation, 6.0 s.\[ \omega = \frac{2\pi}{6.0} \approx 1.047 \text{ rad/s} \]
04

Determine the Centripetal Acceleration

Centripetal acceleration (\( a_c \)) is calculated using:\[ a_c = r \omega^2 \]Substituting the given radius (6.0 m) and calculated \( \omega \approx 1.047 \text{ rad/s} \):\[ a_c = 6.0 \times (1.047)^2 \approx 6.58 \text{ m/s}^2 \]
05

Calculate the Required Coefficient of Static Friction

The frictional force is \( f_s = \mu_s \cdot m \cdot g \) (where \( \mu_s \) is the coefficient of static friction, \( m \) is the mass of the cat, and \( g \approx 9.81 \text{ m/s}^2 \)). This force must at least equal the centripetal force needed:\[ m \cdot a_c = \mu_s \cdot m \cdot g \]Cancelling mass from both sides gives:\[ \mu_s = \frac{a_c}{g} \]Substitute \( a_c \approx 6.58 \text{ m/s}^2 \) and \( g \approx 9.81 \text{ m/s}^2 \):\[ \mu_s = \frac{6.58}{9.81} \approx 0.67 \]
06

Conclusion

The least coefficient of static friction required to keep the cat from sliding off the merry-go-round is approximately 0.67.

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Key Concepts

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

Centripetal Force
Centripetal force is crucial for maintaining any object in a circular path. Imagine a cat lounging on a moving merry-go-round. For the cat to stay put, it must deal with a force that constantly draws it toward the center of the roundabout. This force is known as the centripetal force. Without this, the cat would likely skid off, akin to how a car might slide off a curve if not enough traction is applied. In our given exercise involving the merry-go-round, the centripetal force is provided by the friction between the cat and the surface. It's this force that keeps the cat moving in a circle rather than flying off in a straight line. Here, it's vital to match the centripetal force with the frictional force. Otherwise, the cat won't just sit still. Essentially, the centripetal force ensures the cat's motion remains tethered to the rotating system. When figuring out how much force is needed, one must consider the speed and radius of the circle, leading us into our next topic: angular velocity.
Angular Velocity
Angular velocity is about how fast something spins or rotates. It measures how swiftly an object travels along a circular path. To calculate angular velocity, use the formula \[ \omega = \frac{2\pi}{T} \] where \( \omega \) represents angular velocity in radians per second, and \( T \) is the period it takes for one full cycle, measured in seconds. In the merry-go-round scenario, if the ride completes one full rotation every 6 seconds, you would calculate angular velocity as \[ \omega = \frac{2\pi}{6.0} \approx 1.047 \text{ rad/s} \]. This value tells us how the position of the cat changes over time during the motion. Since the rotation speed directly affects the centripetal force, understanding angular velocity is key. It forms the foundation for calculating the centripetal acceleration, which we'll explore next.
Centripetal Acceleration
Centripetal acceleration describes how rapidly an object changes direction as it moves in a circular path. While linear acceleration talks about change in speed, centripetal acceleration is all about changing direction constantly. To calculate, use the formula: \[ a_c = r \omega^2 \] where \( a_c \) is the centripetal acceleration, \( r \) is the radius of the circle, and \( \omega \) is the angular velocity. In our merry-go-round example, where \( r = 6.0 \text{ m} \) and \( \omega = 1.047 \text{ rad/s} \), \[ a_c = 6.0 \times (1.047)^2 \approx 6.58 \text{ m/s}^2 \]. This calculation gives the acceleration "pushing" the cat towards the center, ensuring it stays on track. Understanding centripetal acceleration is vital because it ties into determining the friction needed to prevent slipping. It's what allows the cat to enjoy the ride without taking an unintended slide off the merry-go-round! Each of these concepts interlinks, helping us figure out how the forces at play keep the ride's furry passenger safe and snug.

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Most popular questions from this chapter

Suppose the coefficient of static friction between the road and the tires on a car is 0.60 and the car has no negative lift.What speed will put the car on the verge of sliding as it rounds a level curve of 32.0 m radius?

A worker pushes horizontally on a 35 kg crate with a force of magnitude 110 N.The coefficient of static friction between the crate and the floor is 0.37. (a) What is the value of fs,max under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker’s 110 N push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

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. 6-16). 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 kg stone (a typical mass) to maintain the stone’s motion once a gust has started it moving? (Story continues with Problem 37.)

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.65, what minimum force magnitude is required from the rope to start the crate moving? (b) If mk 0.35, what is the magnitude of the initial acceleration of the crate?

A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 12 m at a constant speed of 5.5 m/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?

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