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Merry-Go-Round. One December identical twins Jena and Jackie are playing on a large merry-go-round (a disk mounted parallel to the ground, on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has mass 30.0 \(\mathrm{kg}\) . The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting 1.80 \(\mathrm{m}\) from the center of the merry- go-round, must hold on to one of the metal posts attached to the merry- go-round with a horizontal force of 60.0 \(\mathrm{N}\) to keep from sliding off. Jackie is sitting at the edge, 3.60 \(\mathrm{m}\) from the center. (a) With what horizontal force must Jackie hold on to keep from falling off? (b) If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

Short Answer

Expert verified
(a) Jackie needs a force of 120 N to stay on.\n(b) Her velocity would be 3.78 m/s if she falls off.

Step by step solution

01

Analyze Jena's Situation for Centripetal Force

Jena is sitting 1.80 m from the center and has to exert a force of 60 N to hold herself which acts as the centripetal force required to keep her in circular motion. Given Jena's mass is 30 kg, we use the formula for centripetal force: \( F = m \times a_c \), where \( a_c \) is the centripetal acceleration. From this, we solve for \( a_c \) as \( a_c = \frac{60}{30} = 2 \, \text{m/s}^2 \).
02

Relate Jena's Centripetal Acceleration to Angular Velocity

To find the centripetal acceleration in terms of angular velocity \( \omega \), use the relation \( a_c = \omega^2 \times r \). For Jena, this becomes \( 2 = \omega^2 \times 1.80 \). Solving for \( \omega^2 \), we have \( \omega^2 = \frac{2}{1.80} \approx 1.11 \, \text{rad}^2/\text{s}^2 \).
03

Calculate Force Required for Jackie

Jackie sits at 3.60 m from the center. The centripetal force for Jackie is calculated using \( F = m \times \omega^2 \times r \). Substituting the values, \( F = 30 \times 1.11 \times 3.60 \). This results in \( F = 119.88 \, \text{N} \). Thus, Jackie must exert approximately 120 N to stay on the merry-go-round.
04

Determine Jackie's Horizontal Velocity if She Falls Off

If Jackie falls off, the linear (tangential) velocity \( v \) can be found using \( v = \omega \times r \). Using \( \omega^2 = 1.11 \) previously found, we calculate \( \omega = \sqrt{1.11} \approx 1.05 \, \text{rad/s} \).Jackie's velocity \( v = 1.05 \times 3.60 \approx 3.78 \, \text{m/s} \) would be her horizontal velocity.

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Key Concepts

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

Angular Velocity
Angular velocity is like the speed at which an object spins around a circle. Imagine holding a spinning top. How fast it spins is its angular velocity. For Jena and Jackie on the merry-go-round, angular velocity tells us how quickly they are moving in a circle.

In mathematics, angular velocity is represented by the symbol \( \omega \). It relates to how many radians an object covers in one second. Radians are a way to measure angles, much like degrees. When Jena is just 1.80 meters from the center, we find out her centripetal acceleration to link it with angular velocity. We use this formula:
  • \( a_c = \omega^2 \times r \).
This tells us that the more centrally Jena sits, the smaller \( r \), and hence angular velocity can be determined to maintain her balance without sliding off. For example, Jena has a centripetal acceleration of \( 2 \, \text{m/s}^2 \), which helps us calculate the angular velocity as about \( 1.05 \, \text{rad/s} \).

Simply put, angular velocity shows how fast an object spins, which is vital for Jena and Jackie to stay safe on their ride.
Circular Motion
Circular motion occurs when an object moves in a circle at a constant speed. On a merry-go-round, this is the kind of motion enjoyed by Jena and Jackie. Even though it seems like they are not changing speed, they are constantly changing direction, which means acceleration occurs.

This circular motion requires a special kind of force to keep objects moving along the curve of a circle, rather than flying off in a straight line. That's why Jena and Jackie have to hold on tight. The forces making them cling to the metal posts are called centripetal forces.

Key points about circular motion are:
  • It seems like a consistent speed but involves a continuous change in direction.
  • It requires centripetal force directed towards the center of the circle.
  • The further you are from the center, the greater the force needed to keep you there.
In Jackie's case, sitting farther from the center increases the circle's radius, thus requiring a stronger grip, at almost 120 N, to stay on during the ride.
Centripetal Acceleration
Centripetal acceleration is a fascinating concept that keeps things spinning around in circles, such as Jena and Jackie on their merry-go-round adventure. It refers to the change in direction of velocity for an object moving in a circular path.

Because circular motion results in continuous direction changes, objects need an inward acceleration to keep them on the track, much like a race car in a turn. This inward pull is what we term centripetal acceleration, symbolized as \( a_c \).

To find centripetal acceleration, we use the formula:
  • \( a_c = \frac{F}{m} \) where \( F \) is the force applied and \( m \) is mass.
For Jena, the centripetal acceleration is \( 2 \, \text{m/s}^2 \), calculated by dividing the force \( 60 \, \text{N} \) by her mass of \( 30 \, \text{kg} \). This acceleration gives us insight into how snugly she moves around without skidding off.

Centripetal acceleration ensures riders like Jena and Jackie stay bound within the path of their circular motion, preventing any straight-line paths that could result from a lack of this accelerating force.

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Most popular questions from this chapter

You find it takes 200 \(\mathrm{N}\) of horizontal force to move an empty pickup truck along a level road at a speed of 2.4 \(\mathrm{m} / \mathrm{s}\) . You then load the pickup and pump up its tires so that its total weight increases by 42\(\%\) while the coefficient of rolling friction decreases by 19\(\%\) . Now what horizontal force will you need to move the pickup along the same road at the same speed? The speed is low enough that you can ignore air resistance.

A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal surface. The coefficient of static friction between the box and the sur- face is \(0.40,\) and the coefficient of kinetic friction is 0.20 . (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{N},\) what is the magnitude of the friction force and what is the box's acceleration?

An \(85-N\) box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 \(\mathrm{m} / \mathrm{s}\) each second. The push force has a horizontal component of 20 \(\mathrm{N}\) and a vertical component of 25 \(\mathrm{N}\) downward. Calculate the coefficient of kinetic friction between the box and floor.

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

A \(70-\mathrm{kg}\) person rides in a \(30-\mathrm{kg}\) cart moving at 12 \(\mathrm{m} / \mathrm{s}\) at the top of a hill that is in the shape of an are of a circle with a radius of 40 \(\mathrm{m}\) . (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

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