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Chapter 6: Problem 65

(a) A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is \(1.25 \mathrm{m}\) from its center? (b) What centripetal force is exerted if the merry-go-round rotates at 3.00 rev/min and he is 8.00 m from its center? (c) Compare each force with his weight.

Short Answer

Expert verified
For part (a), the centripetal force is \(F_{c_a} = 772.57 \thinspace\mathrm{N}\) and its ratio to the child's weight is \(\frac{F_{c_a}}{W} = 3.54\). For part (b), the centripetal force is \(F_{c_b} = 24.48 \thinspace\mathrm{N}\) and its ratio to the child's weight is \(\frac{F_{c_b}}{W} = 0.112\).

Step by step solution

01

Convert speed from rev/min to rad/s

In part (a), the speed is given in rev/min (40.0 rev/min). We will convert it to rad/s by using the conversion factor: 1 rev = \(2\pi\) rad So, \[\omega_a = 40.0 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{\text{rev}} \times \frac{1 \text{ min}}{60 \text{s}}\] For part (b), we have to do the same but with 3.00 rev/min: \[\omega_b = 3.00 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{\text{rev}} \times \frac{1 \text{ min}}{60 \text{s}}\]
02

Calculate centripetal acceleration

Now, we will calculate the centripetal acceleration using the formula: \[a_c = r\omega^2\] where \(r\) is the distance from the center, and \(\omega\) is the angular velocity in rad/s. For part (a): \[a_{c_a} = 1.25 \mathrm{m} \times (\omega_a)^2\] For part (b): \[a_{c_b} = 8.00 \mathrm{m} \times (\omega_b)^2\]
03

Calculate centripetal force

Next, we will calculate the centripetal force using the centripetal acceleration and the child's mass: For part (a): \[F_{c_a} = m \times a_{c_a}\] For part (b): \[F_{c_b} = m \times a_{c_b}\]
04

Calculate the child's weight

We'll now calculate the child's weight using the formula: \[W = m \times g\] where \(W\) is the weight, \(m\) is the mass, and \(g\) is the gravitational acceleration (approximately 9.81 m/s虏).
05

Compare centripetal force and weight

Finally, we will compare each centripetal force with the child's weight by calculating their ratio: For part (a): \[\frac{F_{c_a}}{W}\] For part (b): \[\frac{F_{c_b}}{W}\]

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Key Concepts

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

Angular Velocity
Imagine a child spinning happily on a playground merry-go-round. The speed at which the merry-go-round rotates is crucial in understanding its movement and is defined by the concept of angular velocity. Angular velocity, represented by the Greek letter \( \omega \), describes the rate of rotation, specifying how many radians the object turns through per second.

It is different from linear velocity, which describes how fast the object travels in a straight path. For rotation, the angular velocity is given in radians per second (rad/s). This conversion is important because it allows us to calculate the circular motion of the child using a common unit 鈥� especially when dealing with physics equations that require consistency in measurement units.
Centripetal Acceleration
As the merry-go-round spins, something interesting happens 鈥� even though the child is moving at a constant speed, there's still an acceleration acting on them. This non-linear acceleration is what we call centripetal acceleration. It always points towards the center of the circle.

The formula for centripetal acceleration is \( a_c = r\omega^2 \), where \( r \) is the radius of the circular path and \( \omega \) is the angular velocity. This acceleration is not due to an increase in speed but due to the direction of velocity changing as the child moves in a circle. Centripetal acceleration is crucial for keeping the child moving along the curved path of the merry-go-round 鈥� it's the reason they don't fly off in a straight line.
Rotational Motion
Rotational motion is what occurs when an object spins or rotates about an axis. In the case of our merry-go-round and the child, this axis is the center point that the merry-go-round rotates around.

The principles governing rotational motion are rooted in the interplay between angular velocity and centripetal force. The rotation of the merry-go-round is not as simple as moving from one point to another; instead, it involves constantly changing direction, which requires the use of radial (centripetal) acceleration to maintain this circular path. Understanding rotational motion helps us analyze the 'how' and 'why' of objects that move in circles, allowing for safe and fun playground equipment designs, among many other practical applications.

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Most popular questions from this chapter

A banked highway is designed for traffic moving at \(90.0 \mathrm{km} / \mathrm{h} .\) The radius of the curve is \(310 \mathrm{m} .\) What is the angle of banking of the highway?

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of \(110.0 \mathrm{m}\). (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell's velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

Modern roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g}\) ? (b) How high above the top of the loop must the roller coaster start from rest, assuming negligible friction? (c) If it actually starts \(5.00 \mathrm{m}\) higher than your answer to (b), how much energy did it lose to friction? Its mass is \(1.50 \times 10^{3} \mathrm{kg}\)

Using Stokes' law, verify that the units for viscosity are kilograms per meter per second.

If centripetal force is directed toward the center, why do you feel that you are 'thrown' away from the center as a car goes around a curve? Explain.
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