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Chapter 6: Problem 96

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

Short Answer

Expert verified
The speed is 0.962 m/s; the minimum coefficient of static friction is 0.021.

Step by step solution

01

Calculate the Circumference of the Merry-Go-Round

The first step is to determine the distance a point on the rim travels in one complete revolution. Since the path is circular, the distance is the circumference of the merry-go-round. Use the formula for the circumference of a circle: \( C = 2\pi r \). Here, \( r = 4.6 \text{ m} \), so \( C = 2\pi \times 4.6 = 9.2\pi \) meters.
02

Determine the Speed of a Point on the Rim

The speed of a point on the rim is the distance traveled per unit time. From Step 1, the circumference is \( 9.2\pi \text{ m} \). Given that the merry-go-round revolves once every \( 30 \text{ s} \), the speed \( v \) is \( v = \frac{9.2\pi}{30} \). Calculate this to find \( v \approx 0.962 \text{ m/s} \).
03

Determine the Centripetal Force Required

The centripetal force needed to keep the basket from sliding off is determined by the formula \( F_c = m\cdot a_c = m\cdot \frac{v^2}{r} \). Use the speed \( v \) from Step 2 and radius \( r = 4.6 \text{ m} \). Replacing, \( F_c = m \frac{(0.962)^2}{4.6} \).
04

Relate the Friction Force to the Centripetal Force

The static friction force must be at least equal to the needed centripetal force to prevent the basket from slipping. The static friction is given by \( F_f = \mu_s m g \), where \( \mu_s \) is the coefficient of static friction and \( g \approx 9.8 \text{ m/s}^2 \) is the gravitational acceleration.
05

Solve for the Coefficient of Static Friction \( \mu_s \)

Set the centripetal force equal to the friction force for the basket not to slip: \( \mu_s m g = m \frac{(0.962)^2}{4.6} \). Cancel the mass \( m \) from both sides, as it appears in both terms: \( \mu_s g = \frac{0.962^2}{4.6} \). Solving for \( \mu_s \), \( \mu_s = \frac{(0.962)^2}{4.6 \times 9.8} \approx 0.021 \).

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Key Concepts

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

Circumference
The circumference of a circle is the total distance around its edge. In this exercise, it's important because it tells us the distance a point travels in one complete circle on the merry-go-round. To find the circumference, we use the formula:
  • \( C = 2\pi r \)
Here, \( r \) represents the radius of the circle, which is given as \( 4.6 \text{ m} \) in the problem. Thus, the circumference \( C \) is calculated as:
  • \( C = 2\pi \times 4.6 = 9.2\pi \text{ m} \)
The circumference is essential to determine how far a point on the rim moves in each full rotation. This forms the base for further calculations like speed.
Coefficient of Static Friction
The coefficient of static friction is a crucial factor that determines how much grip one surface has on another without slipping. In the context of this exercise, it's all about keeping the basket from sliding off the merry-go-round.
To understand it better, consider that the frictional force is what provides the necessary grip.
  1. It is calculated by the formula \( F_f = \mu_s m g \).
  2. Here, \( \mu_s \) is the coefficient of static friction.
  3. \( m \) is the mass of the basket, and \( g \approx 9.8 \text{ m/s}^2 \) is the gravitational force.
The main idea here is that the frictional force should be strong enough to act as the centripetal force that keeps the basket on the ride. In the step-by-step solution, by equating the needed centripetal force with the static friction force, we find that the lowest value of \( \mu_s \) needed is approximately \( 0.021 \). This value ensures that the basket remains in place as the merry-go-round spins.
Static Friction
Static friction is the force that keeps an object at rest when a force is applied until the point when sliding begins. It works against potential movement between two surfaces.In our scenario, static friction is what prevents the picnic basket from skidding off when the merry-go-round spins. When the merry-go-round rotates, the basket experiences a force pulling it outward鈥攖his is where static friction comes into play to counterbalance it.The effectiveness of static friction is determined by:
  • The texture and roughness of the surfaces involved.
  • The coefficient of static friction, \( \mu_s \).
Static friction reaches its maximum just before the object starts moving. Beyond this point, kinetic friction takes over, which is usually lower than static friction. In calculations, we use static friction to ensure stability, as it acts to prevent motion. By calculating static friction, we ascertain that the basket can securely remain on the merry-go-round without sliding off due to insufficient grip.

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Most popular questions from this chapter

Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N}\), are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10\), and the coefficient between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (Fig. 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the coefficient of static friction between two such layers is \(0.5\), what is the least angle \(\phi\) through which the present slope should be reduced to prevent slippage?

A \(1.5 \mathrm{~kg}\) box is initially at rest on a horizontal surface when at \(t=0\) a horizontal force \(\vec{F}=(1.8 t) \hat{\mathrm{i}} \mathrm{N}\) (with \(t\) in seconds) is applied to the box. The acceleration of the box as a function of time \(t\) is given by \(\vec{a}=0\) for \(0 \leq t \leq 2.8 \mathrm{~s}\) and \(\vec{a}=(1.2 t-2.4) \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}\) for \(t>\) \(2.8 \mathrm{~s}\). (a) What is the coefficient of static friction between the box and the surface? (b) What is the coefficient of kinetic friction between the box and the surface?

The terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spreadeagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.

ssm A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68\), and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N}\), (c) \(445 \mathrm{~N}\), and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?
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