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Chapter 5: Problem 52

A child is twirling a \(0.0120-\mathrm{kg}\) plastic ball on a string in a horizontal circle whose radius is \(0.100 \mathrm{m}\). The ball travels once around the circle in 0.500 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

Short Answer

Expert verified
(a) The centripetal force is approximately 0.189 N. (b) The force increases by a factor of 4, not 2.

Step by step solution

01

Find the velocity

First, we need to find the velocity of the ball. The circumference of the circle is given by \( C = 2\pi r \). Plugging in the radius, we get \[ C = 2\times 3.1416 \times 0.100 \approx 0.628 \text{ m} \].The ball completes this circle in 0.500 seconds, so the velocity \( v \) is \[ v = \frac{C}{T} = \frac{0.628 \text{ m}}{0.500 \text{ s}} \approx 1.256 \text{ m/s} \].
02

Calculate centripetal acceleration

Using the formula for centripetal acceleration \( a_c \), which is \( a_c = \frac{v^2}{r} \), substitute the values:\[ a_c = \frac{(1.256 \, \text{m/s})^2}{0.100 \, \text{m}} \approx 15.786 \, \text{m/s}^2 \].
03

Determine the centripetal force

Centripetal force \( F_c \) is given by \( F_c = m \times a_c \), where \( m \) is the mass of the ball. Given \( m = 0.012 \, \text{kg} \), we can calculate:\[ F_c = 0.012 \, \text{kg} \times 15.786 \, \text{m/s}^2 = 0.1894 \, \text{N} \].
04

Analyze the effect of doubling the speed

If the speed is doubled, the new speed would be \( 2v = 2 \times 1.256 \, \text{m/s} = 2.512 \, \text{m/s} \). The new centripetal force \( F_c' \) can be found using the formula \( F_c' = m \times \frac{(2v)^2}{r} \). Thus, \( F_c' = m \times \frac{(2 \times 1.256)^2}{r} = m \times \frac{4 \times v^2}{r} = 4 \times F_c \). This means the centripetal force increases by a factor of 4, not 2.

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Key Concepts

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

Centripetal Acceleration
When an object moves in a circle, it constantly changes direction. This change requires an inward force, directed towards the center of the circle.
This is where centripetal acceleration comes into play. Centripetal acceleration ( a_c ) is crucial for circular motion, as it keeps the object moving in a curved path rather than a straight line. In mathematical terms, centripetal acceleration can be defined as \( a_c = \frac{v^2}{r} \), where \( v \) is the object's velocity and \( r \) is the radius of the circle.
  • The direction of \( a_c \) is always towards the center of the circle.
  • It is proportional to the square of the velocity, meaning any change in speed greatly affects it.
  • Larger circles require less acceleration for the same speed, as \( r \) increases.
In the problem, the centripetal acceleration was calculated as \( 15.786 \, \text{m/s}^2 \) based on the given velocity and radius.
Uniform Circular Motion
Uniform circular motion refers to the movement of an object in a circle at a constant speed. Despite maintaining a constant speed, the object's velocity is not constant due to the continuous change in direction. This scenario exhibits unique characteristics:
  • The path is always circular, creating a loop.
  • The speed remains constant throughout, though the velocity (which includes direction) changes.
  • Centripetal force is required to maintain the motion, ensuring the object doesn鈥檛 drift.
In the exercise, the ball revolves around a circle uniformly, with consistent looping periods of half a second per cycle. This steady motion emphasizes how uniform circular motion sustains a balance between speed and continuous directional change.
Physics Problem Solving
Approaching physics problems effectively involves understanding concepts and applying formulas methodically. Let's look at how this was tackled in the exercise. First, identifying given values and what needs to be found is crucial. Here, we were given the radius, mass of the ball, and the time taken for a complete revolution. From that, the problem was decomposed into logical steps:
  • Calculate the velocity using the circumference and time.
  • Find centripetal acceleration using the velocity and radius.
  • Compute the centripetal force with mass and acceleration.
  • Analyze the impact of changing variables, like doubling speed.
Such structured problem-solving helps ensure comprehension and accuracy, as seen by calculating a centrifugally driven increase by a factor of 4 when the speed is doubled.
Mass and Velocity Relationship
Understanding the relationship between mass and velocity is crucial in problems involving forces in circular motion. Here, the relationship shows how changes in velocity significantly affect force requirements. Key points include:
  • The centripetal force \( F_c \) is expressed as \( m \times a_c \). This means force is directly proportional to the mass.
  • A change in velocity impacts centripetal force due to its square relation. Doubling velocity quadruples the force, as shown by \( F_c' = m \times \frac{(2v)^2}{r} \), simplifying to \( 4 \times F_c \).
  • Mass indirectly influences force changes since it stays constant while the object's speed varies.
This deeper look at mass and velocity interplay highlights their importance in determining the necessary centripetal force for circular motion and adjusting for any speed changes. In our exercise, this is why the force multiplied by four instead of two when speed doubled.

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Most popular questions from this chapter

A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop- the-loop ride is in the vertical plane and its radius is \(21 \mathrm{m}\). Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads \(770 \mathrm{N}\). At the top of the loop, the rider is upside down and moving, and the sensor reads \(350 \mathrm{N}\). What is the speed of the rider at the top of the loop?

A \(0.20-\mathrm{kg}\) ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the stick tension is 16 N. Find the tensions in the stick when the ball is at the twelve o'clock and at the six o"clock positions.

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of \(45.0 \mathrm{m} .\) Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is \(12 \mathrm{cm}\) When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
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