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Chapter 4: Problem 59

Music is still available on vinyl records that are played on turntables. Such a record rotates with a period of \(1.8 \mathrm{~s}\). For a record with a radius of \(16 \mathrm{~cm}\), find the centripetal accceleration of a point on the edge of the record.

Short Answer

Expert verified
The centripetal acceleration is approximately 194.79 cm/s虏.

Step by step solution

01

Determine the Circumference of the Record

First, we need to calculate the distance a point on the edge of the record travels in one rotation. This is the circumference of the record, given by the formula: \[ C = 2\pi r \]where \( r = 16 \, \text{cm} \). Therefore, \[ C = 2 \times \pi \times 16 = 32\pi \text{ cm} \].
02

Calculate the Speed of the Point

The speed \( v \) of a point on the edge is the circumference divided by the period \( T \). Thus, \[ v = \frac{C}{T} = \frac{32\pi}{1.8} \text{ cm/s} \].
03

Use the Formula for Centripetal Acceleration

The formula for centripetal acceleration \( a_c \) is: \[ a_c = \frac{v^2}{r} \].Substitute the speed from the previous step and the radius \( r = 16 \, \text{cm} \) to get \[ a_c = \frac{\left(\frac{32\pi}{1.8}\right)^2}{16} \].
04

Perform the Necessary Calculations

Calculate the value of the speed \( v \) and then use it to find \( a_c \). Start with:\[ v = \frac{32\pi}{1.8} \approx 55.77 \, \text{cm/s}\]Then calculate:\[ a_c = \frac{(55.77)^2}{16} \approx 194.79 \, \text{cm/s}^2 \].

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Key Concepts

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

Centripetal Force
The concept of centripetal force is essential when discussing objects moving in a curved path, like the point on the edge of a rotating vinyl record. Centripetal force is the force that keeps the object moving in a circle or along a curved path instead of flying off in a straight line.

It always points towards the center of the circle, helping to keep the object from veering off the edge. Imagine whirling a ball tied to a string in a circle around you. The tension in the string provides the centripetal force, pulling the ball towards the center.

There are a few key points to remember:
  • The centripetal force is not a new or separate force; instead, it results from other forces acting upon the object.
  • The formula for centripetal force involves mass, velocity, and the radius of the circle: \( F_c = \frac{mv^2}{r} \).
  • The force can come from various sources, such as tension, gravity, or friction.
By understanding the nature of centripetal force, one can better analyze objects in circular motion, such as the vinyl records on a turntable.
Circular Motion
Circular motion is a term used in physics to describe the movement of an object along a circular path. In the context of vinyl records, a point on the edge of the record undergoes circular motion as it spins around the turntable.

The key characteristic of circular motion is that even though the object moves at a constant speed, its velocity is not constant because its direction is continually changing. This change in velocity is what results in centripetal acceleration.

Circular motion can be either uniform or non-uniform:
  • Uniform circular motion: The object moves along the circle with constant speed.
  • Non-uniform circular motion: The speed of the object varies as it goes around the circle.
The formula of centripetal acceleration for uniform circular motion is \( a_c = \frac{v^2}{r} \), where \( v \) is the speed, and \( r \) is the radius. This formula shows that even if the speed is constant, the acceleration keeps the object moving along the circle rather than off on a tangent.
Physics Problem Solving Steps
When tackling physics problems, especially those involving centripetal force and circular motion, it's crucial to follow a clear problem-solving strategy. Here鈥檚 a simple process:

  • Understand the problem: Carefully read the problem description, noting all given values and what you need to find.
  • Identify relevant formulas: Focus on the applicable formulas, like those for speed \( v = \frac{C}{T} \) and centripetal acceleration \( a_c = \frac{v^2}{r} \).
  • Substitute known values: Plug the given numbers into the formulas you identified.
  • Carry out the calculations: Solve step by step, ensuring each calculation is correct before moving on.
By following these structured steps, you'll have an easier time solving physics problems systematically and accurately. This methodology is particularly helpful in exercises like finding the centripetal acceleration of a point on a vinyl record's edge.

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Most popular questions from this chapter

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \((4.00\) \(\left.\mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

The position vector for an electron is \(\vec{r}=(6.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((4.0 \mathrm{~m}) \hat{\mathrm{j}}+(3.0 \mathrm{~m}) \hat{\mathrm{k}}\). (a) Find the magnitude of \(\vec{r}\). (b) Sketch the vector on a right-handed coordinate system.

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of \(v_{0}=6.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta_{0}=35.0^{\circ}\), what percentage of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height)?

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of \(v_{0}=30.0\) \(\mathrm{m} / \mathrm{s}\) and at an angle of \(\theta_{0}=40.0^{\circ}\). What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?

A watermelon seed has the following coordinates: \(x=-5.0 \mathrm{~m}\), \(y=9.0 \mathrm{~m}\), and \(z=0 \mathrm{~m}\). Find its position vector (a) in unit- vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the \(x\) axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the \(x y z\) coordinates \((3.00 \mathrm{~m}\), \(0 \mathrm{~m}, 0 \mathrm{~m}\) ), what is its displacement (e) in unit-vector notation and as (f) a magnitude and \((\mathrm{g})\) an angle relative to the positive \(x\) direction?
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