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A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g} .\) Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) linear speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) angular speed?

Short Answer

Expert verified
(a) 0.532 m/s² (b) 2.136 s (c) 1.137 m/s (d) 0.0774 J (e) 6.00 J (f) 355.3 rad/s

Step by step solution

01

Define Initial Parameters

Before starting the calculations, let's convert all units to SI units. The rotational inertia \( I = 950 \text{ g} \cdot \text{cm}^2 \) becomes \( I = 9.5 \times 10^{-5} \text{ kg} \cdot \text{m}^2 \). The mass \( m = 120 \text{ g} \) becomes \( m = 0.120 \text{ kg} \). The axle radius \( r = 3.2 \text{ mm} \) becomes \( r = 3.2 \times 10^{-3} \text{ m} \). Length of the string \( L = 120 \text{ cm} \) becomes \( L = 1.20 \text{ m} \).
02

Calculate the Linear Acceleration (a)

Using the rotational inertia and radius, we apply the equation for the combined linear and angular acceleration: \( a = \frac{mg}{m + \frac{I}{r^2}} \). Substitute the values: \[ a = \frac{0.120 \times 9.81}{0.120 + \frac{9.5 \times 10^{-5}}{(3.2 \times 10^{-3})^2}} \]. Calculating this gives approximately \( a = 0.532 \text{ m/s}^2 \).
03

Time to Reach the End of the String (b)

We use the kinematic equation \( s = ut + \frac{1}{2}at^2 \) where initial velocity \( u = 0 \) and displacement \( s = L \). Solving \( L = \frac{1}{2}at^2 \) gives \( t = \sqrt{\frac{2L}{a}} \). Substitute the values: \[ t = \sqrt{\frac{2 \times 1.20}{0.532}} \] which calculates to \( t \approx 2.136 \text{ s} \).
04

Calculate Linear Speed at String's End (c)

Using \( v = at \), substitute the values: \( v = 0.532 \times 2.136 \approx 1.137 \text{ m/s} \).
05

Translational Kinetic Energy (d)

The translational kinetic energy is \( KE_{trans} = \frac{1}{2}mv^2 \). Substitute the values: \[ KE_{trans} = \frac{1}{2} \times 0.120 \times (1.137)^2 \]. This calculates to approximately \( KE_{trans} \approx 0.0774 \text{ J} \).
06

Rotational Kinetic Energy (e)

Rotational kinetic energy is \( KE_{rot} = \frac{1}{2}I\omega^2 \), and \( \omega = \frac{v}{r} \). First, calculate \( \omega = \frac{1.137}{3.2 \times 10^{-3}} \approx 355.3 \text{ rad/s} \). Use it to find \( KE_{rot} = \frac{1}{2} \times 9.5 \times 10^{-5} \times (355.3)^2 \approx 6.00 \text{ J} \).
07

Calculate Angular Speed (f)

Angular speed \( \omega = \frac{v}{r} \). Substitute the values: \( \omega = \frac{1.137}{3.2 \times 10^{-3}} \approx 355.3 \text{ rad/s} \).

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Key Concepts

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

Linear Acceleration
Linear acceleration in our context refers to the rate of change of velocity of the yo-yo as it descends along the string. This is crucial in understanding motion because it tells us how quickly the yo-yo speeds up from rest.
To find the linear acceleration, we use the formula: \[a = \frac{mg}{m + \frac{I}{r^2}}\] where:
  • \( m \) is the mass of the yo-yo,
  • \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
  • \( I \) is the rotational inertia,
  • and \( r \) is the radius of the axle.
This equation arises because the yo-yo's motion must account for both its linear acceleration along the string and its rotation around its axle.
By substituting the known values, we calculate the linear acceleration as approximately \( 0.532 \, \text{m/s}^2 \). This result helps us predict how quickly the yo-yo gains speed along the string, and it's fundamental for calculating subsequent motions.
Translational Kinetic Energy
Translational kinetic energy is the energy that an object possesses because of its motion through space, specifically relating to the linear, or straight-line, movement. For the yo-yo, this involves its progress down the string.
The formula for translational kinetic energy is: \[KE_{\text{trans}} = \frac{1}{2}mv^2\] where:
  • \( m \) is the mass of the object,
  • \( v \) is the linear speed at which the object is moving.
Plugging in the values, we find that \( KE_{\text{trans}} \approx 0.0774 \, \text{J} \).
This tells us how much energy the yo-yo has due to its straight movement at the end of the string. Understanding this concept is important because it gives insight into the forces at work and helps gauge which part of the energy budget is "spent" moving the yo-yo in a straight line.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation around an axis. For the yo-yo, this means considering how it spins as it descends.
To calculate rotational kinetic energy, we use: \[KE_{\text{rot}} = \frac{1}{2}I\omega^2\] where:
  • \( I \) is the rotational inertia,
  • \( \omega \) is the angular speed.
The angular speed \( \omega \) is determined by the ratio of linear speed to the radius of the axle: \[\omega = \frac{v}{r}\] By substituting the given values, we calculate \( KE_{\text{rot}} \approx 6.00 \, \text{J} \).
This indicates how much energy the yo-yo uses up in spinning about its axle. Understanding rotational kinetic energy is key to getting a full picture of how energy is transferred and transformed in systems involving rotational motion, like our yo-yo here.
Angular Speed
Angular speed describes how quickly an object rotates or spins, expressed in radians per second. For a yo-yo, angular speed is crucial to determine how fast the yo-yo is spinning as it reaches the end of the string.
The relationship between linear speed and angular speed is given by: \[\omega = \frac{v}{r}\] where:
  • \( v \) is the linear speed,
  • \( r \) is the radius of the axle.
Substituting these values provides \( \omega \approx 355.3 \, \text{rad/s} \).
Understanding angular speed is important as it shows how the linear motion of the string translates into the spinning motion of the yo-yo. Knowing a yo-yo's angular speed serves not only in practical applications but also supports the study of dynamics and rotational mechanics.

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Most popular questions from this chapter

A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the bricks the man decreases the rotational inertia of the system to \(2.0\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

Two disks are mounted (like a merry-go-round) on lowfriction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia \(3.30 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at 450 rev/min. The second disk, with rotational inertia \(6.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at 900 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at \(900 \mathrm{rev} / \mathrm{min}\), what are their (b) angular speed and (c) direction of rotation after they couple together?

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\) and a rotational inertia of \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\) Putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

A particle is acted on by two torques about the origin: \(\vec{\tau}_{1}\) has a magnitude of \(2.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the positive direction of the \(x\) axis, and \(\vec{\tau}_{2}\) has a magnitude of \(4.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the negative direction of the \(y\) axis. In unit-vector notation, find \(d \vec{\ell} / d t\), where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

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