Problem 17 - Fundamentals of Physics

Question

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?

Step-by-Step Solution

Answer

Linear acceleration is 6.48 m/s², time is 0.61 s, linear speed is 3.95 m/s, translational KE is 0.937 J, rotational KE is 0.905 J, angular speed is 1234.38 rad/s.

Step 1: Calculate Linear Acceleration

To find the linear acceleration, use the formula for a yo-yo: \[ a = \frac{g}{1 + \frac{I}{m r^2}} \]where \( I = 950 \, \mathrm{g} \cdot \mathrm{cm}^2 \) (converted to \( \mathrm{kg} \cdot \mathrm{m}^2 \)), \( m = 120 \, \mathrm{g} = 0.12 \, \mathrm{kg} \), and \( r = 3.2 \, \mathrm{mm} = 0.0032 \, \mathrm{m} \). Substituting in the values, we get:First convert rotational inertia:\[ I = 950 \, \mathrm{g} \cdot \mathrm{cm}^2 = 950 \times 10^{-7} \, \mathrm{kg} \cdot \mathrm{m}^2 \]Now substitute into the formula:\[ a = \frac{9.8}{1 + \frac{950 \times 10^{-7}}{0.12 \times (0.0032)^2}} \]By calculating the above expression, we find that:\[ a \approx 6.48 \, \mathrm{m/s^2} \]

Step 2: Calculate Time to Reach the End of the String

To calculate the time, use the kinematic equation for constant acceleration:\[ d = \frac{1}{2} a t^2 \]where \( d = 120 \, \mathrm{cm} = 1.2 \, \mathrm{m} \) and \( a = 6.48 \, \mathrm{m/s^2} \). Rearranging the equation for time, we find:\[ t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 1.2}{6.48}} \approx 0.61 \, \mathrm{s} \]

Step 3: Calculate Linear Speed at the End

The final speed can be calculated by the formula:\[ v = a t \]where \( a = 6.48 \, \mathrm{m/s^2} \) and \( t = 0.61 \, \mathrm{s} \). Substitute these values:\[ v = 6.48 \times 0.61 \approx 3.95 \, \mathrm{m/s} \]

Step 4: Calculate Translational Kinetic Energy

The translational kinetic energy is computed using:\[ KE_t = \frac{1}{2} m v^2 \]where \( m = 0.12 \, \mathrm{kg} \) and \( v = 3.95 \, \mathrm{m/s} \). Substitute these values:\[ KE_t = \frac{1}{2} \times 0.12 \times (3.95)^2 \approx 0.937 \mathrm{~J} \]

Step 5: Calculate Rotational Kinetic Energy

Rotational kinetic energy is given by:\[ KE_r = \frac{1}{2} I \omega^2 \]First, calculate \( \omega \) using \( \omega = \frac{v}{r} \):\[ \omega = \frac{3.95}{0.0032} \approx 1234.38 \, \mathrm{rad/s} \]Then calculate \( KE_r \) using \( I = 950 \times 10^{-7} \, \mathrm{kg} \cdot \mathrm{m}^2 \):\[ KE_r = \frac{1}{2} \times 950 \times 10^{-7} \times (1234.38)^2 \approx 0.905 \, \mathrm{J} \]

Step 6: Calculate Angular Speed

The angular speed when reaching the end of the string is calculated as described earlier:\[ \omega = \frac{v}{r} \approx \frac{3.95}{0.0032} \approx 1234.38 \, \mathrm{rad/s} \]

Key Concepts

Linear Acceleration

Linear acceleration is the rate of change of velocity of an object in a straight line. When analyzing the motion of a yo-yo, which experiences both translational and rotational motion, the linear acceleration can be calculated by considering the forces acting on it. For a yo-yo, the formula to calculate linear acceleration is:
\[ a = \frac{g}{1 + \frac{I}{m r^2}} \]where:

• \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{~m/s^2}\))
• \(I\) is the rotational inertia
• \(m\) is the mass of the yo-yo
• \(r\) is the radius of the axle

In our example, the yo-yo's rotational inertia \(I\) is given as \(950 \, \mathrm{g} \cdot \mathrm{cm}^2\). This value needs to be converted into \(\mathrm{kg} \cdot \mathrm{m}^2 \) for consistency with other SI units used in equations. After conversion, we use the formula to find the linear acceleration, which comes out to approximately \(6.48 \, \mathrm{m/s^2}\). This result tells us how quickly the yo-yo speeds up along the string.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They're especially handy for predicting how an object moves along a path without directly considering forces at each point in time. For the yo-yo, we used the equation:
\[ d = \frac{1}{2} a t^2 \]
This equation relates the distance \(d\), linear acceleration \(a\), and time \(t\). Here, the distance is the length of the string (\(120 \mathrm{~cm}\) or \(1.2 \mathrm{~m}\)), and the linear acceleration \(a\) is already known.To find the time \(t\) it takes for the yo-yo to reach the end of the string, rearrange the equation to:
\[ t = \sqrt{\frac{2d}{a}} \]Substituting the known values gives us \(t \approx 0.61 \, \mathrm{s}\). This tells us that the yo-yo takes just over half a second to reach the end of its string.

Translational Kinetic Energy

Translational kinetic energy represents the energy of an object moving along a straight path. It's calculated using the formula:
\[ KE_t = \frac{1}{2} m v^2 \]
where \(m\) is the mass and \(v\) is the linear speed of the object. For the yo-yo, with a mass of \(0.12 \, \mathrm{kg}\) and a speed of \(3.95 \, \mathrm{m/s}\), the translational kinetic energy is:
\[ KE_t = \frac{1}{2} \times 0.12 \times (3.95)^2 \approx 0.937 \, \mathrm{J} \]
This calculation shows how much energy the yo-yo has because of its linear motion. Understanding this energy helps in analyzing the conversion of potential energy as the yo-yo unwinds and moves downward.

Rotational Kinetic Energy

Rotational kinetic energy is associated with the spinning motion of an object. This type of energy can be found using the formula:
\[ KE_r = \frac{1}{2} I \omega^2 \]
where \( I \) is the rotational inertia and \( \omega \) is the angular speed. For a yo-yo, the angular speed \( \omega \) can be calculated using:
\[ \omega = \frac{v}{r} \]The rotational inertia \( I \) is given, and the linear speed \( v \) is known.Calculating these values for a yo-yo results in a rotational kinetic energy of approximately \(0.905 \, \mathrm{J}\). This energy reflects the yo-yo's spinning motion as it descends, complementing its translational kinetic energy.

Angular Speed

Angular speed measures how quickly an object spins around an axis and is critical in the study of rotational dynamics. For a yo-yo, it's calculated using:
\[ \omega = \frac{v}{r} \]
where \(v\) is the linear speed and \(r\) is the radius of the axle, the point about which it rotates.In this example, the linear speed \(v\) was found to be \(3.95 \, \mathrm{m/s}\), and the axle radius \(r\) is \(0.0032 \, \mathrm{m}\). Therefore, the angular speed \(\omega\) is approximately \(1234.38 \, \mathrm{rad/s}\).
Knowing the angular speed helps to understand how fast the yo-yo is spinning, which is vital in assessing both its kinetic energies when it reaches the end of the string.