91Ó°ÊÓ

The string in a yo-yo is wound around an axle of radius \(0.500 \mathrm{cm} .\) The yo-yo has both rotational and translational motion, like a rolling object, and has mass \(0.200 \mathrm{kg}\) and outer radius \(2.00 \mathrm{cm} .\) Starting from rest, it rotates and falls a distance of \(1.00 \mathrm{m}\) (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of $1.00 \mathrm{m} ?$ (b) How long does it take to fall? [Hint: The transitional and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]

Step by step solution

01

Define known variables and equations

We know the following variables and equations (where `R` denotes the outer radius, `r` is the axle's radius, `m` is the mass of the yo-yo, and `h` is the height it falls): - Mass of yo-yo (m): 0.200 kg - Outer radius of yo-yo (R): 2.00 cm = 0.0200 m - Axle radius (r): 0.500 cm = 0.00500 m - Distance fallen (h): 1.00 m Moment of inertia (I) of a uniform disk about its center: \(I=\frac{1}{2}mR^2\)

02

Apply conservation of energy

Since the yo-yo starts at rest, its initial energy is only potential: \(U_i = mgh\). At the end, it has both translational and rotational kinetic energy. If \(v_f\) is the final linear speed, then: - Translational kinetic energy (TKE): \(\frac{1}{2}mv_f^2\) - Rotational kinetic energy (RKE): \(\frac{1}{2}I\omega^2\) Since the yo-yo rolls without slipping, \(\omega = \frac{v_f}{r}\). Therefore, the final energy state is given by \(\frac{1}{2}mv_f^2 + \frac{1}{2}I(\frac{v_f}{r})^2 = mgh\).

03

Solve for the final linear velocity (v_f)

By applying the conservation of energy, we get the following equation: \(\frac{1}{2}mv_f^2 + \frac{1}{2}I(\frac{v_f}{r})^2 = mgh\). Substitute the moment of inertia of a uniform disk, \(I=\frac{1}{2}mR^2\), into the equation: \(\frac{1}{2}mv_f^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v_f}{r})^2 = mgh\) Now, simplify and solve for \(v_f\): \(v_f^2(1 + \frac{1}{4}R^2\frac{1}{r^2}) = 2gh\) \(v_f = \sqrt{\frac{2gh}{1 + \frac{1}{4}\frac{R^2}{r^2}}}\) Plug in the given values: \(v_f = \sqrt{\frac{2(9.81)(1)}{1 + \frac{1}{4}(\frac{0.02}{0.005})^2}} \approx 2.025 \ \text{m/s}\) So, the final speed of the yo-yo is approximately 2.025 m/s.

04

Find the time it takes to fall (t_f)

We can use the following kinematic equation: \(h = v_i t_f + \frac{1}{2}at_f^2\) Since the initial velocity (\(v_i\)) is 0, the equation becomes: \(h = \frac{1}{2}at_f^2\) Solving for \(t_f\): \(t_f = \sqrt{\frac{2h}{a}}\) Using acceleration due to gravity (a = 9.81 m/s²) and height (h = 1m): \(t_f = \sqrt{\frac{2(1)}{9.81}} \approx 0.451 \ \text{s}\) So, it takes approximately 0.451 seconds for the yo-yo to fall. #Answer#: (a) The yo-yo's final speed is approximately 2.025 m/s. (b) It takes approximately 0.451 seconds for the yo-yo to fall.

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