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Chapter 10: Problem 23

With center and spokes of negligible mass, a certain bicycle wheel has a thin rim of radius \(36.3 \mathrm{~cm}\) and mass \(3.66 \mathrm{~kg}\); it can turn on its axle with negligible friction. A man holds the wheel above his head with the axis vertical while he stands on a turntable free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of \(57.7 \mathrm{rad} / \mathrm{s}\), and the turntable is initially at rest. The rotational inertia of wheel-plus-man-plus-turntable about the common axis of rotation is \(2.88 \mathrm{~kg} \cdot \mathrm{m}^{2} .(a)\) The man's hand suddenly stops the rotation of the wheel (relative to the turntable). Determine the resulting angular velocity (magnitude and direction) of the system. (b) The experiment is repeated with noticeable friction introduced into the axle of the wheel, which, starting from the same initial angular speed \((57.7 \mathrm{rad} / \mathrm{s})\), gradually comes to rest (relative to the turntable) while the man holds the wheel as described above. (The turntable is still free to rotate without friction.) Describe what happens to the system, giving as much quantitative information as the data permit.

Short Answer

Expert verified
The resulting angular velocity of the system is roughly 9.7 rad/s clockwise. When friction is introduced to the wheel's axle and makes the wheel come to a stop, the rest of the system gains angular momentum to conserve the total angular momentum of the entire system.

Step by step solution

01

Calculate Initial Angular Momentum

Firstly, we will calculate the angular momentum of the wheel before it is stopped. The angular momentum of a rotating object is given by \(L = Iω \), where \(L\) is the angular momentum, \(I\) is the moment of inertia, and \(ω\) is the angular velocity. The moment of inertia \(I\) of a thin rim is given by \(I = mR^2\), where \(m\) is the mass of the rim, and \(R\) is the radius. Substituting these values, we obtain \(I = (3.66 kg) * (0.363 m)^2 \approx 0.483 kgm^2\). Hence, the initial angular momentum (\(L_i\)) of only the wheel is \(L_i = Iω = (0.483 kgm^2)(57.7 rad/s) \approx 27.9 kgm^2/s\).
02

Calculate Final Angular Momentum

After the wheel is stopped, according to the law of conservation of angular momentum, the angular momentum of the system (wheel + man + turntable) will remain same. The final moment of inertia is given in the problem as \(2.88 kgm^2\) and final angular velocity is 'ωf'. Therefore, Final angular momentum (\(L_f\)) = \(2.88 kgm^2 * ωf\). But, we know that \(L_f = L_i\) (From law of conservation of angular momentum), so, we can equate the two and solve for \(ωf\) = \(L_i / 2.88 kgm^2 = 27.9 kgm^2/s / 2.88 kgm^2 \approx 9.7 rad/s\)
03

Explain Direction of Final Angular Velocity

Now, since the wheel was rotating clockwise initially and no external torques have been applied, the final angular velocity will also be in the clockwise direction according to conservation of angular momentum.
04

Anticipate Impact of Friction

When there is friction in the axle of the wheel, the wheel will gradually lose its angular momentum as it comes to rest. As the angular momentum of the wheel decreases due to friction, the angular momentum of the rest of the system (man + turntable) would increase to conserve the total angular momentum. Due to increased friction, the man, wheel, and turntable will all rotate together, until the wheel loses all of its angular momentum.

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Most popular questions from this chapter

A particle of mass \(13.7 \mathrm{~g}\) is moving with a constant velocity of magnitude \(380 \mathrm{~m} / \mathrm{s}\). The particle, moving in a straight line, passes within \(12 \mathrm{~cm}\) of the origin. Calculate the angular momentum of the particle about the origin.

A girl of mass \(50.6 \mathrm{~kg}\) stands on the edge of a frictionless merry- go-round of mass \(827 \mathrm{~kg}\) and radius \(3.72 \mathrm{~m}\) that is not moving. She throws a \(1.13-\mathrm{kg}\) rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(7.82 \mathrm{~m} / \mathrm{s}\). Calculate \((a)\) the angular speed of the merry-go-round and \((b)\) the linear speed of the girl after the rock is thrown. Assume that the merry-go-round is a uniform disk.

A man stands on a frictionless platform that is rotating with an angular speed of \(1.22 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is \(6.13 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the weights the man decreases the rotational inertia to \(1.97 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what is the resulting angular speed of the platform?

A uniform stick has a mass of \(4.42 \mathrm{~kg}\) and a length of \(1.23 \mathrm{~m}\). It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse \(12.8 \mathrm{~N} \cdot \mathrm{s}\) at a distance of \(46.4 \mathrm{~cm}\) from the center. Determine the subsequent motion of the stick.

Let \(\overrightarrow{\mathbf{r}}_{\mathrm{cm}}\) be the position vector of the center of mass \(C\) of a system of particles with respect to the origin \(O\) of an inertial reference frame, and let \(\overrightarrow{\mathbf{r}}_{i}^{\prime}\) be the position vector of the \(i\) th particle, of mass \(m_{i}\), with respect to the center of mass \(C\). Hence \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\) (see Fig. \(\left.10-22\right)\). Now define the total angular momentum of the system of particles relative to the center of mass \(C\) to be \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right), \quad\) where \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i}^{\prime} / d t\) (a) Show that \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i} / d t-\) \(m_{i} d \overrightarrow{\mathbf{r}}_{\mathrm{cm}} / d t=\overrightarrow{\mathbf{p}}_{i}-m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}\) (b) Show next that \(d \overrightarrow{\mathbf{L}}^{\prime} / d t=\) \(\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times d \overrightarrow{\mathbf{p}}_{i}^{\prime} / d t\right) .(c)\) Combine the results of \((a)\) and \((b)\) and, using the definition of center of mass and Newton's third law, show that \(\vec{\tau}_{\mathrm{ext}}^{\prime}=d \overrightarrow{\mathbf{L}}^{\prime} / d t\), where \(\vec{\tau}_{\mathrm{ext}}^{\prime}\) is the sum of all the \(\mathrm{ex}\) - ternal torques acting on the system about its center of mass.
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