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Chapter 11: Problem 49

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?

Short Answer

Expert verified
(a) 78.57 rad/s; (b) -47.07 rad/s; (c) Clockwise.

Step by step solution

01

Convert Revolutions Per Minute to Radians Per Second

First, we need to convert the initial angular velocities from revolutions per minute to radians per second. \(1\text{ rev} = 2\pi\text{ rad}\) and we know \(1\text{ min} = 60\text{ s}\). For the first disk: \[ \omega_1 = 450 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 47.12 \text{ rad/s} \]For the second disk: \[ \omega_2 = 900 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 94.25 \text{ rad/s} \]
02

Use Conservation of Angular Momentum for Both Disks Counterclockwise

Since both disks are rotating in the counterclockwise direction before coupling, the total angular momentum before and after coupling is the same.Angular momentum before coupling:\[ L_{\text{before}} = I_1 \omega_1 + I_2 \omega_2 \]\[ L_{\text{before}} = 3.30 \cdot 47.12 + 6.60 \cdot 94.25 \]\[ L_{\text{before}} = 155.496 + 621.3 = 776.796 \text{ kg} \cdot \text{m}^2/\text{s}\]Angular momentum after coupling:\[ L_{\text{after}} = (I_1 + I_2) \omega_f \]Equating the two and solving for \(\omega_f\):\[ \omega_f = \frac{776.796}{3.30 + 6.60} = \frac{776.796}{9.90} = 78.57 \text{ rad/s}\]
03

Use Conservation of Angular Momentum with Opposite Direction

If the second disk spins clockwise, it has a negative angular velocity. Therefore, the initial angular momentum expression changes:For clockwise second disk:\[ L_{\text{before}} = I_1 \omega_1 - I_2 \omega_2 \]\[ L_{\text{before}} = 3.30 \cdot 47.12 - 6.60 \cdot 94.25 \]\[ L_{\text{before}} = 155.496 - 621.3 = -465.804 \text{ kg} \cdot \text{m}^2/\text{s}\]After coupling:\[ L_{\text{after}} = (I_1 + I_2) \omega_f \]Equating the two and solving for \(\omega_f\):\[ \omega_f = \frac{-465.804}{9.90} = -47.07 \text{ rad/s}\]
04

Determine Direction of Rotation

Since the final angular velocity \(\omega_f = -47.07 \text{ rad/s}\) is negative, the direction of rotation after coupling is clockwise.

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Key Concepts

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

Rotational Inertia
Rotational inertia, sometimes called the moment of inertia, is a measure of an object's resistance to changes in its rotation. Picture trying to spin a merry-go-round. The heavier and more widely spread out the weight is, the harder it is to start or stop spinning. That's the essence of rotational inertia. It depends on both the mass of the object and how that mass is distributed relative to the axis of rotation.
  • Heavier objects with mass located farther from the axis have higher rotational inertia.
  • In our problem, Disk 1 has a rotational inertia of 3.30 kg路m虏, while Disk 2's inertia is 6.60 kg路m虏.
Greater rotational inertia means the object wants to keep rotating if it's already spinning, or stay still if it's not. Understanding this concept is essential when dealing with multiple rotating bodies, as it dictates how they will react when influenced to rotate as one unit.
Angular Speed
Angular speed indicates how quickly an object is rotating. Instead of measuring in meters per second like linear speed, angular speed is measured in units like radians per second. In our exercise, we initially convert revolutions per minute (rpm) into radians per second for easy calculation. Converting between these units is crucial:
  • For the first disk, which rotates at 450 rpm, the conversion yields approximately 47.12 rad/s.
  • For the second disk at 900 rpm, we find about 94.25 rad/s.
After coupling the disks, we use conservation of angular momentum to determine the new angular speed. Angular momentum, a product of rotational inertia and angular speed, remains constant in the absence of external torques. By combining both disks' inertia and initial speeds, we calculate the system's final angular speed after coupling. Understanding these transformations between units and the application of conservation principles is fundamental in physics.
Direction of Rotation
Direction of rotation determines whether an object is spinning clockwise or counterclockwise. In physics problems like this one, direction is vital, as it affects the calculations we make, especially in angular momentum conservation. Consider the scenario where both disks initially spin counterclockwise. Combining them in the same direction merely sums up their angular momenta to find the final state. However, when one disk spins clockwise:
  • We consider its angular speed as negative, affecting the total angular momentum.
  • This determines that the final angular velocity might be negative, indicating a direction change.
In the exercise, when the second disk spins clockwise, we find that the system rotates clockwise after coupling because the net angular momentum becomes negative. So, not only do we deduce the speed but also clearly identify the direction. This clear understanding of direction helps solve complex rotational dynamics problems efficiently.

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Most popular questions from this chapter

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

At time \(t\), the vector \(\vec{r}=4.0 t^{2} \hat{\mathrm{i}}-\left(2.0 t+6.0 t^{2}\right) \hat{\mathrm{j}}\) gives the position of a \(3.0 \mathrm{~kg}\) particle relative to the origin of an \(x y\) coordinate system ( \(\vec{r}\) is in meters and \(t\) is in seconds). (a) Find an expression for the torque acting on the particle relative to the origin. (b) Is the magnitude of the particle's angular momentum relative to the origin increasing, decreasing, or unchanging?

A solid sphere of weight \(36.0 \mathrm{~N}\) rolls up an incline at an angle of \(30.0^{\circ}\). At the bottom of the incline the center of mass of the sphere has a translational speed of \(4.90 \mathrm{~m} / \mathrm{s}\). (a) What is the kinetic energy of the sphere at the bottom of the incline? (b) How far does the sphere travel up along the incline? (c) Does the answer to (b) depend on the sphere's mass?

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a time-dependent torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m} .\) At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s}\) ?
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