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Chapter 11: Problem 36

Shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks \(A\) and \(C\). Another belt runs around a central hub on disk \(A\) and the rim of disk \(B .\) The belts move smoothly without slippage on the rims and hub. Disk \(A\) has radius \(R ;\) its hub has radius \(0.5000 R ;\) disk \(B\) has radius \(0.2500 R ;\) and disk \(C\) has radius \(2.000 R .\) Disks \(B\) and \(C\) have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk \(C\) to that of disk \(B\) ?

Short Answer

Expert verified
The ratio of the angular momentum of disk C to disk B is 128.

Step by step solution

01

Understanding Angular Velocity Relationships

Since the belts do not slip, the linear velocities at the points where the belts contact the disks must be equal. For disks A and C, the linear velocity is the product of angular velocity and radius. Thus, \( \omega_A R = \omega_C (2R) \), giving \( \omega_C = \frac{\omega_A}{2} \). For disks A and B, the linear velocities at the hub and rim connection yield \( \omega_B (0.25R) = \omega_A (0.5R) \), so \( \omega_B = 2\omega_A \).
02

Calculating Moment of Inertia

The moment of inertia \( I \) for a disk can be given by \( I = \frac{1}{2}MR^2 \). Since disks B and C have the same density and thickness, their masses can be determined in terms of their radii \( r \), using \( M = \text{density} \times \pi r^2 \times \text{thickness} \). Thus, \( M_B = k(0.25R)^2 \) and \( M_C = k(2R)^2 \), where \( k \) incorporates density and thickness.
03

Finding Angular Momentum of Each Disk

The angular momentum \( L \) of a disk is \( L = I \omega \). For disk B, \( L_B = \frac{1}{2} M_B (0.25R)^2 \cdot 2\omega_A \). Substituting \( M_B = k(0.25R)^2 \), we have \( L_B = \frac{1}{2} k (0.25R)^4 \cdot 2\omega_A \). For disk C, \( L_C = \frac{1}{2} M_C (2R)^2 \cdot \frac{\omega_A}{2} \). Substituting \( M_C = k(2R)^2 \), we have \( L_C = \frac{1}{2} k (2R)^4 \cdot \frac{\omega_A}{2} \).
04

Determine the Ratio of Angular Momentum

Using the expressions for \( L_B \) and \( L_C \) we found earlier, \( \frac{L_C}{L_B} = \frac{\frac{1}{2} k (2R)^4 \cdot \frac{\omega_A}{2}}{\frac{1}{2} k (0.25R)^4 \cdot 2\omega_A} \). Simplifying inside the fraction, \( \frac{(2)^4 \cdot \frac{1}{2}}{(0.25)^4 \cdot 2} \). The ratio is \( \frac{16}{0.0625 \cdot 2} = \frac{16}{0.125} = 128 \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins. It is represented by the symbol \( \omega \). The relationship between angular velocity and linear velocity is critical when objects rotate without slipping.
For rotating disks connected by belts, like in the problem, the condition of no slippage means equal linear velocities at the points of connection.
  • For Disk A and Disk C, this condition leads to the equation \( \omega_A R = \omega_C (2R) \).
  • Solving for \( \omega_C \), we get \( \omega_C = \frac{\omega_A}{2} \).
  • For Disk A and Disk B, the equation is \( \omega_B (0.25R) = \omega_A (0.5R) \), giving \( \omega_B = 2\omega_A \).
These equations show how different radii affect the angular velocities when connected by belts without slipping.
Moment of Inertia
The moment of inertia \( I \) is a measure of an object's resistance to changes in its rotational motion. In simpler terms, it tells us how hard it is to change the rotation of an object. For a solid disk, the formula for moment of inertia is \( I = \frac{1}{2}MR^2 \), where \( M \) is the mass and \( R \) is the radius.
In the given exercise:
  • Both disks \( B \) and \( C \) have the same material properties, which means they have the same density and thickness.
  • The mass for each disk can be expressed relative to its radius as \( M = k \times \text{radius}^2 \), where \( k \) is a constant representing density and thickness multiplied by \( \pi \).
  • This simplifies calculations and helps in understanding how radius differences impact the inertia of each disk.
Linear Velocity
Linear velocity is the distance an object travels per unit of time in a given direction. In rotational systems, linear velocity can be related to angular velocity and radius. The key equation linking them is \( v = \omega \times r \), where \( v \) is linear velocity, \( \omega \) is angular velocity, and \( r \) is the radius of the rotation path.
In the context of the exercise:
  • Belts ensure that the parts of the disks they contact move with the same linear velocity.
  • This relationship is crucial for connecting the disks' different angular and linear characteristics.
  • The linear velocity remains constant at the contact points, determining how one disk's angular velocity affects another's.
Understanding linear velocity in rotating systems is vital for solving problems involving rotational motion with interconnected components.
Rotational Motion
Rotational motion describes the motion of an object around a center or axis. It encompasses both angular and linear velocities as well as moment of inertia. In physics, it's important to know how different objects will rotate based on external and internal forces.
In the exercise provided, the rotational motion is understood through:
  • Angular momentum \( L = I \times \omega \), which combines moment of inertia and angular velocity.
  • Comparisons between disks based on their radii, which affect both their inertia and angular velocities.
  • The final calculation of angular momentum ratio, determining the relationship between the rotating disks.
Being able to dissect rotational motion into its components is essential for solving mechanical problems and understanding the dynamics of systems with interrelated parts.

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Most popular questions from this chapter

A \(30 \mathrm{~kg}\) child stands on the edge of a stationary merry-go-round of radius \(2.0 \mathrm{~m}\). The rotational inertia of the merry-go-round about its rotation axis is \(150 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The child catches a ball of mass \(1.0 \mathrm{~kg}\) thrown by a friend. Just before the ball is caught, it has a horizontal velocity \(\vec{v}\) of magnitude \(12 \mathrm{~m} / \mathrm{s}\), at angle \(\phi=37^{\circ}\) with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with posi- tion vector \(\vec{r}=(0.50 \mathrm{~m}) \hat{\mathrm{j}}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unitvector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) ?

Two skaters, each of mass \(50 \mathrm{~kg}\), approach each other along parallel paths separated by \(3.0\) \(\mathrm{m}\). They have opposite velocities of \(1.4 \mathrm{~m} / \mathrm{s}\) each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by \(1.0 \mathrm{~m}\). What then are (d) their angular speed and (e) the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy?

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 \(2.50 \mathrm{~kg}\) 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?
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