/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 In 1980 , over San Francisco Bay... [FREE SOLUTION] | 91Ó°ÊÓ

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In 1980 , over San Francisco Bay, a large yo-yo was released from a crane. The \(116 \mathrm{~kg}\) yo-yo consisted of two uniform disks of radius \(32 \mathrm{~cm}\) connected by an axle of radius \(3.2 \mathrm{~cm} .\) What was the magnitude of the acceleration of the yo-yo during (a) its fall and (b) its rise? (c) What was the tension in the cord on which it rolled? (d) Was that tension near the cord's limit of \(52 \mathrm{kN}\) ? Suppose you build a scaled-up version of the yo-yo (same shape and materials but larger). (e) Will the magnitude of your yo-yo's acceleration as it falls be greater than, less than, or the same as that of the San Francisco yo-yo? (f) How about the tension in the cord?

Short Answer

Expert verified
(a) and (b) Acceleration is 0.62 m/s² during fall and rise. (c) Tension is 1064.88 N. (d) No, it's within limits. (e) Acceleration is the same. (f) Tension is greater with increased size.

Step by step solution

01

Find Yo-Yo's Moment of Inertia

The moment of inertia for a cylindrical disk is given by \( I = \frac{1}{2} m r^2 \). Since there are two disks, \( I_{disks} = 2 \times \frac{1}{2} m r_d^2 = m r_d^2 \). The yo-yo's axle also has a moment of inertia but is much less and can be neglected here assuming a focus on the significant parts. Where \( m = 116 \, \text{kg} \), \( r_d = 0.32 \, \text{m} \), the moment of inertia is \[ I = 116 \, \text{kg} \times (0.32 \, \text{m})^2 = 11.8784 \, \text{kg} \cdot \text{m}^2. \]
02

Find Acceleration During Fall

Using the equation \( au = I \alpha \) (torque and angular acceleration relation) and knowing that torque \( \tau = r_a \times T \), the equation becomes \( T \times r_a = I \times \alpha \). We know \( a = \alpha \times r_a \), which becomes \( T = m(g - a) \). Solving these simultaneous equations gives the acceleration during the fall:\[ a = \frac{m g}{m + \frac{I}{r_a^2}} \]Using \( r_a = 0.032 \, \text{m} \):\[ a = \frac{116 \times 9.8}{116 + \frac{11.8784}{(0.032)^2}} \approx 0.62 \, \text{m/s}^2 \]
03

Find Acceleration During Rise

The scenario is symmetrical but reversed; hence the magnitude of acceleration remains the same, as the same calculations apply to a different direction of motion. Therefore:\[ a = 0.62 \, \text{m/s}^2 \]
04

Calculate Tension in the Cord

Using the expression \( T = m(g - a) \), with \( a = 0.62 \, \text{m/s}^2 \), calculate the tension during fall (or rise):\[ T = 116 \times (9.8 - 0.62) = 116 \times 9.18 \approx 1064.88 \, \text{N} \]
05

Compare Tension to Cord's Limit

Convert the known cord tension limit into Newtons:\( 52 \, \text{kN} = 52000 \, \text{N} \). The calculated tension \( \approx 1064.88 \, \text{N} \) is well under the limit, thus safe.
06

Scaled-Up Yo-Yo Acceleration

For a scaled-up yo-yo, acceleration depends on the ratio \( \frac{r_a}{r_d} \). The ratio remains consistent due to scaling, implying the acceleration magnitude remains unchanged as mass, radius relationships are maintained.
07

Scaled-Up Yo-Yo Cord Tension

The tension in the cord will scale with mass, which increases; hence it will be greater than the original \( 1064.88 \, \text{N} \) though always ensures not exceeding the cord’s given capacity like the larger 52 kN capacity.

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Key Concepts

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

Torque
Torque is a fundamental concept in rotational dynamics. It is the rotational equivalent of linear force, causing an object to rotate around a pivot point or axis. Imagine twisting a door knob or yanking the hand of a clock; you're applying torque in these scenarios. Mathematically, torque \( \tau \) is defined as the product of a force \( F \) and the distance \( r \) from the pivot point, expressed as:
  • \( \tau = F \times r \)
This means if you push a door far from its hinges (the pivot), the door turns more easily due to the increased torque. In the context of a yo-yo, torque is created by the tension in the string acting at a radius. It is essential for determining the yo-yo's angular acceleration as it rolls and unrolls.
Moment of Inertia
The moment of inertia, often symbolized as \( I \), needs a bit of understanding as it is akin to mass in linear motion but for rotation. It represents how easily an object with a particular mass distribution will begin rotating about an axis. For a cylindrical disk, the calculation simplifies to:
  • \( I = \frac{1}{2} m r^2 \)
Here, \( m \) is the mass of the disk and \( r \) is the radius. The combination of the two disks of a yo-yo would directly influence how it spins down and up the string. More mass or a larger radius means a larger moment of inertia, making it harder to start or stop spinning.
Angular Acceleration
Angular acceleration \( \alpha \) is all about how quickly an object's rotational speed changes. Much like how linear acceleration describes an increase in linear speed, angular acceleration captures the rate of change in rotational velocity. The relationship between torque \( \tau \) and angular acceleration \( \alpha \) is rooted in Newton’s second law in rotational form:
  • \( \tau = I \alpha \)
This helps us calculate how quickly or slowly the yo-yo begins to rotate as influenced by the torque from the string tension. Given the moment of inertia from a previous section, angular acceleration is crucial in determining the yo-yo’s movement characteristics.
Cylindrical Disk
A cylindrical disk is a simple yet beautiful geometry favoring rotation, which the yo-yos famously utilize. As seen in our yo-yo problem, each half of the yo-yo is a uniform cylindrical disk. This shape allows for easy calculation of properties such as the moment of inertia, important for understanding rotational dynamics. For a disk:
  • It generally has a uniform mass distribution.
  • It rotates easily about a central axis.
  • It is defined by parameters such as radius and thickness.
These characteristics ensure that all the formulas and dynamics of rotational movement apply neatly to the yo-yo's disks.
Tension Force
Tension force in the context of a yo-yo relates to the force propagated along the cord that’s holding or rotating the disk. This force arises when the cord is pulled taught; as the yo-yo falls or rises, tension adjusts to balance gravitational forces and promote rotation. The significant elements include:
  • Tension provides the radial force needed for spinning.
  • Determines torque as it acts along the radius of the axle.
  • Calculated by balancing gravitational force against net force required for acceleration.
Understanding the tension in the cord allows one to predict the yo-yo’s behavior under stress and ensure safe operation, especially when considering scaled yo-yo versions within the limits of tensile strength.

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Most popular questions from this chapter

Figure 11-51 shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

Aparticle moves through an \(x y z\) coordinate system while a force acts on the particle. When the particle has the position vector \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}+(2.00 \mathrm{~m}) \hat{\mathrm{k}}\), the force is given by \(\vec{F}=F_{x} \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}-(6.00 \mathrm{~N}) \hat{\mathrm{k}}\) and the corresponding torque about the origin is \(\vec{\tau}=(4.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{i}}+(2.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{j}}-(1.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{k}}\). Determine \(F_{x^{*}}\)

A body of radius \(R\) and mass \(m\) is rolling smoothly with speed \(v\) on a horizontal surface. It then rolls up a hill to a maximum height \(h\). (a) If \(h=3 v^{2} / 4 g\), what is the body's rotational inertia about the rotational axis through its center of mass? (b) What might the body be?

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