/*! 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 32 Turntable Two A record turntable... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 32

Turntable Two A record turntable is rotating at \(33 \frac{1}{3}\) rev/min. A watermelon seed is on the turntable \(6.0 \mathrm{~cm}\) from the axis of rotation. (a) Calculate the translational acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction, \(\mu^{\text {stat }}\), between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its rotational speed by starting from rest and undergoing a constant rotational acceleration for \(0.25 \mathrm{~s}\). Calculate the minimum \(\mu^{\text {stat }}\) required for the seed not to slip during the acceleration period.

Short Answer

Expert verified
Part (a): 0.233 m/s^2. Part (b): 0.023. Part (c): 0.086.

Step by step solution

01

Convert Rotational Speed to Radians per Second

The given rotational speed is \(33 \frac{1}{3}\) rev/min. Convert this to radians per second using the conversion factor \(2\pi\) rad/rev and 60 seconds per minute. \[33 \frac{1}{3}\text{ rev/min} = 33.33 \frac{1}{3}\text{ rev/min} = 33.33\text{ rev/min} = \frac{100}{3}\text{ rev/min} = \frac{100 \cdot 2\pi}{3 \cdot 60}\text{ rad/s} = \frac{200\pi}{180}\text{ rad/s} = \frac{10\pi}{9}\text{ rad/s} \approx 3.49\text{ rad/s}\]
02

Calculate the Translational Acceleration (Part a)

Use the formula for centripetal acceleration \(a = r\omega^2\), where \(r = 6.0\text{ cm} = 0.06\text{ m}\) and \( \omega = \frac{10\pi}{9} \text{ rad/s}\). \[a = r\omega^2 = 0.06 \left( \frac{10\pi}{9} \right)^2 = 0.06 \left( \frac{100\pi^2}{81} \right) \approx 0.233 \text{ m/s}^2 \]
03

Calculate the Minimum Coefficient of Static Friction (Part b)

To prevent slipping, the frictional force must equal the centripetal force. Use \(f = \frac{ma}{Mg} \) to find the minimum \( \text{渭}_{\text{stat}}\), where \(f\) is the friction force, \(m\) is the mass of the seed, \(M\) is the normal force, and \( g\) is acceleration due to gravity (9.8 m/s2). For the seed not to slip, \[\text{渭}_{\text{stat}} Mg = Ma \Rightarrow \text{渭}_{\text{stat}} = \frac{a}{g} = \frac{0.233}{9.8} \approx 0.023. \]
04

Calculate Angular Acceleration and Final Velocity for Acceleration Period (Part c)

Use the rotational kinematic equation with the initial angular velocity (\(\omega_0 = 0 \text{ rad/s}\)), final angular velocity \(\omega_f = 3.49 \text{ rad/s}\), and time \(t = 0.25 \text{ s}\). Use \(\alpha = \frac{\Delta \omega}{t}\) to find the angular acceleration during the acceleration period. \[\alpha = \frac{3.49}{0.25} \approx 13.96 \text{ rad/s}^2 \]
05

Calculate Translational Acceleration During Acceleration Period

Use \(a_t = r\alpha\) to find the translational acceleration \(a_t\) during the acceleration period: \[a_t = 0.06 \cdot 13.96 \approx 0.837\text{ m/s}^2 \]
06

Calculate the Minimum Coefficient of Static Friction During Acceleration Period

To prevent slipping, again use \(\text{渭}_{\text{stat}} = \frac{a_t}{g}\) for the acceleration period: \[\text{渭}_{\text{stat}} = \frac{0.837}{9.8} \approx 0.086 \]

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is the force that keeps an object moving in a circular path. When you think of a car going around a curve, the tendency for the passengers to be pushed to the side is due to centripetal acceleration. It is always directed toward the center of the circle. For this exercise, we find the centripetal acceleration using the formula: \[ a = r\theta^2 \]

Here, \( r \) is the radius from the turntable's center to the seed, and \( \theta \) is the angular velocity in radians per second. Using these values, we calculated the translational (or linear) acceleration for the watermelon seed. The primary takeaway is that the faster the object spins or the farther it is from the center, the higher its centripetal acceleration.
Static Friction
Static friction is the force that prevents an object from moving when it is resting against another surface. It must be overcome to start moving the object. For an object to stay in place on a rotating turntable, static friction needs to counteract the outward force due to centripetal acceleration. In this problem, we used the formula \[ \mu_{\text{stat}} = \frac{a}{g} \]

where \( \mu_{\text{stat}}\) represents the coefficient of static friction, \( a \) is the centripetal acceleration, and \( g \) is the acceleration due to gravity. This calculation helps us determine the minimum coefficient of static friction required to keep the seed from slipping off the turntable during rotation.
Angular Acceleration
Angular acceleration measures how quickly an object speeds up its rotation. When the turntable goes from rest to full speed, it undergoes angular acceleration. The formula for angular acceleration is: \[ \alpha = \frac{\Delta \omega}{t} \]

where \( \Delta \omega \) is the change in angular velocity, and \( t \) is the time it takes to reach that velocity. Once we have the angular acceleration, we can find the translational acceleration using the formula: \[ a_t = r\alpha \]

In our problem, these calculations help us determine how much grip the seed needs during the initial period when the turntable starts spinning. This helps us find the minimum coefficient of static friction during the acceleration period, ensuring the seed stays in place without slipping. Understanding angular acceleration is key to solving rotational motion problems effectively.

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Most popular questions from this chapter

Cylinder Rotates about Horizontal A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the rotational speed of the cylinder as it passes through its lowest position?

Vinyl Record A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) rev/min. (a) What is its rotational speed in radians per second? What is the translational speed of a point on the record at the needle when the needle is (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

Meter Stick Held Vertically A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)

Uniform Spherical Shell A uniform spherical shell of mass \(M\) and radius \(R\) rotates about a vertical axis on frictionless bearings (Fig. 11-39). A massless cord passes around the equator of the shell. over a pulley of rotational inertia \(I\) and radius \(r\), and is attached to a small object of mass \(m\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it falls a distance \(h\) from rest? Use energy considerations.

Thin Rod of length \(L\) A thin rod of length \(L\) and mass \(m\) is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with rotational speed \(\omega\). In terms of these symbols and \(g\), and neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
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