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An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

Short Answer

Expert verified
The final angular velocity is [Result from Step 1] in rad/s. The turntable has spun approximately [Result from Step 2] revolutions. The tangential speed at any point on the rim at t=0.200s is [Result from Step 3] m/s. The magnitude of the resultant acceleration on the rim at t=0.200s is [Result from Step 4] m/s\(^2\).

Step by step solution

01

Compute the Final Angular Velocity

The final angular velocity \( \omega_f \) after a certain time \( t \) can be obtained from the formula \( \omega_f = \omega_i + \alpha t \), where \( \omega_i \) is the initial angular velocity and \( \alpha \) is the angular acceleration. Substituting the given values \( \omega_i = 0.250 \) rev/s, \( \alpha = 0.900 \) rev/s\(^2\) and \( t = 0.200 \) s results in \( \omega_f = 0.250 + 0.900(0.200) \). Note that the angular velocities are in rev/s and should be converted to rad/s (\(1 rev = 2 \pi \) rad). The conversion gives the final result \( \omega_f \) in rad/s.
02

Compute Total Revolutions

The total angular displacement \( \theta \) or total number of revolutions can be calculated using the formula \( \theta = \omega_i t + 0.5 \alpha t^2 \), substituting the given values \( \omega_i = 0.250 \) rev/s, \( \alpha = 0.900 \) rev/s\(^2\) and \( t = 0.200 \) s gives \( \theta = 0.250(0.200) + 0.5(0.900)(0.200)^2 \). The answer gives the total number of revolutions the turntable has spun in \(0.200 \) s.
03

Compute Tangential Speed

Tangential speed \(v_t \) of a point on the rim of the turntable can be calculated using the formula \( v_t = r \omega \), where \( r \) is the radius of the turntable and \( \omega \) is the angular velocity. Substituting the given radius \( r = 0.750/2\) m (half the diameter) and the calculated angular velocity from Step 1 gives \( v_t = r \omega \). The answer will be in m/s.
04

Compute Resultant Acceleration

The resultant acceleration (\( a \)) of a point on the rim is given by the formula \(a = \sqrt{(r \alpha)^2 + (v_t)^2} \), where \( r \) is the radius, \( \alpha \) is the angular acceleration and \( v_t \) is the tangential speed. Substituting the earlier found values from steps 1 and 3, we get \( a = \sqrt{(r \alpha)^2 + (v_t)^2} \) which gives the resultant acceleration in m/s\(^2\).

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Key Concepts

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

Angular Acceleration
Imagine you're on a merry-go-round that’s starting to pick up speed. The rate at which the spinning gets quicker is a concept we call angular acceleration, symbolized by the Greek letter alpha (\( \alpha \)). In physics, angular acceleration is a measure of how fast the rotational speed of an object is changing over time. This is analogous to linear acceleration in straight-line motion, but instead of speed increasing, it's the number of rotations per unit time.

To put it mathematically, angular acceleration is the change in angular velocity (\( \Delta \omega \)) over the change in time (\( \Delta t \)), expressed as \( \alpha = \frac{\Delta \omega}{\Delta t} \). This concept is essential when analyzing the behavior of rotating systems, like an electric turntable in our exercise, which has a constant angular acceleration. It allows us to predict future motion and calculate other attributes like tangential speed or the total number of revolutions.
Tangential Speed
Now let's focus on the tip of a record player as it spins. The speed at which this point moves in a straight line is the tangential speed. It's the linear speed of a point situated at a radius (\( r \)), along the circumference of a circle that is in rotational motion.

The formula to calculate tangential speed (\( v_t \)) is given by multiplying the radius (\( r \)) of the circular path with the angular velocity (\( \omega \)): \( v_t = r \omega \). This is invaluable when analyzing a point on the rim of the turntable, as it tells us how fast the point is moving along its circular path. The larger the radius, or the faster the angular velocity, the greater the tangential speed.
Resultant Acceleration
Consider a car rounding a curve. We have two factors at play: the change in the speed of the car and the change in direction. Similarly, a point on a spinning object experiences a change in speed from tangential acceleration and a change in direction from radial, or centripetal, acceleration. The combination of these two, at any given moment, gives us the resultant acceleration.

Mathematically, we find this by considering both the radial (\( r \alpha \)) and tangential components (\( v_t \)) of acceleration. The resultant acceleration (\( a \)) is calculated using the Pythagorean theorem: \( a = \sqrt{(r \alpha)^2 + (v_t)^2} \). It defines the total acceleration of a point on the object, which encapsulates both the increasing speed in the direction of motion and the changing direction as the point revolves.
Angular Displacement
If we trace the path of a location on the turntable, starting at one point and coming back to the same angular position, we've created a circle. Angular displacement is this angle, which a rotating object has moved through during its motion, and is usually measured in radians or revolutions.

The formula often used for angular displacement (\( \theta \)) when there is constant angular acceleration is \( \theta = \omega_i t + 0.5 \alpha t^2 \), where \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time interval. This equation takes into account both the initial movement and the increase in rotation due to acceleration. In the scenario with our turntable, it helps determine the number of revolutions it completes in a given time frame.

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Most popular questions from this chapter

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{~m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are \(25.0 \mathrm{~mm}\) and \(58.0 \mathrm{~mm}\), respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). (a) What is the angular speed of the \(\mathrm{CD}\) when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 min playing time? Take the direction of rotation of the disc to be positive.

A disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)

A flywheel with radius \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2} .\) For a point on the rim of the flywheel, what are the magnitudes of the tangential, radial, and resultant accelerations after \(2.00 \mathrm{~s}\) of acceleration?

A thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M,\) find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

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