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\(\cdot\) A turntable that spins at a constant 78.0 rpm takes 3.50 s to reach this angular speed after it is turned on. Find (a) its angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2} ),\) assuming it to be constant, and (b) the number of degrees it turns through while speeding up.

Short Answer

Expert verified
Angular acceleration is approximately 2.333 rad/s², and it turns about 819 degrees while speeding up.

Step by step solution

01

Convert RPM to Radians per Second

First, convert the turntable's speed from revolutions per minute (RPM) to radians per second (rad/s). The formula to convert RPM to rad/s is:\[ \omega = \left(\frac{2\pi \times \text{RPM}}{60}\right) \]Substitute 78 for RPM:\[ \omega = \left(\frac{2\pi \times 78}{60}\right) = \frac{156\pi}{60} = \frac{78\pi}{30} \approx 8.168 \text{ rad/s} \]
02

Calculate Angular Acceleration

Use the angular kinematic equation for constant acceleration to find angular acceleration \(\alpha \).The equation is:\[ \omega = \omega_0 + \alpha t \]Since the turntable starts from rest, \(\omega_0 = 0\).Therefore,\[ 8.168 = 0 + \alpha \times 3.50 \]Solve for \(\alpha\):\[ \alpha = \frac{8.168}{3.50} \approx 2.333 \text{ rad/s}^2 \]
03

Calculate Number of Radians Turned

Use another kinematic equation to calculate the total angle \(\theta\) in radians the turntable turns through:The equation is:\[ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \]Since \(\omega_0 = 0\), it simplifies to:\[ \theta = \frac{1}{2}\times 2.333 \times (3.5)^2 \]\[ \theta = \frac{1}{2}\times 2.333 \times 12.25 \]\[ \theta = \frac{1}{2}\times 28.591 = 14.295 \text{ radians} \]
04

Convert Radians to Degrees

Finally, convert \(\theta\) from radians to degrees:Since there are \(180/\pi\) degrees in a radian,\[ \theta_{degrees} = \theta \times \frac{180}{\pi} \]\[ \theta_{degrees} = 14.295 \times \frac{180}{\pi} \approx 819.164 \text{ degrees} \]

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Key Concepts

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

Angular Kinematics
Angular kinematics deals with the motion of rotational bodies. It is similar in concept to linear kinematics, but instead of focusing on straight-line motion, it focuses on rotational or circular motion. In the context of our problem, we analyze how the turntable's angular velocity changes over time.

The relationship connecting angular velocity, angular acceleration, and time is crucial here. The equation \( \omega = \omega_0 + \alpha t \) is used to determine how quickly the turntable reaches its target angular speed, where \( \omega \) represents the final angular velocity, \( \omega_0 \) is the initial angular velocity, and \( \alpha \) is the angular acceleration.

Understanding these relationships allows us to predict how a rotating object behaves over time, including how much time it takes to speed up or slow down.
Radians to Degrees Conversion
Converting between radians and degrees is a common task in angular measurements. Radians and degrees are two different units for measuring angles. One full revolution around a circle is \( 2\pi \) radians, equivalent to 360 degrees. This means that 1 radian is approximately 57.2958 degrees.

The conversion formula is:
  • Degrees to radians: \( \text{radians} = \text{degrees} \times \frac{\pi}{180} \)
  • Radians to degrees: \( \text{degrees} = \text{radians} \times \frac{180}{\pi} \)
In our exercise, after calculating the angle in radians, we use this conversion to express the rotation in degrees, making it easier for us to grasp the actual rotation extent in more common terms.
Constant Acceleration
Constant acceleration implies that an object's rate of change of velocity is steady over time. For objects in rotational motion, constant angular acceleration indicates that the object's angular velocity increases evenly in a linear fashion.

In mathematical terms, if a turntable has a constant angular acceleration, it means \( \alpha \) remains the same throughout the turntable’s speed-up process. This simplifies calculations significantly, as it allows us to use straightforward kinematic equations to determine velocity and displacement over time, just as we did in the exercise when calculating the angular acceleration with the given formula.
Revolutions per Minute (RPM)
Revolutions per minute (RPM) is a unit of rotational speed or the number of turns completed by a rotating object in one minute. It is one of the most common units used to describe how fast something is spinning.

The first task in our problem involved converting RPM to radians per second to use in further calculations. RPM is an intuitive measure because it directly tells us how many full turns happen in a minute. However, when doing physics calculations involving rotational dynamics, we prefer radians per second since radian is the standard angular measurement in physics.

To convert RPM into radians per second, we use the formula \( \omega = \left(\frac{2\pi \times \text{RPM}}{60}\right) \), which brings the rotational speed into the realm of standard units for easier application of angular kinematic equations.

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Most popular questions from this chapter

A passenger bus in Zurich, Switzerland, derived its motive power from the energy stored in a large flywheel. Whenever the bus was stopped at a station, the wheel was brought up to speed with the use of an electric motor that could then be attached to the electric power lines. The flywheel was a solid cylinder with a mass of 1000 \(\mathrm{kg}\) and a diameter of \(1.80 \mathrm{m} ;\) its top angular speed was 3000 \(\mathrm{rev} / \mathrm{min.}\) At this angular speed, what was the kinetic energy of the flywheel?

A flywheel in a motor is spinning at 500.0 \(\mathrm{rpm}\) when a power failure suddenly occurs. The flywheel has mass 40.0 \(\mathrm{kg}\) and diameter 75.0 \(\mathrm{cm} .\) The power is off for 30.0 \(\mathrm{s}\) , and during this time the flywheel slown down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 200.0 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

\(\bullet\) (a) A cylinder 0.150 \(\mathrm{m}\) in diameter rotates in a lathe at 620 \(\mathrm{rpm} .\) What is the tangential speed of the surface of the cylinder? (b) The proper tangential speed for machining cast iron is about 0.600 \(\mathrm{m} / \mathrm{s} .\) At how many revolutions per minute should a piece of stock 0.0800 \(\mathrm{m}\) in diameter be rotated in a lathe to produce this tangential speed?

A flexible straight wire 75.0 \(\mathrm{cm}\) long is bent into the arc of a circle of radius 2.50 \(\mathrm{m} .\) What angle (in radians and degrees) will this arc subtend at the center of the circle?

A \(\mathrm{A} 392\) -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 J. Calculate \(h .\)

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