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A turntable rotating at 78 rev/min slows down and stops in \(32 \mathrm{~s}\) after the motor is turned off. \((a)\) Find its (constant) angular acceleration in rev \(/ \mathrm{min}^{2} .(b)\) How many revolutions does it make in this time?

Short Answer

Expert verified
The angular deceleration is -146.17 rev/min^2 and the turntable makes approximately 42 revolutions before coming to a stop.

Step by step solution

01

Convert Units

Firstly, convert the time from seconds to minutes, so it is the same unit as given in rev/min. Thus, the time will be 32/60 = 0.533 min.
02

Find Angular Acceleration

Use the formula for acceleration \( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{\omega_{f} - \omega_{i}}{\Delta t} \). From the problem, we know that \( \omega_{i} = 78 \ rev/min \) and \( \omega_{f} = 0 \ rev/min \). Substitute the values and we have \( \alpha = \frac{0 - 78}{0.533} = -146.17 \ rev/min^2 \). The acceleration is negative, indicating a slowdown.
03

Find Total Number of Revolutions

Use the displacement formula \( \Delta \theta = \omega_{i} \Delta t+0.5 \alpha \Delta t^{2} \). Substituting the values we have \( \Delta \theta = 78 * 0.533 + 0.5*(-146.17) * (0.533)^2 = 41.51 \ rev \). The negative term in the formula comes from the slowing down of the turntable. Hence, the turntable makes approximately 42 revolutions before coming to a stop.

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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 is spinning around an axis. It tells us the rate at which the angle changes with respect to time. For a circular object like a turntable, angular velocity is often given in revolutions per minute (rev/min). This unit tells us how many complete circles the object makes in one minute.
  • Initial Angular Velocity: When the motor of a turntable is turned off, it might have started at a high angular velocity. In our example, the initial angular velocity is 78 rev/min. This means that initially, the turntable was making 78 full rotations every minute.
  • Final Angular Velocity: This is the angular velocity of the turntable when it comes to a complete stop. Since it stops, the final angular velocity becomes 0 rev/min.
Understanding angular velocity is crucial for determining how fast or slow something spins, and it's a key factor in solving problems involving rotational motion.
Revolutions
A revolution is one complete turn around a circle. When we say that something rotates, revolutions are the measure of how many complete circles it makes. In the context of a turntable, it is crucial to know how many revolutions are completed over a period of time.
  • Calculating Revolutions: To find out how many revolutions a turntable makes while slowing down, we use the displacement formula. This formula accounts for both the initial rate of spinning and the deceleration after the motor is turned off.
  • Total Revolutions: In the example given, the turntable completes around 42 full rotations before it stops completely.
Revolutions are important because they help us quantify rotational motion, especially in mechanical systems like motors and gears.
Time Conversion
Time conversion is an essential step when dealing with unit-based calculations, particularly when different units of time are involved. In rotational motion problems, you often need to convert time into consistent units to ensure formulas function correctly.
  • From Seconds to Minutes: When dealing with angular velocity in rev/min, we must convert the time from seconds to minutes to keep the units consistent. In the example, 32 seconds was converted to approximately 0.533 minutes by dividing 32 by 60.
  • Maintaining Consistent Units: Always ensure the time units used in calculations match those used for velocity or displacement. This prevents discrepancies in the final results.
Converting units properly is crucial for accurate calculations, as it ensures that all parts of your equations align through the same units of measurement.
Displacement Formula
The displacement formula in rotational dynamics helps us calculate the total angle traversed by an object in motion. It is similar to how we compute distances in linear motion but applied to rotation.
The formula used is:\[ \Delta \theta = \omega_{i} \Delta t + 0.5 \alpha \Delta t^{2} \]
  • \(\omega_{i}\): This is the initial angular velocity. It sets the initial rate of spin for calculations.
  • \(\alpha\): Represents the angular acceleration, which in the given problem, is negative due to deceleration.
  • \(\Delta \theta\): Total angular displacement, which tells us the total number of revolutions made.
This formula considers both the initial speed and the change in speed over time to accurately calculate how much rotation occurs. It's particularly essential for complex calculations where both velocity and acceleration play significant roles.

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Most popular questions from this chapter

A good baseball pitcher can throw a baseball toward home plate at \(85 \mathrm{mi} / \mathrm{h}\) with a spin of 1800 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 -ft trajectory is a straight line.

A rigid body, starting at rest, rotates about a fixed axis with constant angular acceleration \(\alpha\). Consider a particle a distance \(r\) from the axis. Express \((a)\) the radial acceleration and \((b)\) the tangential acceleration of this particle in terms of \(\alpha, r\), and the time \(t .(c)\) If the resultant acceleration of the particle at some instant makes an angle of \(57.0^{\circ}\) with the tangential acceleration, through what total angle has the body rotated from \(t=0\) to that instant?

A certain wheel turns through 90 rev in \(15 \mathrm{~s}\), its angular speed at the end of the period being \(10 \mathrm{rev} / \mathrm{s} .(a)\) What was the angular speed of the wheel at the beginning of the 15 -s interval, assuming constant angular acceleration? \((b)\) How much time had elapsed between the time the wheel was at rest and the beginning of the 15 -s interval?

An object moves in the \(x y\) plane such that \(x=R \cos \omega t\) and \(y=R \sin \omega t\). Here \(x\) and \(y\) are the coordinates of the object, is the time, and \(R\) and \(\omega\) are constants. (a) Eliminate \(t\) between these equations to find the equation of the curve in which the object moves. What is this curve? What is the meaning of the constant \(\omega ?(b)\) Differentiate the equations for \(x\) and \(y\) with respect to the time to find the \(x\) and \(y\) components of the velocity of the body, \(v_{x}\) and \(v_{y}\). Combine \(v_{x}\) and \(v_{y}\) to find the magnitude and direction of \(v .\) Describe the motion of the object. ( \(c\) ) Differentiate \(v_{x}\) and \(v_{y}\) with respect to the time to obtain the magnitude and direction of the resultant acceleration.

A speedometer on the front wheel of a bicycle gives a reading that is directly proportional to the angular speed of the wheel. Suppose that such a speedometer is calibrated for a wheel of diameter \(72 \mathrm{~cm}\) but is mistakenly used on a wheel of diameter \(62 \mathrm{~cm} .\) Would the linear speed reading be wrong? If so, in what sense and by what fraction of the true speed?

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