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Chapter 10: Problem 22

An astronaut is tested in a centrifuge with radius \(10 \mathrm{~m}\) and rotating according to \(\theta=0.30 t^{2} .\) At \(t=5.0 \mathrm{~s},\) what are the magnitudes of the (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?

Short Answer

Expert verified
(a) 3.0 rad/s, (b) 30 m/s, (c) 0.60 m/s虏, (d) 90 m/s虏.

Step by step solution

01

Find Angular Velocity

The angular position is given by \( \theta = 0.30t^2 \). To find the angular velocity \( \omega \) at a specific time, we take the derivative of \( \theta \) with respect to time \( t \): \[\omega = \frac{d\theta}{dt} = \frac{d}{dt}(0.30t^2) = 0.60t.\]Thus, at \( t = 5.0 \text{ s} \), the angular velocity is \[\omega = 0.60 \times 5.0 = 3.0 \text{ rad/s}.\]
02

Calculate Linear Velocity

The linear velocity \( v \) is related to angular velocity \( \omega \) by the formula:\[v = r\omega,\]where \( r = 10 \text{ m} \) is the radius of the centrifuge. Substituting the values:\[v = 10 \times 3.0 = 30 \text{ m/s}.\]
03

Determine Tangential Acceleration

Tangential acceleration \( a_t \) is given by the derivative of angular velocity with respect to time:\[a_t = \frac{d\omega}{dt} = \frac{d}{dt}(0.60t) = 0.60 \text{ m/s}^2.\]At any time including \( t = 5.0 \text{ s} \), the tangential acceleration is \( 0.60 \text{ m/s}^2 \).
04

Evaluate Radial (Centripetal) Acceleration

Radial or centripetal acceleration \( a_r \) is given by:\[a_r = r\omega^2.\]Substitute \( r = 10 \text{ m} \) and \( \omega = 3.0 \text{ rad/s} \):\[a_r = 10 \times (3.0)^2 = 90 \text{ m/s}^2.\]

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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 quickly an object rotates or spins around an axis. It is represented by the symbol \( \omega \) and is typically measured in radians per second (rad/s). In this exercise, the angular position \( \theta \) of the centrifuge is given by the equation \( \theta = 0.30t^2 \). To find the angular velocity, you need to determine the rate of change of this angle over time.
  • The formula to find angular velocity is \( \omega = \frac{d\theta}{dt} \).
  • By differentiating, we obtain \( \omega = 0.60t \), which shows how the velocity changes with time.
At time \( t = 5.0 \) seconds, substitute \( t = 5.0 \) into this equation to get \( \omega = 3.0 \text{ rad/s} \). This indicates how fast the centrifuge is spinning at that specific moment.
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of rotation. It is crucial for maintaining an object鈥檚 circular motion. In this context, it鈥檚 represented as \( a_r \). You can calculate it using the formula:
  • \( a_r = r\omega^2 \).
  • This formula tells us that centripetal acceleration depends on both the radius \( r \) and the square of the angular velocity \( \omega \).
With a radius of 10 meters and \( \omega = 3.0 \text{ rad/s} \), the centripetal acceleration of the centrifuge is \( a_r = 90 \text{ m/s}^2 \). Remember, this acceleration is always directed towards the center of the circle to keep the object moving in a circular path.
Linear Velocity
Linear velocity refers to the rate at which an object moves along a path. In circular motion, it is related to the angular velocity. Linear velocity \( v \) is determined by the relationship:
  • \( v = r\omega \), where \( r \) is the radius of the circle, and \( \omega \) is the angular velocity.
For the centrifuge with a radius of 10 meters and \( \omega = 3.0 \text{ rad/s} \), the linear velocity is \( v = 30 \text{ m/s} \). This calculation shows how fast the astronaut is moving along the circular path of the centrifuge. Linear velocity gives the actual speed along the path, unlike angular velocity which is a measure of rotational speed.
Tangential Acceleration
Tangential acceleration denotes the rate of change of the linear velocity concerning time along the circular path. It is particularly significant when the speed of the rotating object is changing. Represented by \( a_t \), it is derived from angular velocity:
  • \( a_t = \frac{d\omega}{dt} \).
  • In this exercise, \( a_t \) is calculated as a constant value of 0.60 m/s2, indicating uniform change over time.
This implies that as the centrifuge continues rotating, the linear velocity of the astronaut will increase at this constant rate of acceleration. Tangential acceleration is crucial for understanding how an object's speed changes along the tangent to the circle, especially when accounting for any increases in rotational speed.

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Most popular questions from this chapter

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0,\) what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s} ?\) (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. \(10-47) .\) A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm},\) and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\) There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

A flywheel turns through 40 rev as it slows from an angular speed of \(1.5 \mathrm{rad} / \mathrm{s}\) to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?

Attached to each end of a thin steel rod of length \(1.20 \mathrm{~m}\) and mass \(6.40 \mathrm{~kg}\) is a small ball of mass \(1.06 \mathrm{~kg}\). The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at \(39.0 \mathrm{rev} / \mathrm{s}\). Because of friction, it slows to a stop in \(32.0 \mathrm{~s}\). Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the \(32.0 \mathrm{~s}\). (e) Now suppose that the retarding torque is known not to be constant. If any of the quantities (a), (b), (c), and (d) can still be computed without additional information, give its value.

A drum rotates around its central axis at an angular velocity of \(12.60 \mathrm{rad} / \mathrm{s} .\) If the drum then slows at a constant rate of \(4.20 \mathrm{rad} / \mathrm{s}^{2}\) (a) how much time does it take and (b) through what angle does it rotate in coming to rest?
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