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Chapter 10: Problem 22

An astronaut is being tested in a centrifuge. The centrifuge has a radius of \(10 \mathrm{~m}\) and, in starting, rotates according to \(\theta=0.30 t^{2}\), where \(t\) is in seconds and \(\theta\) is in radians. When \(t=5.0 \mathrm{~s}\), what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?

Short Answer

Expert verified
Angular velocity: 3.0 rad/s; Linear velocity: 30 m/s; Tangential acceleration: 0.60 m/s虏; Radial acceleration: 90 m/s虏.

Step by step solution

01

Find Angular Velocity

The angular displacement given is \(\theta=0.30 t^{2}\). The angular velocity \(\omega\) can be found by differentiating \(\theta\) with respect to \(t\). Thus: \[\omega = \frac{d\theta}{dt} = \frac{d}{dt}(0.30t^2) = 0.60t\]At \(t = 5.0\) seconds, \[\omega = 0.60 \times 5.0 = 3.0\, \text{rad/s}\].
02

Calculate Linear Velocity

The linear velocity \(v\) is related to angular velocity by \(v = r\omega\). Given \(r = 10\, m\), and \(\omega = 3.0\, \text{rad/s}\):\[ v = 10 \times 3.0 = 30\, \text{m/s} \].
03

Determine Tangential Acceleration

Tangential acceleration \(a_t\) is found by differentiating \(\omega\) with respect to \(t\):\[ a_t = \frac{d\omega}{dt} = \frac{d}{dt}(0.60t) = 0.60\, \text{m/s}^2 \].
04

Calculate Radial Acceleration

Radial acceleration \(a_r\) is given by the formula \(a_r = r\omega^2\). Using \(r = 10\, m\) 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, often denoted by \( \omega \), measures how quickly an object is rotating. Think of it as the speed of rotation, with a specific focus on how many radians are covered in a given amount of time. Radians are the standard unit of angular measurement, similar to degrees but very suitable for calculations in physics.
  • In the given problem, the angular displacement is expressed as \( \theta=0.30 t^{2} \).
  • To find angular velocity, we take the derivative of \( \theta \) with respect to time \( t \), resulting in \( \omega = 0.60 t \).
At \( t = 5.0 \) seconds, substituting the time into our velocity formula gives \( \omega = 3.0 \) rad/s. That means, even though \( t \) is changing, \( \omega \) depends linearly on \( t \), showing a consistent increase in rotational speed over time.
Tangential Acceleration
Tangential acceleration \( a_t \) represents the change in linear speed along the path of a rotating object. It's directly tied to the change in angular velocity over time.
  • Since angular velocity \( \omega \) for our exercise is \( 0.60 t \), finding \( a_t \) involves differentiating \( \omega \) with respect to time.
  • The calculation yields \( a_t = 0.60 \) m/s\(^2\).
This means that every second, the tangential speed experienced by the astronaut increases by 0.60 m/s. Think of \( a_t \) as how much faster the astronaut feels like he's being pushed along the circle's edge over time.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, keeps an object moving along a circular path. It acts towards the center of the circle and isn't responsible for changes in speed but direction. The formula used is: \( a_r = r\omega^2 \).
  • In this setup, \( r = 10 \) m and previously found \( \omega = 3.0 \) rad/s.
  • Substituting values, we compute \( a_r = 10 \times (3.0)^2 = 90 \) m/s\(^2\).
A high radial acceleration signifies a strong force needed to turn the object. It's the reason why astronauts in centrifuges feel so much pressure towards one side.
Linear Velocity
Linear velocity \( v \) is the straight-line speed of an object moving along its path in circular motion. It is derived directly from angular velocity through the formula: \( v = r\omega \).
  • With \( r = 10 \) m and \( \omega = 3.0 \) rad/s, \( v \) is calculated to be \( v = 30 \) m/s.
For a person in the centrifuge, linear velocity describes how fast they are moving if the path was a straight line. Imagine the path unwinding to a straight trajectory; 30 m/s indicates quite a speed, like going from a curve to a fast straight drive!

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