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Chapter 5: Problem 58

A biologist is studying plant growth and wants to simulate a gravitational field twice as strong as Earth's. She places the plants on a horizontal rotating table in her laboratory on Earth at a distance of \(12.5 \mathrm{cm}\) from the axis of rotation. What angular speed will give the plants an effective gravitational field \(\overrightarrow{\mathrm{g}}_{\mathrm{eff}},\) whose magnitude is \(2.0 \mathrm{g} ?\) \([\) Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the apparent weight. \(]\)

Short Answer

Expert verified
The angular speed that will give the plants an effective gravitational field of 2.0g is approximately √78.48 rad/s.

Step by step solution

01

Write the expression for centripetal acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is given by the formula \(a_c = \omega^2 r\), where \(\omega\) is the angular speed and \(r\) is the radius, which is the distance from the axis of rotation. For the given problem, we have \(r = 12.5 \ \mathrm{cm} = 0.125 \ \mathrm{m}\).
02

Find the gravitational field on Earth

We are given that the artificial gravity should be twice the strength of Earth's gravity. We know that the gravitational field strength on Earth is approximately \(g = 9.81 \ \mathrm{m/s^2}\). So, we want the plants' effective gravitational field to be \(g_{\text{eff}} = 2 \times g = 2 \times 9.81 = 19.62 \ \mathrm{m/s^2}\).
03

Find centripetal acceleration of the plants

The effective gravitational field caused by the artificial gravity and Earth's gravity can be determined through centripetal acceleration. We are given that \(g_{\text{eff}} = 19.62 \ \mathrm{m/s^2}\). From this, we want to find the centripetal acceleration, given that \(a_c = g_{\text{eff}} - g\). Plugging in the values, we get: $$ a_c = 19.62 - 9.81 = 9.81 \ \mathrm{m/s^2} $$
04

Solving for angular speed

Now, we know the centripetal acceleration and the radius; we need to find the angular speed. Using the formula \(a_c = \omega^2 r\), we can solve for \(\omega\). Rearranging for \(\omega\) gives us: $$ \omega = \sqrt{\frac{a_c}{r}} $$ Plugging in the values we have: $$ \omega = \sqrt{\frac{9.81}{0.125}} = \sqrt{78.48} \ \mathrm{rad/s} $$ So, the angular speed that will give the plants an effective gravitational field of \(2.0g\) is approximately \(\omega = \sqrt{78.48} \ \mathrm{rad/s}\).

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