/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 126 A partially full can of soda is ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 126

A partially full can of soda is placed at the outer edge of a child's merry- go-round, located \(R=5\) ft from the axis of rotation. The can diameter and height are 2.5 in. and 5 in. respectively. The can is half full, and the soda has specific gravity \(\mathrm{SG}=1.05 .\) Evaluate the slope of the liquid surface in the can if the merry-go-round spins at 20 rpm. Calculate the spin rate at which the can would spill, assuming no slippage between the can bottom and the merry-go-round. Would the can most likely spill or slide off the merry-go-round?

Short Answer

Expert verified
The slope of the liquid surface in the can when the merry-go-round spins at 20 rpm is approximately 0.688. The can would start to spill the soda if the rotation speeds reach 12.24 rpm. Without further information, we can hypothesize that the soda would most likely to spill before the can slides off the merry-go-round.

Step by step solution

01

Conversion of Units

Firstly, to have consistency, let's convert all measurements to SI units. Given that 1 foot equals 0.3048 meters, \(R=5\) ft can be converted to \(R=5*0.3048 = 1.52\) meters. The dimensions of the can, which has a diameter of 2.5 inches and height of 5 inches, need to be converted to meters as well. Using a conversion factor of 0.0254 meters per inch, the diameter becomes \(0.0635\) meters and the height becomes \(0.127\) meters.
02

Calculation of Slope of Liquid

It's given that the merry-go-round is rotating at 20 rpm. We will convert it to rad/s (the SI unit for angular velocity) using the equation \( \omega = \frac{2\pi*n}{60} \), where \(n\) is the rotational speed in rpm. This conversion gives us \( \omega = \frac{2\pi*20}{60} = 2.09 \, rad/s\). The slope of the liquid surface in the can (which will not be horizontal due to the rotation) can be determined using the equation for the angle of tilt \( \tan \theta = \frac{R \omega^2 }{ g} \), where \(g\) is the acceleration due to gravity (\(9.81 \, m/s^2\)). Plugging in the given values gives us \( \tan \theta = \frac{1.52 * (2.09)^2 }{ 9.81} = 0.688 \). Therefore, the slope of the liquid's surface is approximately 0.688.
03

Calculation of Spilling Rate

The spinning rate at which the soda would spill can be calculated using the dimensions of the can. The can will start to spill when the angle of tilt equals the angle of the can which is \( \tan \beta = \frac{1}{2}*\frac{diameter}{height} = \frac{1}{2}*\frac{0.0635}{0.127} = 0.25 \). To calculate the spin rate we can rearrange the equation for angle of tilt to \( \omega = \sqrt{\frac{g \tan \beta}{R}} \). Substituting the values gives \( \omega = \sqrt{\frac{9.81 * 0.25}{1.52}} = 1.28 \, rad/s \). This angular speed can be converted back to rpm using the transformation \( n = \frac{\omega * 60}{2 \pi} \), resulting in \( n = \frac{1.28 * 60}{2 \pi} = 12.24 \, rpm \). Therefore, the can would spill at a speed of 12.24 rpm.
04

Prediction of Spill or Slide

The final task is to check whether the can would spill or slide off the merry-go-round. The can will slide off if the frictional force is less than the centrifugal force. This frictional force depends on the coefficient of friction, which we don't have enough information to calculate. Therefore, we can't definitively answer this part of the problem. In this scenario, with the given information, it is most probable to assume that the can would spill, rather than slide.

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