This text is meant to accompany class discussions. It is not everything there is to know about uniform circular motion. It is meant as a prep for class. More detailed notes and examples are given in the class notes, presentations, and demonstrations (click here.)

The acceleration of gravity, near the surface of the Earth, is . (In other words, if you dropped a rock, its speed would increase 22 mph every second it is in the air.) The term, "g," is used in lay magazines and scientific journals to compare the acceleration or the weight to what it feels like on the Earth's surface. For example, by applying the brakes, you car might be able to decelerate your car at 1.3 g's This means that your car's velocity would change by 1.3(9.80 m/s^{2} or 12.74 m/s^{2}.)

The term "g" is also used to describe the feeling of weight. For example if you were to ride the loop-the-loop on a roller coaster you would probably around experience 4g's as you entered the loop and 2.5 g's when you were upside down at the top of the loop. This means that a 100 pound person would feel as if he weighed 400 pounds when he entered the loop and 250 pounds when he is upside down at the top of the loop.

In kinematics problems a, "g," refers to the acceleration. However, g's cannot be used in the formulae because, "g's," are not part of the S.I. system. Therefore they must be converted to m/s^{2}. This is easy. 1 g is 9.80 m/s^{2}; 2 g's is 2(9.80 m/s^{2}) = 19.60 m/s^{2}; 3g's is 3(9.80 m/s^{2}) = 29.4 m/s^{2}, etcetera. To convert to the S.I. unit of m/s^{2} , multiply the number of g's times 9.80. To convert from m/s^{2} to g's divide by 9.80 m/s^{2}.

Example #1

When throwing a baseball, the pitcher moves the ball with an acceleration of 49 m/s^{2}. How many g's s this?

Solution

Use dimensional analysis where g = 9.80 m/s^{2}.

Example #2

Question

Hint

Solution

Video Solution

When the Space Shuttle returns to Earth, it can experience as much as 7.8 g's. If the space shuttle endured this deceleration while changing speed from 75,997 m/s to 51,657 m/s, then how much time did this part of their journey take?

When the Space Shuttle returns to Earth, it can experience as much as 7.8 g's. If the space shuttle endured this deceleration while changing speed from 75,997 m/s to 51,657 m/s, then how much time did this part of their journey take?

Hint:

Look for units clues.

Don't use g's for the acceleration. You need to use all S.I. units.

The shuttle is decellerating so the acceleration is negative in the formulas.

When the Space Shuttle returns to Earth, it can experience as much as 7.8 g's. If the space shuttle endured this deceleration while changing speed from 75,997 m/s to 51,657 m/s, then how much time did this part of their journey take?

Choose the best answer. It does not matter how many questions you miss. What matters is that you learn from any mistakes.

Convert 35.7 ^{m}/_{s}^{2} to g's

3.64 g's

349.86 g's

Convert 5.6g's to ^{m}/_{s}^{2}.

54.9 ^{m}/_{s}^{2}

0.571 ^{m}/_{s}^{2}

Convert 14.5 ^{m}/_{s}^{2} to g's.

142.1 gs

1.48 g's

Convert 0.76 g's to ^{m}/_{s}^{2}.

7.45 ^{m}/_{s}^{2}

0.0776 ^{m}/_{s}^{2}

In sprint a runner, may exert 11.4 g's before reaching a speed of 13.0 m/s. Convert the acceleration to ^{m}/_{s}^{2}.

1.33 ^{m}/_{s}^{2}

127 ^{m}/_{s}^{2}

112 ^{m}/_{s}^{2}

^{m}/_{s}^{2}

by Tony Wayne ...(If you are a teacher, please feel free to use these resources in your teaching.)

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