anonymous
  • anonymous
Centripetal Force Tutoring
Physics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Centripetal Force; Force considered to be acting towards the center of circle. Ex. Ferris Wheel
anonymous
  • anonymous
The direction of velocity is constantly changing in a typical centripetal force because at every different point of circumference there is a centripetal force at work perpendicular to the center of the circle.
anonymous
  • anonymous
The formula to calculate for the centripetal acceleration is as follows. The following equations make an assumption that centripetal force is acting towards the center of the circle. a=v^2/R Where v is the velocity of the moving object, and R is the radius of circle in question.

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anonymous
  • anonymous
Alternatively, you can calculate the direction and magnitude of centripetal force using the following, a=4pi^2R/T^2 Where pi=constant for pi R=radius of the circle T=period of rotation in seconds (how many seconds it takes for the circle to complete a cycle)
anonymous
  • anonymous
Consider the following problem, While riding the Ferris wheel of radius 8.5 m, your Physics teacher notices that she is making 5.4 rotations in 1.0 minutes. What is her centripetal acceleration while riding the ferris wheel?
anonymous
  • anonymous
First, you would need to convert the minute to seconds, as velocity per second is far more common and handy. 1min=60s Radius=8.5m We know that the Ferris Wheel makes 5.4 rotations every 60 seconds. We also know, by division 60s/5.4=11.11 s (notice that I calculate the period with one extra sig fig). Using the second equation a=4pi^2R/T^2 Where a is the acceleration, pi is the pi constant R is the radius T is the period per rotation a=4(3.14)^2(8.5m)/11.1s^2 a=2.720772665 After adjusting for sig figs, The centripetal acceleration acting on the Ferris Wheel towards the center is 2.7 m/s^2.
anonymous
  • anonymous
*p.s Always remember that centripetal force is acting towards the center, hence making the Ferris Wheel possible, as well as the circular motion. This concept applies to Earth orbiting around the Sun as well. The reason Earth continues to move around the Sun is due to Centripetal Force acting towards the center of the Sun, otherwise, if you think about it, the Earth will be in relative linear motion from Newton's first law to the sun forever.
anonymous
  • anonymous
In the meanwhile let's attempt another question that tackles on the first equation, which is a=v^2/R... Quite a straight forward one. A car cruising around a circular track of radius 32 m, has a centripetal acceleration of 5.2 m/s2. Calculate the speed of the car.
anonymous
  • anonymous
All you have to do in this question is plug in the values for acceleration and radius, in which case you would rearrange the formula a=v^2/R to v^2=aR v^2=(5.2m/s^2)(32m) v^2=166.4m^s/s^s v=sqrt(166.4m^2/s^) v=12.8996m/s Therefore the velocity towards perpendicular to the center is 13m/s P.S Velocity is to exactly perpendicular to the center of the circle.
anonymous
  • anonymous
This problem involves a bit of twist. A ball on the end of a string is moving in a horizontal circle. If the centripetal acceleration of the ball is 12 m/s2, while it is travelling at 3.0 m/s, what would the acceleration be if its speed increased to 4.0 m/s. Assume that the radius of rotation does not change. Since we already know the acceleration 12mm/s^2, as well as the velocity 3.0m/s, we can identify the radius of the first using a=v^2/R 12m/s^2=(3.0m/s)^2/R R=(3.0m/s)^2/12m/s^2 R=0.75m Since we know that the radius of the circle in question is 0.75m, we can now use the equation a=v^2/R again to account for the acceleration of the ball if the velocity is 4.0m/s. a=(4m/s)^2/0.75m a=16m^2/s^2/0.75m=21.33333m/s^2 After accounting for the sig figs of 2, Acceleration at v=4.0m/s in the same circle is 2.1*10^2m/s^2
anonymous
  • anonymous
Input& Additional questions regarding centripetal acceleration are always welcome!
arindameducationusc
  • arindameducationusc
Awesome Examples @Robert136 ! and also awesome tutorial. Its good to have you in open study physics forum......!
arindameducationusc
  • arindameducationusc
I might drop in some examples too... if you don't mind @Robert136 .... :D
anonymous
  • anonymous
Yeah please pitch in too;)
anonymous
  • anonymous
In our previous tutoring we did the centripetal acceleration but at this time just to get you to understand the concept of acceleration in changing directions further, take a look at the following questions.
anonymous
  • anonymous
A baseball is thrown by a professional pitcher at 45 m/s [S] and is hit by a batter at 55 m/s [N 45° W]. If the ball was in contact with the bat for 0.01 s, determine the acceleration of the ball. Note that within 0.01 s the velocity of the ball changes from 45m/s to the South to 55m/s to the N 45 degrees W.. In a question like this you need to use a vector diagram.|dw:1440650555945:dw|
anonymous
  • anonymous
Got it? Using the cosine law for hypotenuse a^2=(55m/s)^2+(45m/s)^2-2(55m/s)(45m/s)(cos 135)=92.5m/s Now you need to calculate the direction using the sine law, so Sine(135)/92.5m/s=SineA/45m/s=20.1 degrees So the instantaneous acceleration of the ball when hit by the baseball player is 93m/s [N20.1W]

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