IB 11 Question: Circular Acceleration Problem Involving Forces

In summary, the conversation is about a student seeking advice on how to solve a question in their IB 11 SL course involving concepts of force, acceleration, and gravity. The student shares their initial attempt at solving the question and receives feedback from their classmates. They then clarify their understanding and correct a calculation error.
  • #1
JerusalemLuver
11
0
Hi, this question is a part of an 'online homework packet I received through my IB 11 SL course (ie. International Baccalaureate 11 Standard Leval). I submitted my first attempt to answer this question and had the wonderful "sorry your answer is incorrect" screen pop up. I was wondering if I could get some advice on howe to solve the question. Am I understanding the concept but doing something wrong when it comes to calculations? Or do I need a push in the right direction with concept?
Thanks:smile:
JL

1.
question10.jpg



2.

trinaglews.jpg


  • (Total Force) = mass * acceleration
  • (Centrepital Acceleration) = (Velocity squared) / (Radius)
  • (Force of Gravity) = (mass)*(9.8)
  • (Radius) = (Diameter)/2



3.
  • Using trig. Radius = (sin35.5 *1.35) + (diameter/2) =6.43394899 m
  • (Force of Gravity) = 9.8 * 10 kg = 98 N
  • Using trig (Total Force) = mass*acceleration = (tan 35.5 *Force of Gravity)= 69.990272065
  • (Acceleration) = (Total Force)/mass = 69.990272065/10 =6.9990272065 m/s^2

I first tried to calculate the radius adding the bottom part of the first triangle to half of the diameter.I'm fairly certain of the radius but if anyone thinks up another method or thinks that this method is incorrect please let me know.
I secondly calculated the Force of gravity on the 10 kg seat
Using the second triangle I used trig to solve for the total force
I then substituted the Total force (ie. Mass * Acceleration) and mass to solve for Acceleration.


 
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  • #2
As you can all pretty much see that was my first post, if I'm doing anything wrong, (asking the wrong kind of question, not clear, whatever it is) let me know, thx
 
  • #3
Everything seems to be fine. Although you could have made life a bit easier for yourself:

[tex]m a = m g \tan(\theta)[/tex]

[tex]\frac{v^2}{r} = g \tan(\theta)[/tex]

What value do you then get for the speed?
 
Last edited:
  • #4
actually sumone in my class told me that i should have used 90(degrees) minus theta in my second triangle , it works but i don't know why??
 
  • #5
wait no i got it, my bad it was a calculation error!:blushing:
 
  • #6
?
 

1. What is circular acceleration?

Circular acceleration is the change in velocity of an object moving along a circular path. It is a vector quantity that includes both a magnitude and direction.

2. How is circular acceleration different from linear acceleration?

Circular acceleration occurs when an object is moving in a circular path, and its velocity is constantly changing due to the direction of its motion. Linear acceleration, on the other hand, occurs when an object's velocity is changing in a straight line.

3. How do forces affect circular acceleration?

Forces can cause circular acceleration by changing the direction of an object's velocity, which results in a change in its direction of motion. This change in direction causes the object to accelerate towards the center of the circle.

4. What is the relationship between circular acceleration and centripetal force?

Circular acceleration and centripetal force are directly proportional to each other. This means that as the magnitude of the centripetal force increases, so does the circular acceleration of the object.

5. How do you calculate circular acceleration?

Circular acceleration can be calculated using the formula a = v^2/r, where a is the circular acceleration, v is the velocity of the object, and r is the radius of the circular path. Alternatively, it can also be calculated by taking the derivative of the object's angular velocity with respect to time.

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