Circular motion (magnitude of radial acceleration)

Click For Summary
SUMMARY

The discussion focuses on calculating the magnitude of radial acceleration for a ball swinging in a vertical circle with a rope length of 1.5m. At an angle of 26.9 degrees from the lowest point, the total acceleration is given as -22.5i + 20.2j. Participants clarify that radial acceleration is derived from the tension in the rope and the tangential component of gravity, emphasizing the need to separate the total acceleration into radial and tangential components using trigonometric functions. The correct approach involves resolving the acceleration vector into components aligned with the radial and tangential directions.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Knowledge of vector decomposition in physics
  • Familiarity with trigonometric functions and their applications in physics
  • Concept of radial and tangential acceleration
NEXT STEPS
  • Study the principles of circular motion and centripetal acceleration
  • Learn how to decompose vectors into components using trigonometry
  • Explore the relationship between tension, gravity, and acceleration in circular motion
  • Practice problems involving radial and tangential acceleration calculations
USEFUL FOR

Physics students, educators, and anyone interested in mastering concepts related to circular motion and acceleration analysis.

mohdhm
Messages
42
Reaction score
0
a ball swings in a vertical circle at the end of a rope 1.5m long. when the ball is 26.9 degrees past the lowest point on its way up, its total acceleration is -22.5i+20.2j. at that instant (b) determine the magnitude of its radial acceleration

isnt the magnitude of Ar... +20.2m/s^2 ? The answer key says otherwise.

i really need someone to explain this concept to me... radial acceleration causes a change in magnitude... and A = Ar + At right? so why isn't the Magnitude of Ar... the vertical component of A in this question .

Please explain how i can find the magnitude of radial acceleration using trig.

thanks in advance
 
Physics news on Phys.org
The vertical component isn't radial, so that can't be right. There are only two forces providing acceleration i) the tension in the rope (purely radial) and ii) gravity (partly radial, partly tangential). So if you find the tangential component of gravity (using trig) and subtract it from the total acceleration then what is left should be radial.
 
here is what i tried. Tried to use basic trig to solve it and i failed because i still cannot conceptualize it.

shouldnt the x in cos(53.1)=22.5/x be the same as sin (53.1)=20.2/x where x is supposedly the tangental acceleration?

all i know is that the direction of total acceleration isn't the same as the rope because i used arctan on its vertical and horizontal component and got a completely different answer.

edit: the tangental component is the hypotenuse right?
?
 
Try drawing yourself a diagram that shows the components of the acceleration that you've been given (vertical and horizontal) as they relate to the radial direction. (The radial direction is towards the center--use the angle given to relate it to x and y axes.)

To find the radial component of total acceleration, you can find the radial components of the x & y components and then add them.
 
mohdhm said:
here is what i tried. Tried to use basic trig to solve it and i failed because i still cannot conceptualize it.

shouldnt the x in cos(53.1)=22.5/x be the same as sin (53.1)=20.2/x where x is supposedly the tangental acceleration?

No.

all i know is that the direction of total acceleration isn't the same as the rope because i used arctan on its vertical and horizontal component and got a completely different answer.

edit: the tangental component is the hypotenuse right?
?

Hypotenuse of what? Looking back at my hint - I don't think its valid to assume only gravity and tension are acting on the ball. The numbers just don't add up. So assume nothing. The radial direction is parallel to the rope and the tangential direction is perpendicular to that. So you will need to split the acceleration vector up into two parts - one part in the radial direction and one part in the tangential direction. Can you draw the picture? One way to do this is to rotate the acceleration vector 26.9 degrees clockwise. Then the radial direction becomes the same as the vertical direction. So then you can take the j component of that vector. Does that make sense from your picture?
 
that doesn't change anything .. e axis on which ur values are taken from would change .. so u still have to resolve them ..
 
this problem is pretty simple i think. you just have to know which of the two acceleration components is which, that is whether i stands for radial or tangential acceleration. i don't believe gravity is a factor this being based solely on the assumption that my physics professor assigned this homework problem and I'm pretty sure he doesn't have us messing with gravity in these types of problems just yet. but if anyone can be certain on what i and j stand for please do reply.
 
note: the tangential and radial accelerations are always perpendicular much like the x and y components of velocity. if drawn correctly, trig can be used to find the magnitude and direction of the total acceleration.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
813
  • · Replies 6 ·
Replies
6
Views
2K
Replies
55
Views
3K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K