Circular Motion Equation homework

Click For Summary
SUMMARY

The discussion centers around solving a problem related to the Circular Motion Equation, specifically calculating acceleration in a system involving tension. The initial calculations yielded an acceleration of 6 m/s², which was questioned due to discrepancies with provided answer choices. Participants clarified that the tension in the string must be considered, leading to the suggestion of using torque equations (τ = Id²θ/dt²) and the moment of inertia (I = mR²) for a more accurate analysis. Ultimately, the conclusion emphasizes that the original approach was valid, as the tension vector does not contribute to tangential acceleration.

PREREQUISITES
  • Circular Motion Equation
  • Newton's Second Law (F = ma)
  • Torque and Moment of Inertia concepts
  • Trigonometric functions in physics (e.g., sin 37°)
NEXT STEPS
  • Study the application of torque in rotational dynamics
  • Learn about the relationship between tension and acceleration in circular motion
  • Explore the derivation and implications of the Circular Motion Equation
  • Investigate the role of trigonometric functions in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for clarification on common misconceptions in problem-solving related to tension and acceleration.

Sarah00
Messages
64
Reaction score
1

Homework Statement


2015_06_27_6_20_40.gif


Homework Equations


Circular Motion Equation

The Attempt at a Solution


I solved the problem and got the result to be 6 m/s2 but it is not one of the answers. I was pretty sure of my solution. So, are all choices wrong?

F = ma
mg sin 37 = ma
a = g sin 37 = 6.01 m/s2
 
Physics news on Phys.org
Yep I agree with you.
 
I got 5.9 ms-2 so same answer if g = 9.8 ms-2 instead of 10.
BUT - I don't like your rationale.
The force on the mass includes the tension from the string which you have ignored.
I suggest instead τ = Id2θ/dt2
I = mR2
etc.
Answer is the same though.
EDIT: on the other hand, maybe you were quick enough to realize that the tension vector, being orthogonal to the tangent vector, cannot impart force, therefore acceleration, in the tangential direction, making what you originally did 100% justified.
 
Last edited:

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K