Why Does a Toy Airplane on a String Tilt When It Flies in Circles?

  • Thread starter Thread starter kirafreedom
  • Start date Start date
  • Tags Tags
    Airplane
Click For Summary
SUMMARY

The discussion focuses on the physics of a toy airplane suspended from a string, analyzing the forces acting on it as it moves in a circular path. The tension in the string, T, is derived using two methods: balancing vertical forces and applying circular motion equations. The expressions for T are equated to find the velocity, v, in terms of the angle θ, gravitational acceleration g, and the string length l. Additionally, the relationship between the period of rotation and velocity is explored, emphasizing the importance of understanding these concepts for accurate calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with circular motion equations
  • Knowledge of trigonometric functions and their applications
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of tension in circular motion scenarios
  • Learn about the relationship between period and tangential velocity in circular motion
  • Explore the application of trigonometric identities in physics problems
  • Investigate the effects of varying mass and string length on circular motion dynamics
USEFUL FOR

Physics students, educators, and hobbyists interested in understanding the mechanics of circular motion and the forces acting on objects in motion.

kirafreedom
Messages
10
Reaction score
0

Homework Statement



A particular type of toy airplane available in stores have little motors attached to them. Consider hanging one on a string from the ceiling and starting its motor. As it moves around in a circle with constant speed the stringwill make an angle with the vertical as shown in the diagram.
a) Assuming the plane has a mass m, find T, the tension in the string, in terms of m, θ, and g, by balancing the vertical forces on the plane.
b) Now use the equation for circular motion to solve for T in terms of m, θ, and v, (the speed of the plane), and l, (the length of the string).
c) By setting these two expressions for T equal to each other, find v in terms of θ, g, and l.
d) If you were to try this, you would find v difficult to measure directly. Write the period of rotation T, (not to beconfused with the tension), in terms of θ.

Homework Equations


T*cos(θ)=mg
r=l*sin(θ)
T*sin(θ)=m(v^2/r)

The Attempt at a Solution


a) T*cos(θ)=mg
T=mg/cos(θ)

b) T*sin(θ)=m(v^2/r)
T=[m(v^2/r)]/sin(θ)
r= [T*sin(θ)v^2]/m

r=l*sin(θ)
[T*sin(pheta)v^2]=l*sin(θ)
T=[l*sin(θ)m]/v^2

c) [l*sin(θ)m]/v^2=[m(v^2/r)]/sin(θ)
v=squart[(l*(sinθ)^2)/r]

d)
I am not sure
 
Last edited:
Physics news on Phys.org
Ok for a)
BUT: pay attention to the math in b)... I didn't understand what you did but you did something wrong. Hint: it is simpler than that... you got the expression for ##T## right, then just substitute the expression for ##r## in terms of ##l## and ##\theta##
c) Need correct b) first :wink:
d) Use the relation between period and velocity (keep in mind for you ##v## is the speed of the plane, therefore the tangential velocity) http://en.wikipedia.org/wiki/Circular_motion#Uniform_circular_motion
 
kirafreedom said:
b) T*sin(θ)=m(v^2/r)
T=[m(v^2/r)]/sin(θ)
r= [T*sin(θ)v^2]/m
Try that last step again.
 

Similar threads

  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K