Rotational Motion of a toy airplane

AI Thread Summary
The discussion revolves around analyzing the motion of a toy airplane flying in a circular path while attached to a string with a tension of 6.5N at a 50° angle. The user calculated the tangential velocity to be 5.28 m/s and determined the mass of the airplane to be 0.42 kg using Newton's second law. However, they are struggling to find the length of the string, realizing they need to calculate the radius of the circular motion first. They are considering using the equation v = rω but are unsure how to find the radius. The conversation emphasizes the importance of understanding centripetal force and acceleration in solving the problem.
SChiO
Messages
11
Reaction score
0

Homework Statement



We were given a video...I'm not sure if you can post video links otherwise I would...but it's a video of an airplane on a string flying in a circular motion. On the top of the string you can see that the tension of the string is 6.5N. and it is flying at an angle of 50°. We are told to find the angular velocity, mass of plane and length of string.


The Attempt at a Solution



Using ω=2pi(period) I found it to be 5.28 m/s with a period of 1.19s/rev
Using Newton's 2nd law and knowing that the force of gravity is equal to the y-component of the tension force I found the mass to be equal to .42 kg.
However I am stuck on finding the length of the string...I have an idea that I need to find the radius of the circle and then from there I know I can find the length of the string but I'm not sure how to find the radius. I'm thinking to use this equation: v=rω but I'm not sure how to find the linear velocity or if that's even right...HELP PLEASE! I'll see if I can post the video for ease...Thanks
 
Physics news on Phys.org
SChiO said:

Homework Statement



We were given a video...I'm not sure if you can post video links otherwise I would...but it's a video of an airplane on a string flying in a circular motion. On the top of the string you can see that the tension of the string is 6.5N. and it is flying at an angle of 50°. We are told to find the angular velocity, mass of plane and length of string.
You can post links to a video - but it is best to give a written description anyway.
Is that 50deg angle to the horizontal or the vertical?
Using ω=2pi(period) I found it to be 5.28 m/s with a period of 1.19s/rev
Just a note - the units of ##\omega## is radiens per second. Since you were "given the angular velocity" I'm guessing you mean that the tangential velocity was 5.28m/s.
"Period" is usually just given in seconds.

Using Newton's 2nd law and knowing that the force of gravity is equal to the y-component of the tension force I found the mass to be equal to .42 kg.

However I am stuck on finding the length of the string...I have an idea that I need to find the radius of the circle and then from there I know I can find the length of the string but I'm not sure how to find the radius.
Hint: centripetal force and acceleration.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top