Vertical circular motion in uniform circular motion

AI Thread Summary
A pilot is flying an airplane in a vertical circular loop with a radius of 1200 m, reaching a speed of 150 m/s at the bottom. The total acceleration at this point is 2.3g, prompting the need to calculate the tangential acceleration and its direction. The centripetal acceleration is calculated as 18.75 m/s², but the correct tangential acceleration is determined to be 12.5 m/s². The direction of the plane's acceleration at the bottom of the loop is approximately 56.3 degrees. Understanding the relationship between tangential and centripetal acceleration is crucial for solving these types of problems.
seanpk92
Messages
6
Reaction score
0

Homework Statement


A pilot flies an airplane in a vertical circular loop with a radius of R = 1200 m. The plane is gaining speed as the pilot makes a dive, and its speed is measured to be 150 m/s when the plane reaches the bottom of the circle. If the total acceleration of the plane at the bottom is 2.3g, find (a) the tangential acceleration of the plane, (b) the direction of the plane’s acceleration at the bottom of the loop.


Homework Equations


What I can think of doing, is:
an=v2/R
and
a = an + at

The Attempt at a Solution


I tried making a free body diagram with the total force and an pointing toward the center, and the v and the at point to the right. At the bottom of the circle.
an = 1502/1200 = 18.75m/s2
a = 2.3g = 2.3/9.8m/s2 = 0.2346m/s2

but this doesn't give me my answer. The answer is (a) 12.5 m/s2 and (b) beta = 56.3o

(i don't know how to do a, so I can't start b)
 
Physics news on Phys.org
seanpk92 said:

Homework Statement


A pilot flies an airplane in a vertical circular loop with a radius of R = 1200 m. The plane is gaining speed as the pilot makes a dive, and its speed is measured to be 150 m/s when the plane reaches the bottom of the circle. If the total acceleration of the plane at the bottom is 2.3g, find (a) the tangential acceleration of the plane, (b) the direction of the plane’s acceleration at the bottom of the loop.


Homework Equations


What I can think of doing, is:
an=v2/R
yes, good
and
a = an + at
the tangential and centripetal acceleration vectors are at right angles to each other, so you need to add them vectorially to get the total acceleartaion.

The Attempt at a Solution


I tried making a free body diagram with the total force and an pointing toward the center,
yes
]and the v and the at point to the right.
yes, and the tangential force points in the same direction
At the bottom of the circle.
an = 1502/1200 = 18.75m/s2
a = 2.3g = 2.3/9.8m/s2 = 0.2346m/s2
2.3 g's means the acceleration is 2.3 times the gravitational acceleration
but this doesn't give me my answer. The answer is (a) 12.5 m/s2 and (b) beta = 56.3o

(i don't know how to do a, so I can't start b)
Correct you formula for total acceleration
 
Thank you so much PhantomJay!
 
seanpk92 said:
Thank you so much PhantomJay!
(You're) Welcome to PF!:smile:
 
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