Tangential and radial particle acceleration

AI Thread Summary
The discussion revolves around determining the x and y coordinates of a particle moving in a circular path given its velocity and acceleration. The particle's velocity is 4.6j m/s, indicating it is moving vertically, while the acceleration is (2.3i - 2.1j) m/s^2. Participants clarify that the radial acceleration formula ar = -v^2/r requires the correct interpretation of the total acceleration, emphasizing that the y-coordinate must be zero for the velocity to remain tangential to the circle. The confusion arises from the calculation of the radius and the role of the acceleration components. Ultimately, the consensus is that the y-coordinate is zero to align with the particle's motion.
Why-not2007
Messages
3
Reaction score
0

Homework Statement


A particle moves in the xy plane in a circle centered origin. At a certain instant the velocity and acceleration of the particle are 4.6j m/s and (2.3i - 2.1j) m/s^2. What are the x and y coordinates of the particle at this moment


Homework Equations


ar=-v^2/r


The Attempt at a Solution


I took the magnitude of the radial acceleration and got 3.11 m/s^2 and plugged it into the equation. I got -6.8 for my radius and was told that was the wrong answer for the x coordinate. Then I tried plugging in 2.3 for the radial acceleration and got -9.2 for my radius. Does the -2.1j m/s^2 not play a factor in this problem? I got x=-9.2, but I don't know how to solve for the y component of the problem.
 
Physics news on Phys.org
Why-not2007 said:
I took the magnitude of the radial acceleration and got 3.11 m/s^2 and plugged it into the equation.
How did you get this? Are you assuming that the particle is moving with a constant speed? (You are given the total acceleration, not the radial acceleration.)
I got -6.8 for my radius and was told that was the wrong answer for the x coordinate. Then I tried plugging in 2.3 for the radial acceleration and got -9.2 for my radius. Does the -2.1j m/s^2 not play a factor in this problem? I got x=-9.2, but I don't know how to solve for the y component of the problem.
The velocity should tell you the y-coordinate, since it moves in a circle.
 
Why-not2007 said:

Homework Statement


A particle moves in the xy plane in a circle centered origin. At a certain instant the velocity and acceleration of the particle are 4.6j m/s and (2.3i - 2.1j) m/s^2. What are the x and y coordinates of the particle at this moment


Homework Equations


ar=-v^2/r


The Attempt at a Solution


I took the magnitude of the radial acceleration and got 3.11 m/s^2 and plugged it into the equation. I got -6.8 for my radius and was told that was the wrong answer for the x coordinate. Then I tried plugging in 2.3 for the radial acceleration and got -9.2 for my radius. Does the -2.1j m/s^2 not play a factor in this problem? I got x=-9.2, but I don't know how to solve for the y component of the problem.
The velocity is given as being in the 'j' direction only. That should give you a hint on the possible location of the y coordinate. Your value of the centripetal acceleration that you must use depends on the value of the y coordinate you must find first.
 
The y coordinate equal 0?
 
Why-not2007 said:
The y coordinate equal 0?
Absolutely. That's the only way to make sense of the given velocity, since that velocity must be tangential to the circle at all times.
 
Thanks, I appreciate the help.
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Radial & Tangential Acceleration of a Car at Indianapolis 500 | Physics Solution
Replies
7
Views
4K
Would particle P be under non-unform circular motion?
Replies
29
Views
2K
Angle between radius and acceleration
Replies
4
Views
1K
Radial and Tangential Acceleration
Replies
14
Views
3K
What is the net acceleration of the coin on a rotating disk?
Replies
2
Views
2K
Angular acceleration problem for a pulley used to raise an elevator
Replies
8
Views
2K
Closest distance of approach between 2 charged particles
Replies
5
Views
1K
Difference between radial and centripetal acceleration?
Replies
9
Views
8K
Tangential acceleration and centripetal acceleration
Replies
5
Views
2K
Finding radial & tangential acceleration at a point
Replies
7
Views
6K

Hot Threads

Recent Insights

Back
Top