How do vectors relate to perpendicularity in circular motion?

AI Thread Summary
The discussion revolves around the relationship between vectors in circular motion, specifically focusing on the perpendicularity of the position, velocity, and acceleration vectors. The user initially finds that both the position vector (r) and the acceleration vector (a) are perpendicular to the velocity vector (v), leading to confusion about how the particle can speed up while maintaining circular motion. The mathematical derivation shows that the dot product of acceleration and velocity is zero, indicating they are indeed perpendicular. The user realizes that the acceleration vector's direction, which points towards the center of the circle, does not contradict the particle's increasing speed due to the time-dependent angular velocity. Ultimately, the discussion clarifies the dynamics of circular motion and the role of vector relationships in understanding acceleration and velocity.
LondonLady
Messages
14
Reaction score
0
Im a bit confused about a question on circular motion that I'm answering. Ill state the entire question and then say what I am confused about.

In class we discussed circular motion for the case

\displaystyle{\frac{d\theta}{dt} = \omega}

Now assume that the circle has radius r and that

\displaystyle{\frac{d\theta}{dt} = 2t}

for t in seconds. Let \theta(t = 0) = 0

(therefore \theta = t^2)

a) Find \vec{r}(t)

b) Find \vec{v}(t). is \vec{v} \perp \vec{r}?

c) Find \vec{a}(t). Express \vec{a} in terms of \vec{r} and \vec{v}. Is \vec{a} \perp \vec{v}?

d) With respect to the circle's centre, sketch \vec{r},\vec{v} and \vec{a} for counter clockwise rotation.


Ok. I have found all the vectors in i-j form. My question is about the perpendicularity questions. Mathematically I have found that \vec{r} \perp \vec{v} and that \vec{a} \perp \vec{v}. I have also found that \vec{a} can be written as -\alpha \vec{r} (where \alpha is a constant. All this implies that the acceleration vector is pointed back into the centre of the circle as some negative multiple of \vec{r}.

If this is so then how is the particle speeding up?? (the rate of change of theta is time dependent)

I would have thought that the acceleration vector would have been at some angle to the position vector. But then it wouldn't move in a circle... I am confused...


Also, the second part of (c) I am finding hard. Anyone any ideas?
 
Physics news on Phys.org
The position vector is perpendicular to the velocity vector, but the acceleration vector is NOT perpendicular to the velocity vector!
 
hmm... thankyou for your reply. I can't agree though. I got

\vec{r}(t) = r\cos (t^2)\vec{i} + r\sin (t^2)\vec{j}

\vec{v}(t) = -2tr\sin (t^2)\vec{i} + 2tr\cos (t^2)\vec{j}

\vec{a}(t) = -4t^2r\cos(t^2)\vec{i} - 4t^2r\sin(t^2)\vec{j}

Then if you find the dot product

\vec{a}.\vec{v} = 8t^3r^2\sin(t^2)\cos(t^2) - 8t^3r^2\sin(t^2)\cos(t^2) = 0

Which implies they are perpendicular... Is my logic wrong?
 
You have not differentiated \vec{v} correctly:
\frac{d\vec{v}}{dt}=(\frac{d}{dt}2tr)(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})+2tr\frac{d}{dt}(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})=
2r(-\sin(t^{2})\vec{i}+\cos(t^{2})\vec{j})+4t^{2}r(-\cos(t^{2})\vec{i}-\sin(t^{2})\vec{j})
 
Ahh! It didnt even occur to me that it might have changed into a product! Thankyou so much!
 
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}...
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...
Back
Top