Circular Motion using polar coordinates - Mechanics

In summary, the conversation discusses a problem involving a block of mass M on a frictionless table attached to a massless string that goes through a hole a distance S from the block. A force F is applied to the string, causing the block to move in a circular motion of radius S with an initial angular velocity w0. As the force F is increased, the distance between the block and the hole decreases according to S-c1t2, where c1 is a known constant. The goal is to find the torque exerted by each force on the block and the angular acceleration of the block as a function of time. The conversation also mentions the use of the equations of acceleration vector and conservation laws for torque and angular momentum. Through
  • #1
fishking2
6
0

Homework Statement


A block of mass M is on a frictionless table that has a hole a distance S from the block. The block is attached to a massless string that goes through the hole. A force F is applied to the string and the block is given an angular velocity w0 , with the hole as the origin, so that it goes in a circle of radius S. The force F is increased, starting at time t = 0, so that the distance between the block and the hole decreases according to S-c1t2 . Here c1 is a known, positive constant. Assuming the string stays straight and can only pull along its length, find the torque about the hole exerted by each force acting on the block and the angular acceleration of the block as a function of time.

Here is the picture.

Homework Equations


acceleration vector = [d2r/dt2 - rw2] ir + [2(dr/dt)(w) + r α] iθ

The Attempt at a Solution


Looking at the system from the top, the torques are:
TN = -mg(S-c1t2) iθ
Tmg = mg(S-c1t2) iθ
TF = 0

given that
r = S-c1*t2
dr/dt=-2c1t
d2r/dt2 = -2c1

Then using F=ma
Fr = m[-2c1 - (S-c1t2)w02]
Fθ = m[(S-c1t2)α - 4(c1t)(w02)]

Assuming that the Fr and Fθ components are correct, I still don't know where to go, especially because it doesn't tell us about this force, which is not a constant.
I am working out review problems for a test, so I know that the final answer is:
α = (4S2w0c1t) / (S - c1t2)3
I am trying to understand how to get the solution, so I will be able to do similar problems on the test. Also, are the torques I wrote above correct?
 
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  • #2
fishking2 said:
the torques are:
TN = -mg(S-c1t2) iθ
Tmg = mg(S-c1t2) iθ
TF = 0
I don't see that any of those torques are interesting.
given that
r = S-c1*t2
dr/dt=-2c1t
d2r/dt2 = -2c1

Then using F=ma
Fr = m[-2c1 - (S-c1t2)w02]
Fθ = m[(S-c1t2)α - 4(c1t)(w02)]
You seem to be assuming that angular velocity is constant (w0). What conservation law are you basing that on?
 
  • #3
haruspex said:
I don't see that any of those torques are interesting.
I only listed the torques because it was part of the question. I don't think they are of any real significance.

haruspex said:
You seem to be assuming that angular velocity is constant (w0). What conservation law are you basing that on?
I guess that is wrong. Would I plug in (w0 + αt) instead?
 
  • #4
fishking2 said:
Would I plug in (w0 + αt) instead?
That will be true if α is constant. Is it?
 
  • #5
No, it isn't a constant. However, if I am understanding the problem, α is what I am trying to solve for. And if that is correct, I am confused as to what I can set the components of force equal to, so that I can solve for alpha.

I also believe I had a mistake in my Fθ component. Where I have (w0)2, which we determined was already incorrect, but it should not be squared. So:
Fr = m[-2c1 - (S-c1t2)(w0+αt)2]
Fθ = m[(S-c1t2)α - 4(c1t)(w0+αt)]

If I set Fθ = 0, because there are no forces acting in the iθ direction (I think). I was able to solve for α, but it came out to an incorrect answer of α = 4c1t*w0 / (S-5c1t2)
 
  • #6
fishking2 said:
So:
Fr = m[-2c1 - (S-c1t2)(w0+αt)2]
No, you've not understood what I told you in the previous post. Because α is not constant you cannot write w =
w0+αt.
If you write that equation correctly you have a differential equation in terms of w. However, solving it is effectively the same as applying a certain conservation law.
Actually, I was wrong to say that none of those torque equations are interesting. What I should have pointed out instead is that a torque is meaningless unless you identify an axis for the torque. Just look at the last one, TF = 0. You seem to be taking a vertical through the hole as the axis, so let's go with that. If there is no torque about that axis, what quantity is conserved about that axis?
 
  • #7
haruspex said:
No, you've not understood what I told you in the previous post. Because α is not constant you cannot write w =
w0+αt.
If you write that equation correctly you have a differential equation in terms of w. However, solving it is effectively the same as applying a certain conservation law.
Actually, I was wrong to say that none of those torque equations are interesting. What I should have pointed out instead is that a torque is meaningless unless you identify an axis for the torque. Just look at the last one, TF = 0. You seem to be taking a vertical through the hole as the axis, so let's go with that. If there is no torque about that axis, what quantity is conserved about that axis?
Angular momentum is conserved.
L0=Lt

L0=r x p = rm(rw)
L0=S2m*w0

Lt= (S-c1t2)(m)(V)
Vθ = r*dθ/dt
Vθ = (S-c1t2)(w)​
S2m*w0=(S-c1t2)(m)((S-c1t2)(w))
S2w0=(S-c1t2)((S-c1t2)(w))

w = S2w0 / (S-c1t2)2α=dw/dt
α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
α= -2*S2w0 / (S-c1t2)3

This would be the answer if it were multiplied by -2c1t, which is dr/dt. I assume this comes from the chain rule when taking the derivative of dw/dt, but I don't quite see why. Also, I just want to take a second to thank you for all your help. I really appreciate it!
 
  • #8
fishking2 said:
α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
α= -2*S2w0 / (S-c1t2)3

This would be the answer if it were multiplied by -2c1t, which is dr/dt. I assume this comes from the chain rule when taking the derivative of dw/dt,
Quite so. Do you not understand how to apply the chain rule here?
 
  • #9
haruspex said:
Quite so. Do you not understand how to apply the chain rule here?
Haha, yes. I am confused at what point the dr/dt is entering into the equation. If I had to guess, it comes from the 'r' term that we plugged in. (S-c1t)

However, since this is only in terms of 't' and the other values are all constants, I didn't think chain rule would apply.
Would you mind explaining where the dr/dt comes from?
 
  • #10
fishking2 said:
Haha, yes. I am confused at what point the dr/dt is entering into the equation. If I had to guess, it comes from the 'r' term that we plugged in. (S-c1t)

However, since this is only in terms of 't' and the other values are all constants, I didn't think chain rule would apply.
Would you mind explaining where the dr/dt comes from?
No, it's just a matter of executing this correctly:
fishking2 said:
d/dt[ 1 / (S-c1t2)2 ]
This takes the form (d/dt)(f(g(t)), where g(t) = S-c1t2 and f(x) = x-2. Can you quote the chain rule in terms of f and g?
 
  • #11
haruspex said:
No, it's just a matter of executing this correctly:

This takes the form (d/dt)(f(g(t)), where g(t) = S-c1t2 and f(x) = x-2. Can you quote the chain rule in terms of f and g?
Wow, I don't know how i missed that! Thanks!
(d/dt)(f(g(t)) = f'(g(x))*g'(x)
(d/dt)(f(g(t)) = -2(S-c1t2)-3*(-2c1t)

So α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
α= (-2S2w0 * (-2c1t)) / (S-c1t2)3
α=4S2w0c1t/ (S-c1t2)3You've been really helpful! Thanks!
 
  • #12
fishking2 said:
Wow, I don't know how i missed that! Thanks!
(d/dt)(f(g(t)) = f'(g(x))*g'(x)
(d/dt)(f(g(t)) = -2(S-c1t2)-3*(-2c1t)

So α= S2w0 * d/dt[ 1 / (S-c1t2)2 ]
α= (-2S2w0 * (-2c1t)) / (S-c1t2)3
α=4S2w0c1t/ (S-c1t2)3You've been really helpful! Thanks!
You're welcome.
 

1. What is circular motion using polar coordinates?

Circular motion using polar coordinates is a way of describing the motion of an object that moves in a circular path using a combination of distance and angle measurements. It is often used in mechanics to understand the motion of objects, such as planets orbiting the sun or a ball rolling in a circular track.

2. How is circular motion using polar coordinates different from using Cartesian coordinates?

In Cartesian coordinates, the position of an object is described using x and y coordinates on a flat plane. However, in polar coordinates, the position is described using a distance (r) from the origin and an angle (θ) from a reference line. This makes it easier to describe circular motion, as the distance and angle can remain constant while the object moves in a circular path.

3. What is the equation for circular motion using polar coordinates?

The equation for circular motion using polar coordinates is r = a, where r is the distance from the origin, and a is the radius of the circular path. This equation can also be written as r = a cos(θ) + a sin(θ) if the object is not moving at a constant speed or if it is moving in an elliptical path.

4. How is angular velocity related to circular motion using polar coordinates?

Angular velocity is a measure of how fast an object is rotating around a fixed point. In circular motion using polar coordinates, the angular velocity is equal to the change in the angle (θ) over time. It is often represented by the symbol ω and is measured in radians per second.

5. What are some real-life examples of circular motion using polar coordinates?

Circular motion using polar coordinates can be observed in many real-life situations, such as the motion of planets around the sun, the rotation of a Ferris wheel, or the motion of a spinning top. It is also commonly used in engineering and physics to describe the motion of objects in circular paths, such as satellites orbiting the Earth or the movement of gears in a machine.

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