Kinematics problem -- Cat chasing a mouse on a rotating disk

Click For Summary
A cat is chasing a mouse on a circular path with a radius of 28 meters, both moving at constant speeds of 4 m/s. The cat maintains its position along the radius connecting it to the mouse, which leads to equal angular velocities for both. The discussion involves calculating the chase duration, which is determined to be 11 seconds. Participants explore concepts like radial and tangential velocities, emphasizing the need for the cat to adjust its trajectory to maintain the chase. Understanding polar coordinates and the decomposition of velocities is crucial for solving the problem.
Satvik Pandey
Messages
591
Reaction score
12

Homework Statement


A mouse starts running on a circular path of Radius = 28m with constant speed u = 4m/s. A cat starts from the center of the path to catch the mouse. The cat always remains on the radius connecting the center of the circle and the mouse and it maintains magnitude of its velocity a constant v = 4m/s. How long (in sec) is the chase?


Homework Equations


ω=d\theta/dt


The Attempt at a Solution


Before solving this problem I tried make a figure of this question.Here it is.
physics figure.png

Hint given in my book says that in this problem angular velocities of cat and mouse are equal.
I know that angular velocity is change in angular displacement upon time.I am not sure about my figure.Please tell me if it is right or wrong and the angular velocity of cat and mouse are equal.It would be nice if you explain it with a figure.







[
 
Physics news on Phys.org
Since you know the angular velocity of the cat, you can compute the radial velocity of the cat as a function of the radius. You then need to relate the radial velocity to how long it will take the cat to reach radius r = 28 m.
 
Hi Satvik...

What is the answer ?
 
Orodruin said:
Since you know the angular velocity of the cat, you can compute the radial velocity of the cat as a function of the radius. You then need to relate the radial velocity to how long it will take the cat to reach radius r = 28 m.
No I can find the angular velocity of rat (1/7) but the radius of the path of cat is not given.Hint given in my book says that in this problem angular velocities of cat and mouse are equal.Could you please tell me why is it so?
 
Tanya Sharma said:
Hi Satvik...

What is the answer ?

The answer is 11 seconds.
 
Satvik Pandey said:
No I can find the angular velocity of rat (1/7) but the radius of the path of cat is not given.Hint given in my book says that in this problem angular velocities of cat and mouse are equal.Could you please tell me why is it so?

This is why:

Satvik Pandey said:
The cat always remains on the radius connecting the center of the circle and the mouse

You also have the speed of the cat, so you can just use Pythagoras' theorem to find the radial velocity (##v^2 = v_r^2 + v_\phi^2##).
 
Satvik Pandey said:
I know that angular velocity is change in angular displacement upon time.I am not sure about my figure.Please tell me if it is right or wrong and the angular velocity of cat and mouse are equal.It would be nice if you explain it with a figure.

Your figure is not correct .

Look at the attached picture .Red dots are location of mouse and blue represents cat .

If you join the blue dots you will get the path traced by the cat.

Let at t=0 mouse be at the top of the circle and cat at the center of the circle .In time t the line joining the mouse to the center makes an angle θ with the vertical .

What angle does the line joining the cat to the center at time t make with the vertical ? What does that tell you about the angular velocity of the two ?
 

Attachments

  • Cat Mouse.GIF
    Cat Mouse.GIF
    5 KB · Views: 831
Tanya Sharma said:
Your figure is not correct .

Look at the attached picture .Red dots are location of mouse and blue represents cat .

If you join the blue dots you will get the path traced by the cat.

Let at t=0 mouse be at the top of the circle and cat at the center of the circle .In time t the line joining the mouse to the center makes an angle θ with the vertical .

What angle does the line joining the cat to the center at time t make with the vertical ? What does that tell you about the angular velocity of the two ?

Yes it also makes the angle \theta.
When I join the blue dots I got the path traced by the cat.It is an arc.This arc will be the part of a circle.Why did you not calculate the angular velocity of cat w.r.t to that center of the path traced by the cat .(Why did you not calculate Angular displacement of cat w.r.t to that center of the path traced by the cat)
physics figure.png

Sorry for the figure it is not clear.
 
Orodruin said:
You also have the speed of the cat, so you can just use Pythagoras' theorem to find the radial velocity (##v^2 = v_r^2 + v_\phi^2##).
What is radial velocity?
 
  • #10
Satvik Pandey said:
Yes it also makes the angle \theta.
When I join the blue dots I got the path traced by the cat.It is an arc.This arc will be the part of a circle.

Right .

But the circle will not be like the one you have shown in your picture .Anyways forget about completing the circle .It is not required .Knowing that the cat will follow a curved path is sufficient for us.

Satvik Pandey said:
Why did you not calculate the angular velocity of cat w.r.t to that center of the path traced by the cat .(Why did you not calculate Angular displacement of cat w.r.t to that center of the path traced by the cat)

This is not the way angular velocity of cat is calculated .We need to consider only one circle i.e I have shown in the figure in earlier post .The reference point is the center and reference line is the vertical line .
 
  • #11
Have you been able to solve the problem given that the cat and mouse have the same angular velocity? I would recommend working in polar coordinates, if that is something you are comfortable with.
 
  • #12
Tanya Sharma said:
This is not the way angular velocity of cat is calculated .We need to consider only one circle i.e I have shown in the figure in earlier post .The reference point is the center and reference line is the vertical line .

Thank you for explanation.
 
  • #13
Satvik Pandey said:

What is radial velocity?
The total velocity of the bodies is composed of a radial part, vr, and a tangential part, vθ in plane polar coordinates. The mouse is constrained to move on a circular path, so what does this tell you about its radial velocity? The constant speed in the question refers to the total velocity v, so it can be decomposed.
 
  • Like
Likes 1 person
  • #14
The radial velocity is the projection of the total velocity on the radial direction, i.e., the velocity with which the r-coordinate of the cat increases. The component orthogonal to this is the velocity in the angular direction. The sum of their squares should equal the square of the total velocity by Pythagoras' theorem, as written in my previous post.
 
  • #15
Could anybody send me the link from where I can study about radial velocity and polar coordinates. Till now I don't have any knowledge about these things.
 
  • #16
Satvik Pandey said:
Could anybody send me the link from where I can study about radial velocity and polar coordinates. Till now I don't have any knowledge about these things.
Plane polar coordinates are useful when the problem admits a circular set up. For a motivation and for descriptions of other coordinate systems see http://www.math.oregonstate.edu/hom...lculusQuestStudyGuides/vcalc/coord/coord.html

The velocity of a particle in polar coordinates is derived in this video:
The terms in the final expression constitute the radial and tangential parts talked about earlier.

Do you have a calculus textbook? That would of course be the best resource for learning about this otherwise the links I give you may seem to be disconnected.
 
Last edited by a moderator:
  • #17
This problem would be much more interesting if the instantaneous velocity vector of the cat was directed along the line between the cat and the mouse. Anyone interested in working on this one?

Chet
 
  • #18
Hello Chet

Chestermiller said:
This problem would be much more interesting if the instantaneous velocity vector of the cat was directed along the line between the cat and the mouse. Anyone interested in working on this one?

Chet

Isn't the velocity of cat always directed towards the mouse in the OP ?
 
  • #19
Tanya Sharma said:
Hello Chet



Isn't the velocity of cat always directed towards the mouse in the OP ?
Hi Tanya. No, in the OP, the radial component of the cat's velocity vector is directed towards the mouse, not the cat's total velocity vector. The cat always remains on the radius connecting the center of the circle to the mouse.

Chet
 
  • #20
OK.

Does the picture correctly represents the situation in OP ? Blue dot represents the cat and red represents the mouse .Green is radial and pink is transverse velocity .

Does the brown vector correctly represents the instantaneous velocity in the OP ?

Chestermiller said:
This problem would be much more interesting if the instantaneous velocity vector of the cat was directed along the line between the cat and the mouse. Anyone interested in working on this one?

Chet

But if the instantaneous velocity vector of the cat was towards the mouse ,that means the velocity has only the radial component and in that case how is it possible for the cat to catch the mouse ?

Doesn't the cat need to have an angular(transverse) component of velocity as well ?
 

Attachments

  • cat.GIF
    cat.GIF
    6.8 KB · Views: 677
Last edited:
  • #21
Tanya Sharma said:
But if the instantaneous velocity vector of the cat was towards the mouse ,that means the velocity has only the radial component
If the cat always directs its velocity towards the mouse then it will not maintain the same angular velocity. It will 'fall behind' in that sense. Thus, in continuing to aim for the mouse it will acquire a tangential velocity.
 
  • #22
haruspex said:
If the cat always directs its velocity towards the mouse then it will not maintain the same angular velocity. It will 'fall behind' in that sense. Thus, in continuing to aim for the mouse it will acquire a tangential velocity.

Right .

But then what did Chet meant in post#17 ?
 
  • #23
CAF123 said:
The total velocity of the bodies is composed of a radial part, vr, and a tangential part, vθ in plane polar coordinates. The mouse is constrained to move on a circular path, so what does this tell you about its radial velocity? The constant speed in the question refers to the total velocity v, so it can be decomposed.

If mouse is moving in circular path then its radial velocity should be zero.
 
  • #24
Satvik Pandey said:
If mouse is moving in circular path then its radial velocity should be zero.

Correct, but the cat's velocity will have both a radial and an angular component.
 
Last edited:
  • #25
Satvik Pandey said:
If mouse is moving in circular path then its radial velocity should be zero.
Yes. So you know that the constant speed of the mouse in the question refers to the tangential speed of the mouse. You correctly calculated somewhere in the thread that the angular speed of the mouse (and therefore the cat since they are equal) is 1/7.

The cat's velocity is not purely radial or tangential. Since it is getting closer and closer to the mouse, its tangential velocity will be increasing (so a function of r). Find this function knowing the cat's angular speed.
 
  • #26
CAF123 said:
Yes. So you know that the constant speed of the mouse in the question refers to the tangential speed of the mouse. You correctly calculated somewhere in the thread that the angular speed of the mouse (and therefore the cat since they are equal) is 1/7.

The cat's velocity is not purely radial or tangential. Since it is getting closer and closer to the mouse, its tangential velocity will be increasing (so a function of r). Find this function knowing the cat's angular speed.

Let 'v' be the tangential velocity at a distance 'x' from the given point.
So the required function is-
v=xω
 
  • #27
Satvik Pandey said:
Let 'v' be the tangential velocity at a distance 'x' from the given point.
So the required function is-
v=xω
Ok, but you know ω too right? And use the more familiar notation vt = rω if that is okay.

The next step is to rewrite vt using the Pythagoras theorem. The motivation for doing this is so that you can make contact with the radial velocity of the cat. Ultimately you will solve a separable differential equation.
 
  • #28
Tanya Sharma said:
Right .

But then what did Chet meant in post#17 ?
In the version presented in the OP, the cat has to anticipate the movement of the mouse in order to remain on the same radial line as the mouse. Therefore, his resultant velocity vector is directed ahead of the radial line, but his radial velocity component is in the direction of the mouse. This is what haruspex was referring to.

In the version I'm proposing, the resultant velocity vector of the cat is always pointing in the direction from the cat to the mouse. This is a different problem from that in the OP. Does anybody have any interest in formulating this problem?

Chet
 
  • #29
CAF123 said:
Ok, but you know ω too right? And use the more familiar notation vt = rω if that is okay.

The next step is to rewrite vt using the Pythagoras theorem. The motivation for doing this is so that you can make contact with the radial velocity of the cat. Ultimately you will solve a separable differential equation.

Let V be the resultant velocity,Vt be the tangential velocity and Vθ be the radial velocity.
So,V=√(Vt^2 +Vθ^2) or Vθ =√(V^2-Vt^2).
 
  • #30
Satvik Pandey said:
Let V be the resultant velocity,Vt be the tangential velocity and Vθ be the radial velocity.
So,V=√(Vt^2 +Vθ^2) or Vθ =√(V^2-Vt^2).
Okay, but that is a peculiar notation since θ is usually associated with the angular direction. Sub in the expression for velocity in tangential direction and reexpress the radial velocity in terms of the time derivative of the radial displacement (that is displacement of cat from the centre of the circle).
 

Similar threads

Replies
5
Views
1K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
28
Views
4K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
30
Views
3K
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K