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

In summary, the problem involves a mouse running on a circular path with a radius of 28m at a constant speed of 4m/s, and a cat chasing it from the center of the path with a constant velocity of 4m/s. The chase lasts for 11 seconds, and the angular velocities of the cat and mouse are equal. The cat follows a curved path, and the calculation of its angular velocity is based on the center of the circle and a reference line. The use of polar coordinates is recommended for solving this problem.
  • #36
Tanya Sharma said:
Hello Chet



Isn't the velocity of cat always directed towards the mouse in the OP ?

What is OP?:confused:
 
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  • #37
Satvik Pandey said:
At a distance 'r' from center tangential velocity is rω or rV/R.
I don't understand the second equation. Yes, the tangential velocity of the cat is ##v_{t, C} = r\omega_{C} = r \omega_{M} = ?## Sub in what you got for the angular speed of the mouse earlier on. {t,C} stands for tangential velocity of the cat and M is a label for the mouse.
So, [itex]\frac{dr}{dt}[/itex]=√(V^2-(rv/R)^2).
Right idea, but make the correction above in this formula.

After some algebraic calculation I found this-
dx/[itex]\sqrt{R^{2}-r^{2}}[/itex]=[itex]/\frac{V}{R}[/itex]dt. Upper and lower limits of LHS is R and 0 respectively. Upper and lower limits of RHS is T and 0 respectively.
You want the time it takes for the cat to reach R=28 having started at R=0. So you know the limits for R numerically. Your bounds for t are right. You just need to fix the integrand and then perform the integration.
 
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  • #38
CAF123 said:
I don't understand the second equation. Yes, the tangential velocity of the cat is ##v_{t, C} = r\omega_{C} = r \omega_{M} = ?## Sub in what you got for the angular speed of the mouse earlier on. {t,C} stands for tangential velocity of the cat and M is a label for the mouse.
I just replaced ω.Here V is the velocity of mouse and R is 28cm(radius).I did this for calculating ω earlier in this discussion.
.
 
  • #39
Satvik Pandey said:
What is OP?:confused:

Original poster/post .I was referring to post#1 i.e to the question.
 
  • #40
Satvik Pandey said:
I just replaced ω.Here V is the velocity of mouse and R is 28cm(radius).I did this for calculating ω earlier in this discussion.
.
Ah okay, I misread the notation. So, V/R = 1/7. Sub this into your expression for the radial speed and the integral.
 
  • #41
Tanya Sharma said:
Thanks Chet .

I am having difficulty in interpreting whether the pursuer’s velocity given in similar type of problems is radial velocity or instantaneous velocity .When the question says that the pursuer is always heading towards the target ,doesn’t it mean that the velocity given is instantaneous velocity ?

Here are two similar type of questions .


Q 1. A boy is on a boat, at a distance H from the shore, when he sees a girl (at the point on the shore where the distance is measured) running with a constant velocity u parallel to the shore. At that time, he moves towards her, with a speed v, in such a way, that the point of the boat is always pointed at the girl (so his vector is always pointing her way). Find the time of their meeting.

Q 2. Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u < v. At the initial moment of time v is perpendicular to u and the points are separated by a distance l. How soon will the points converge?

What is your opinion regarding the velocity v given in the problems ? Do you think that the velocity v given in both the problems are the radial velocity or the instantaneous velocity ?
Maybe it would help if I started to set up the version of the problem that I'm thinking of. Let R and θM be the polar coordinates of the mouse at some particular time t, and let r and ΘC be the polar coordinates of the cat at the same time t. The velocity of the cat is equal to V (a constant speed) times a unit vector in the direction from r,ΘC to R,θM. Note, that in this version of the problem, ΘC≠θM (except initially).

Please tell me if this makes any sense so far.

Chet
 
  • #42
Chestermiller said:
Maybe it would help if I started to set up the version of the problem that I'm thinking of. Let R and θM be the polar coordinates of the mouse at some particular time t, and let r and ΘC be the polar coordinates of the cat at the same time t. The velocity of the cat is equal to V (a constant speed) times a unit vector in the direction from r,ΘC to R,θM. Note, that in this version of the problem, ΘC≠θM (except initially).

Please tell me if this makes any sense so far.

Does the attached picture correctly represents the setup ? The brown vector represents the instantaneous velocity V of cat.Pink radial velocity Vr and green transverse velocity VT .
 

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  • #43
ehild said:
In the OP, v is the magnitude of velocity, that is, speed.

ehild

I would like to know whether in the two problems in post#32 'v' represents the magnitude of radial velocity OR magnitude of instantaneous velocity ?
 
  • #44
Tanya Sharma said:
I am having difficulty in interpreting whether the pursuer’s velocity given in similar type of problems is radial velocity or instantaneous velocity .When the question says that the pursuer is always heading towards the target ,doesn’t it mean that the velocity given is instantaneous velocity ?

Or it is speed given.

Tanya Sharma said:
Here are two similar type of questions .


Q 1. A boy is on a boat, at a distance H from the shore, when he sees a girl (at the point on the shore where the distance is measured) running with a constant velocity u parallel to the shore. At that time, he moves towards her, with a speed v, in such a way, that the point of the boat is always pointed at the girl (so his vector is always pointing her way). Find the time of their meeting.

The speed of the boat is v, and the instantaneous velocity ##\vec v## points towards the girl.

Tanya Sharma said:
Q 2. Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u < v. At the initial moment of time v is perpendicular to u and the points are separated by a distance l. How soon will the points converge?

I do not understand the wording of this problem. What does it mean, moving rectilinearly? If the direction of the velocity changes with time, it is not "uniform". For velocities, you can not say that one is greater than the other unless they are parallel. You can compare the speeds.

Tanya Sharma said:
What is your opinion regarding the velocity v given in the problems ? Do you think that the velocity v given in both the problems are the radial velocity or the instantaneous velocity ?

Velocity is a vector and independent of the coordinate system used. If it changes with time, you can speak about instantaneous velocity.
If you have a planar polar coordinate system, velocity has both radial and azimuthal (tangential) components, both depending on time. At the same time, the speed can be constant.
The radial velocity is the radial component of the velocity vector.

ehild
 
  • #45
Tanya Sharma said:
Does the attached picture correctly represents the setup ?
Yes.
The brown vector represents the instantaneous velocity V of cat.Pink radial velocity Vr and green transverse velocity VT .
I would have drawn it a little differently, with the angles measured from the x - axis, and the mouse running counterclockwise. But this is OK too.

Are you interested in working on this problem? The formulation of the problem should be interesting and instructive; the equations may not have an analytic solution, however.

Chet
 
  • #46
Chestermiller said:
Are you interested in working on this problem? The formulation of the problem should be interesting and instructive; the equations may not have an analytic solution, however.

Sure :smile:

But first I would let Satvik finish his original problem .I have already derailed the thread :shy:.After he completes the problem I will get back to you .

Thanks Chet .
 
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  • #47
Yes I solved it.
sin[itex]^{-1}[/itex][itex]\frac{r}{R}[/itex]=V/R*T.
where the lower limit and upper limit of LHS are 0 and 28 respectively. On substituting the values I got T=11 seconds.THANK YOU ALL FOR GUIDING ME.
 
  • #48
Tanya Sharma said:
Sure :smile:

But first I would let Satvik finish his original problem .I have already derailed the thread :shy:.After he completes the problem I will get back to you .

Thanks Chet .

The above discussion would be helpful for me even.After solving this I will try to go through the above discussion.Thank You.
 

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