# Speed of a point

1. Feb 4, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
I was able to calculate the velocity of point P in x direction (to the left). For the speed, I need to find the velocity in y direction (vertically up). Here is my attempt:
At t=0, the distance of point P from O is l, let at time t, the angle POR be $\theta$. The displacement is $y=l(\sin \theta -1)$. Differentiating this equation, $\frac{dy}{dt}=l \cos \theta \cdot \frac{d \theta}{dt}$. The term dθ/dt is equal to the angular velocity but I don't know the radius of curvature of its path which I have to use in the formula ω=vr, where ω is the angular velocity, v is the velocity in x direction and r is the radius of curvature.

Any help is appreciated. Thanks!

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2. Feb 4, 2013

### tms

How far does P get from O after t =0?

3. Feb 4, 2013

### Saitama

$l \cos \theta$.

4. Feb 4, 2013

### Curious3141

You're overcomplicating things. Let ∠SUT = 2θ (so you're looking for what happens when θ = 45 deg).

Take O as the origin.

Let $u_x, u_y, p_x, p_y$ be (respectively), the co-ordinates of U and P at time t.

Can you find simple expressions for those in terms of l and θ? (except that $u_y$ is simply constant at zero).

Can you find $\frac{du_x}{dt}$ in terms of $\frac{d\theta}{dt}$ using Chain Rule? Hence rearrange to determine a numerical value for $\frac{d\theta}{dt}$ at the instant of interest.

Can you differentiate $p_x, p_y$ wrt t to find the instantaneous horizontal and vertical velocities of P?

Now use Pythagoras theorem to find the resultant speed.

5. Feb 4, 2013

### ehild

Find the coordinates of P in terms of the angle. The velocity components are the time derivatives of coordinates.

ehild

Edit:Curious beat me ...

6. Feb 5, 2013

### Saitama

Thanks a lot Curious, that solved the problem.

The velocity of P in both the directions come out to be same!

7. Feb 5, 2013

### Curious3141

That's because the sine and cosine of 45 deg are the same!

8. Feb 5, 2013

### tms

Not the x coordinate, but the distance from O. I was just trying to point out that P moves in a circle.

9. Feb 5, 2013

### Saitama

Yep, didn't notice that at all. Thanks!