Seemingly simple projectile motion problem.

[tempneff]

1. A projectile motion over a circle with radius r becomes tangent with the circle at 45 degrees. What is the velocity at its highest point/



2. Constant acceleration and energy



3. The angle between the tangent points and the topmost point at the center of the circle is 45 degrees. If the initial velocity is v_i, the tangent velocity is v and the topmost velocity is v_t I know the following: In the x direction all these velocities are equal. Both components of the middle velocity are equal. This is also the total velocity of the highest point. The height of the tangent point is r+rcos45. The height from the tangent to the topmost point is rsin45. The topmost height from the bottom is r+ 2rsin45. Where do I go from here?

 

[Stonebridge]

The velocity at the highest point has no vertical component, just the horizontal. This component is the same as it was at the point the projectile left the circle.
Let the speed of the projectile in the circle = v
Find its horizontal component at the 45 degree position.

 

[tempneff]

I just said that.

 

[Stonebridge]

I just said that.

The question wants the velocity at P or Q.
So the answer is? [=the horizontal component v_x of the velocity at the point where it leaves the circle...]

 

[tempneff]

OVER the circle. More like this.

[PLAIN]http://d.imagehost.org/0966/Untitled_14.jpg

I agree that the velocity at the top will be equal to the velocity in the x direction at any other point (as I said originally) However, I am expected to get a unique numerical value for the velocity at the top.

I know the answer but I can't prove it yet.

At the top: V²= cos 45(appx .71)

 

[Stonebridge]

OK. [I wasn't sure what the question actually meant. Your diagram makes it clearer what you consider to be happening*]

The velocity at the top is just v_x the horizontal component at any point in the trajectory. [we are agreed on that]
At the 45 deg tangent point, the two components, horizontal and vertical, are equal.
v_x = v_y
At the start of the trajectory, the vertical component is u_y say.

So you have a formula for the initial velocity v_i² = v_x² + u_y²

You also know that at the tangent point, the vertical component there is related to the vertical component at the start by
v_y² = u_y² - 2gh

You also know that h = r + r cos 45


*It's not stated explicitly in the question that the projectile is supposed to be tangential at both 45 deg points, on the way up and the way down;
or that the projectile starts from a point level with the bottom of the circle.

It does say "over" the circle, so my trejectory with max height Q can be ignored.

 

[Stonebridge]

I've thought further about this and there is a simpler solution that assumes nothing other than the projectile makes 2 tangents to the circle at P and Q.

Eliminate t between those 2 equations and there is your answer.
It has to include g and r
Just cos 45 = V² would not be dimensionally correct.

 

[tempneff]

This is exactly my problem. I cannot seem to eliminate r and g, however that is the assignment. Others have completed it and found an initial angle of 67.5 degrees, a initial velocity^2 of 4.83 and then top v^2 of 1/root 2. I was told to solve for r in terms of theta (45). g has a value already (assuming we are on earth)

 

[tempneff]

I can now prove that the initial angle is 67.5 degrees. PLease help me to end this madness!