Conservation of Energy in circular motion

[terryds]

Sure, but one step at a time. You cannot start analysing the parabolic arc until you've figured out at what angle the block leaves the track.
You need to obtain the speed as a function of angle (while on the track) using conservation of energy. Then you need think about forces to discover at what combination of speed and angle it will fall off the track.

In my previous post in this thread, I've figured that at the angle sin^-1 2/3 or at the height of R+R sin 2/3 , the block will leave the track.
Please help me.. I really have no idea to obtain the speed as function of the angle.
I'll be very glad if you give me the solution or the hint and explain it :)

 

[haruspex]

In my previous post in this thread, I've figured that at the angle sin^-1 2/3 or at the height of R+R sin 2/3 , the block will leave the track.

Ok, though I still can't see where you did that - but it's the right answer.
So you can work out the KE there, and thus the speed. And you know the 'launch' angle, since it starts tangential to the arc. You've studied uniform acceleration and trajectories, yes? What equations do you know for those (particularly in relation to vertical motion)?

 

[terryds]

Ok, though I still can't see where you did that - but it's the right answer.
So you can work out the KE there, and thus the speed. And you know the 'launch' angle, since it starts tangential to the arc. You've studied uniform acceleration and trajectories, yes? What equations do you know for those (particularly in relation to vertical motion)?

This is how I solved the angle and the tangential velocity: https://www.physicsforums.com/threads/conservation-of-energy-in-circular-motion.795189/#post-4993826

The formulas I know are
Vy(0) = V sin Θ
Vx(0) = V cos Θ
Vy(t) = Vy(0) - g t

The Y max is reached when the velocity in y-axis is zero
Vy(t)^2 = Vy(0)^2 - 2 g Δy
0 = (V sin Θ)^2 - 2 g Δy
0 = (√(g R sin Θ) sinΘ)^2 - 2 g Δy
0 = g R (sin Θ)^3 - 2 g Δy
Δy = 1/2 R(sin Θ)^3
Δy = 1/2 R 8/27
Δy = 8/54 R
So, the max height is R + 2/3 R + 8/54 R = 49/27 R

Actually, I have figured it out in my previous post.

Am I right ? Or do I miss something ?

I'm just doubt on the launch angle value. Is it the same angle as the angle from the horizontal to the position where the block leaves the track ? My intuition says that it's same.

 

[haruspex]

This is how I solved the angle and the tangential velocity: https://www.physicsforums.com/threads/conservation-of-energy-in-circular-motion.795189/#post-4993826

The formulas I know are
Vy(0) = V sin Θ
Vx(0) = V cos Θ
Vy(t) = Vy(0) - g t

The Y max is reached when the velocity in y-axis is zero
Vy(t)^2 = Vy(0)^2 - 2 g Δy
0 = (V sin Θ)^2 - 2 g Δy
0 = (√(g R sin Θ) sinΘ)^2 - 2 g Δy
0 = g R (sin Θ)^3 - 2 g Δy
Δy = 1/2 R(sin Θ)^3
Δy = 1/2 R 8/27
Δy = 8/54 R
So, the max height is R + 2/3 R + 8/54 R = 49/27 R

Actually, I have figured it out in my previous post.

Am I right ? Or do I miss something ?

I'm just doubt on the launch angle value. Is it the same angle as the angle from the horizontal to the position where the block leaves the track ? My intuition says that it's same.

I pointed out in post #28 that you have some errors in that calculation.

 

[terryds]

I pointed out in post #28 that you have some errors in that calculation.

Given your previous equation, that implies V = √(g R sin Θ). Again, as Θ increases, will V increase or decrease? To get the KE at the position in the diagram, calculate how far the block is below its starting point

As Θ increases, V increases since it is sinus function (in 0-90 degree quadrant).

Hmm.. I think I almost figured it out, right ? The only thing I miss is just the launch angle (Let's call this α )
http://www.sumoware.com/images/temp/xzihalbxxfsximxp.png
Could you please tell me what is the relation between α and Θ ?

 

[haruspex]

As Θ increases, V increases since it is sinus function (in 0-90 degree quadrant).

Hmm.. I think I almost figured it out, right ? The only thing I miss is just the launch angle (Let's call this α )
http://www.sumoware.com/images/temp/xzihalbxxfsximxp.png
Could you please tell me what is the relation between α and Θ ?

As theta increases, the block is rising, so must be losing KE and slowing.
Further, the fraction of the speed which is vertical is also decreasing, so it can't be as sin theta.

 

[terryds]

As theta increases, the block is rising, so must be losing KE and slowing.
Further, the fraction of the speed which is vertical is also decreasing, so it can't be as sin theta.

So, it is cos theta ? Right ?

 

[haruspex]

So, it is cos theta ? Right ?

Yes v_y = v cos(theta). Now, what about the value of v as a function of theta?

 

[terryds]

Yes v_y = v cos(theta). Now, what about the value of v as a function of theta?

Do you mean the velocity as a function of theta when it's still on the track ?
E = E'
mg (2R) = mgh' + 1/2 mv^2
g (2R) = g(R+R sinΘ) + 1/2 v^2
2gR - gR - gR sinΘ = 1/2 v^2
gR (1-sin Θ) = 1/2 v^2
v = √(2 gR(1-sinΘ))

Is it right ?

 

[haruspex]

_y

Do you mean the velocity as a function of theta when it's still on the track ?
E = E'
mg (2R) = mgh' + 1/2 mv^2
g (2R) = g(R+R sinΘ) + 1/2 v^2
2gR - gR - gR sinΘ = 1/2 v^2
gR (1-sin Θ) = 1/2 v^2
v = √(2 gR(1-sinΘ))

Is it right ?

Right. So v_y is?

 

[terryds]

_y
Right. So v_y is?

v_y = v cos Θ
v_y = cos Θ √(2 gR(1-sinΘ))

The sin of angle θ is 2/3
So, the cos of angle Θ is √5 / 3
v_y = √5 / 3 √(2 gR(1-(2/3)))
v_y = √((10/27) gR)

Then,
V_y(t) ^2 = V_y(0) ^2 - 2 g Δy
0 = (10/27) gR - 2g Δy
Δy = 5/27 R

Then, the maximum height is R+(2/3)R+(5/27)R = (50/27) R
Right ?
Please tell me if I missed something

 

[haruspex]

v_y = v cos Θ
v_y = cos Θ √(2 gR(1-sinΘ))

The sin of angle θ is 2/3
So, the cos of angle Θ is √5 / 3
v_y = √5 / 3 √(2 gR(1-(2/3)))
v_y = √((10/27) gR)

Then,
V_y(t) ^2 = V_y(0) ^2 - 2 g Δy
0 = (10/27) gR - 2g Δy
Δy = 5/27 R

Then, the maximum height is R+(2/3)R+(5/27)R = (50/27) R
Right ?
Please tell me if I missed something

That's what I get.