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#### DrunkenOldFool

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- Feb 6, 2012

- 20

- Thread starter DrunkenOldFool
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- Feb 6, 2012

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- Feb 1, 2012

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Simply use the energy conversion law.

- Jan 26, 2012

- 890

1. What is the condition that is satisfied at break-off?A small body $A$ starts sliding from the height $h$ down an inclined groove passing through into a half circle of radius $h/2$. Assuming friction to be negligiable, find the velocity of the body at the highest point of its trajectory(after breaking off the groove).

View attachment 268

2. What is the velocity and height at break-off?

3. After break-off the horizontal component of velocity is constant, and at greatest hight the vertical component of the velocity is zero.

4. The haximum height is determined by conservation of energy. You know the initial energy, it is the potential energy at A. The final energy is the potential energy at the greatest height plus the KE corresponding to the horizontal component of velocity at break-off.

CB

- May 12, 2013

- 84

Here's one approach to framing the problem:

Let $x$ be the height of the ball at its break-off point

Let

We note that an object will leave its circular orbit when the radial component of the force of gravity becomes greater than the centripetal force required to keep the ball in the circle. That is, the ball will stay on track as long as

$$

\frac{mv^2}{r} \geq mg \sin \theta

$$

Where $v$ is the speed of the ball, $m$ is its mass (which we can cancel out), $r$ is the radius of the circle, $g$ is the acceleration due to gravity near earth, and $\theta$ is the angle that a radius pointing to the ball would make with the horizontal. From the above equation, we can deduce that the break-off point will be the point at which

$$

\frac{v^2}{r} = g \sin \theta

$$

Now, we can make the above solvable for $x$ by using the following substitutions:

$$

\textbf{conservation of energy: }

\frac12 mv^2 = g(h-x)\Rightarrow v^2 = 2g(h-x)\\

\textbf{definition of our angle: }

\sin \theta = \frac{(x-h/2)}{h/2}=4\left(1-\frac x h\right) \\

\textbf{the radius given: }

r = \frac h2

$$

You should find $x = \frac56\, h$. Where could you go from there, using CB's approach?

**Hint:**

At the point of break-off, the ball's trajectory is perpendicular to the radius. What is the vertical component of velocity at the time of break-off if the ball makes an angle of $\frac{\pi}2-\theta$ with the horizontal?

Let $x$ be the height of the ball at its break-off point

Let

We note that an object will leave its circular orbit when the radial component of the force of gravity becomes greater than the centripetal force required to keep the ball in the circle. That is, the ball will stay on track as long as

$$

\frac{mv^2}{r} \geq mg \sin \theta

$$

Where $v$ is the speed of the ball, $m$ is its mass (which we can cancel out), $r$ is the radius of the circle, $g$ is the acceleration due to gravity near earth, and $\theta$ is the angle that a radius pointing to the ball would make with the horizontal. From the above equation, we can deduce that the break-off point will be the point at which

$$

\frac{v^2}{r} = g \sin \theta

$$

Now, we can make the above solvable for $x$ by using the following substitutions:

$$

\textbf{conservation of energy: }

\frac12 mv^2 = g(h-x)\Rightarrow v^2 = 2g(h-x)\\

\textbf{definition of our angle: }

\sin \theta = \frac{(x-h/2)}{h/2}=4\left(1-\frac x h\right) \\

\textbf{the radius given: }

r = \frac h2

$$

You should find $x = \frac56\, h$. Where could you go from there, using CB's approach?

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