What is the Velocity of a Body at the Highest Point in Vertical Circular Motion?

[DrunkenOldFool]

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

 

[Unknown008]

Simply use the energy conversion law.

 

[CaptainBlack]

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).

https://www.physicsforums.com/attachments/268

1. What is the condition that is satisfied at break-off?

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

 

[TheBigBadBen]

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:

Spoiler

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?

 

[CellCoree]

Vertical circular motion is a type of motion where an object moves in a circular path in a vertical plane. In this scenario, the small body $A$ is sliding down an inclined groove and enters a half circle of radius $h/2$. The velocity of the body at the highest point of its trajectory, after breaking off the groove, can be determined using the principles of conservation of energy.

Since friction is assumed to be negligible, the only forces acting on the body are its weight and the normal force from the surface of the groove. As the body slides down the groove, its potential energy decreases and is converted into kinetic energy. At the highest point of the trajectory, all of the body's potential energy is converted into kinetic energy, and there is no change in the body's kinetic energy due to the absence of any external forces.

Therefore, using the conservation of energy equation $E_{initial}=E_{final}$, we can write:

$mgh=\frac{1}{2}mv^2$

Where $m$ is the mass of the body, $g$ is the acceleration due to gravity, $h$ is the initial height of the body, and $v$ is the velocity of the body at the highest point of its trajectory.

Solving for $v$, we get:

$v=\sqrt{2gh}$

This is the velocity of the body at the highest point of its trajectory after breaking off the groove. It should be noted that this velocity will be the same regardless of the mass of the body, as long as friction remains negligible.

In conclusion, the velocity of the body $A$ at the highest point of its trajectory can be determined using the principles of conservation of energy and is given by $\sqrt{2gh}$. This type of analysis can be applied to various situations involving vertical circular motion, making it a useful tool for scientists and engineers.