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).
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
#4
TheBigBadBen
79
0
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?
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles.
In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra
Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/
by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation
$$
a^n+b^n=c^n
$$
has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy
"Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation
$$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$
is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
