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Physics Problem & Solutions Thread

Anonymous Mon 03 Aug 20:08:41 2020 No.11970030 View Reply Original Report

Quoted By: >>11970035

ITT we present physics problems we enjoyed, and then reply to our posts with our solutions to these problems.
The hope is that this thread shows us interesting problems and interesting solutions we can all learn from.

> inb4 this is a homework thread
No, anons must present the solutions to the problems they pose.

I'll start.

(Problem is from https://openstax.org/books/university-physics-volume-1/pages/8-additional-problems, #72).

A body of mass $m$ and negligible size starts from rest and slides down the surface of a frictionless solid sphere of radius $R$ . (See image.) Prove that the body leaves the sphere when $\theta = \arccos(\frac{2}{3})$ .

Anonymous

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Anonymous Mon 03 Aug 2020 20:09:57 No.11970035 Report

Quoted By: >>11970039

>>11970030
Solution:
Consider the situation when the mass is just fallen off the sphere. At this point, the only force acting on the mass is gravity. (As the mass has fallen off the sphere, there is no normal force.)

My poorly-drawn FBD can be seen in the image, but the important thing we take away from it is that the force of gravity can be decomposed into two forces, one of which, $mg\cos(\theta)$ , provides a centripetal acceleration with the center of the sphere at the center.

We can thus write the following equation: $mg\cos(\theta) = ma_c[\eqn]<br />We solve for <script type="math/tex">\cos(\theta)$ : $\cos(\theta) = \frac{ma_c}{mg} = \frac{a_c}{g}[\eqn]<br />Recall that centripetal acceleration is equal to <script type="math/tex">\frac{v^2}{R}$ . We substitute this expression for $a_c$ : $\cos(\theta) = \frac{v^2}{gR}$

We now must solve for $v^2$ . To do so, we realize that the velocity (and velocity squared) can be found via applying the conservation of energy principle. We know that in this problem, the potential energy of the mass's height is changed to kinetic energy.

We then write the following equation: $E_{p_{lost}}=E_{k_{gained}}$
And solve: $\frac{1}{2}mv^2 = mgh$ $v^2 = \frac{2mgh}{m} = 2gh[\eqn]<br />(Check my diagram to see where <script type="math/tex">h$ came from.)

We plug this back into the expression we derived for $\cos(\theta)$ : $\cos(\theta) = \frac{v^2}{gR} = \frac{2gh}{gR} = \frac{2h}{R}$

We now must get rid of $h$ . To do so, take a look at my diagram and notice that $\cos(\theta) = \frac{R - h}{R}$ . Thus, $\frac{2h}{R} = \frac{R-h}{R}$ , and so $h = \frac{R}{3}$ .
Therefore, $\cos(\theta) = \frac{2h}{R} = \frac{2\frac{R}{3}}{R} = \frac{2}{3}$
This means $\theta = \arccos(\cos(\theta)) = \arccos(\frac{2}{3})$

It is my first time using \LaTeX on /sci/, so I apologize for any formatting issues.

Anonymous

Anonymous Mon 03 Aug 2020 20:11:40 No.11970039 Report

Quoted By: >>11970481

>>11970035
Argh, let me fix the simple errors in the Latex:

Solution:
Consider the situation when the mass is just fallen off the sphere. At this point, the only force acting on the mass is gravity. (As the mass has fallen off the sphere, there is no normal force.)

My poorly-drawn FBD can be seen in the image, but the important thing we take away from it is that the force of gravity can be decomposed into two forces, one of which, $mg\cos(\theta)$ , provides a centripetal acceleration with the center of the sphere at the center.

We can thus write the following equation: $mg\cos(\theta) = ma_c$
We solve for $\cos(\theta)$ : $\cos(\theta) = \frac{ma_c}{mg} = \frac{a_c}{g}$
Recall that centripetal acceleration is equal to $\frac{v^2}{R}$ . We substitute this expression for $a_c$ : $\cos(\theta) = \frac{v^2}{gR}$

We now must solve for $v^2$ . To do so, we realize that the velocity (and velocity squared) can be found via applying the conservation of energy principle. We know that in this problem, the potential energy of the mass's height is changed to kinetic energy.

We then write the following equation: $E_{p_{lost}}=E_{k_{gained}}$
And solve: $\frac{1}{2}mv^2 = mgh$ $v^2 = \frac{2mgh}{m} = 2gh$
(Check my diagram to see where $h$ came from.)

We plug this back into the expression we derived for $\cos(\theta)$ : $\cos(\theta) = \frac{v^2}{gR} = \frac{2gh}{gR} = \frac{2h}{R}$

We now must get rid of $h$ . To do so, take a look at my diagram and notice that $\cos(\theta) = \frac{R - h}{R}$ . Thus, $\frac{2h}{R} = \frac{R-h}{R}$ , and so $h = \frac{R}{3}$ .
Therefore, $\cos(\theta) = \frac{2h}{R} = \frac{2\frac{R}{3}}{R} = \frac{2}{3}$
This means $\theta = \arccos(\cos(\theta)) = \arccos(\frac{2}{3})$

Anonymous

Anonymous Mon 03 Aug 2020 22:43:12 No.11970481 Report

Quoted By:

>>11970039
Have you tried the case with the surface having friction? I remember doing these types of problems before uni

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