TutorMe homepage

SIGN IN

Start Free Trial

Bryan S.

B.S. in Physics, Lehigh University

Tutor Satisfaction Guarantee

Physics (Newtonian Mechanics)

TutorMe

Question:

A person holds a string that is firmly attached to a rock. The person is standing on even ground. Underneath the person is a picture of a compass face. The person begins spinning clockwise. After accelerating to a certain speed, the string is completely taut, so that the string makes a 180 degree angle with the ground. At an instant, the rope breaks - the rope also breaks while the rock is directly over the North part of the compass face. Describe the motion of the rock the instant the rock breaks.

Bryan S.

Answer:

Since the person is spinning, we must immediately identify this is a circular motion problem. We know that in circular motion, a force perpendicular to the motion of the force - a normal force - is required to continue the circular motion. In this case, there is only one possible source - the string! We see how this makes sense, because if we take the velocity of the rock at any moment, it will be tangent to the circular path of the rock, and therefore perpendicular to the string. Taking this knowledge, we imagine the moment the string breaks. The first thing we realize is that there is no longer a normal force acting on the string - it can't be tense, and therefore act as the normal force if it's broken! However - we do need to note there is still a force acting on the rock - gravity! So as far as the vertical motion, we know what will happen - the rock will start to fall. The vertical component of the tension is what counteracted gravity. However, now we must decide what will happen from the horizontal motion. But our earlier analysis of the problem tells us exactly what will happen. At any given moment, the rock is moving tangentially to the normal force. So if there is no force containing the rock in circular motion, then the rock will continue to travel in that tangential direction. But which tangential direction? Since we know that the person is spinning clockwise, whenever the rock is directly over the north direction the rock is moving exactly east. So in conclusion: when the rope breaks, the rock will both fall and move east!

Physics (Electricity and Magnetism)

TutorMe

Question:

A 120 V circuit is connected in series with a 10 ohm resistor, 20 ohm resister, and 30 ohm resistor. The wiring carries no resistance. What is the current through the 20 ohm resistor?

Bryan S.

Answer:

The following problem is a classic simple direct current problem. Immediately when we look at the question, we identify that the problem gave us the voltage of the potential source, and the resistance of the system. Thus, our first thought should be Ohm's law - the simplest of all circuit laws, V=IR. Automatically, we know the voltage - 120 volts. However, the resistance seems to pose a problem - we have three resistors! Luckily for us, we know that in a circuit, if multiple elements are in series, they share the same current. So we can deduce that in this system, the current I is the same for all elements - so if we find the current through at least one element, we have the current for any element. Taking the simplification a step further, another basic circuit fact is that the resistance of two resistors can be added if they are in series - meaning the simplified resistance of the system is 60 ohms. Doing this makes the algebra for our problem very simple, we have one V value, one R value, and then one I value! And since the current of all elements in series is the same, if we have the current for one element - or in this case, all elements - we also have the current for a single resistor! Now for the easy part! We have V and R, so using Ohm's law - plugging in V and R and solving for I, we conclude the current I is 2 amps.

Calculus

TutorMe

Question:

A light post and a person stand at the same point on flat ground. The light post has a height "h" and makes a 90 degree angle with the ground. The person begins to walk to the right at constant speed. A line segment R with two points - one at the top of the post, and one on the ground representing the location of the person - changes at a rate of 100 meters/minute. In terms of the height of the post h, what is the distance between the post and the person (i.e. how far has the person walked from the post in terms of h) after 100 minutes?

Bryan S.

Answer:

Please don't be scared off by the length of this answer - I try to be very thorough, and rather than just tossing a solution at you, I try to develop my thinking so that you can understand how to think about problems instead of churning out pre-memorized solutions! Ultimately, answering this question comes down to very little calculus; in fact, it is almost entirely interpretation of the question. However, in my experience as a student and friend who tutored casually, these are the questions that stump people the most. Step One: Interpret the question. The behavior of the question is very simple: the person starts at the light post and then begins walking at a constant rate. The first important identification of this problem is that geometrically, we are discussing a triangle. You can see this by drawing the post with height h, and then drawing the person at any distance away. The post height and the rate of change of the hypotenuse are given to us in the problem, and the base of the triangle is the distance between the post and the person - the quantity we are looking for! Step two: identify preliminary constants and variables. We are given two constants: the height of the post h, and the rate of change of the hypotenuse dR. The variable we automatically know we want is the distance between the post and the person, let's call it L. Step three: decide strategy. This step has two important realizations. The first is that since we are dealing with a triangle, trigonometric functions and the Pythagorean theorem should be the first tools to come into mind. The second is that the problem wants things in terms of h, so we need to manipulate the height h through the geometric properties of the triangle to get L! For this problem, let's try working backwards. Since we're working with a triangle, the second to last step is very likely to be solving the Pythagorean theorem. Our three side lengths would be R, L, and h. Note we already have h, so this seems like a really good method! So, let's try completing the Pythagorean theorem. For this to work, we need to find an expression for R at t = 100 minutes. Immediately we realize that the problem gave us a constant rate of change dR. Our first instinct might be to integrate, but rather, let's set up and expression for R. Very simply, this expression would be: R = r + dR * t, where r is the value of the hypotenuse at t = 0. At this point I'm sure you're protesting, "but when L = 0, there is no triangle!!!!!". Well, plug 0 into the Pythagorean theorem while L = 0, you'll notice a very nice little fact - when the person is at the same point as the post, the height h and the hypotenuse R are the same! So, we have found that r = h. Plugging this into our equation for R, we see that R = h + dR*t. This is where things get easier. The problem specified the time t and the rate of change dR - so we now know that R at 100 minutes is equal to h + 10,000. And since we have the value of this hypotenuse, we can use the Pythagorean theorem to finish this up. Remembering to foil h+ 10,000, we find that L^2 = (h^2) + (h^2) + 20,000h + 10^8. Note that I put 10,000^2 in scientific notation for ease. The quantity L we want is the square root L^2! And that's it, this problem is finished! You would report that L is equal to the (POSITIVE) square root of the expression for L^2 above, and that is a perfect valid answer - unless we knew the numerical value of h, there isn't much useful simplification to be done.

Send a message explaining your

needs and Bryan will reply soon.

needs and Bryan will reply soon.

Contact Bryan

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2018 TutorMe.com, Inc.