A yo-yo lies on the floor. It’s made out of an inner disc of radius r and an outer disc of radius R. Its mass is m. There’s a (massless) string wound around its inner disc. You pull the string with force F and angle θ with respect to the horizontal direction. What is the acceleration of yo-yo? Assume the yo-yo rolls with slipping.

(If you’ve never done this before, try  it before answering the question and see which direction the yo-yo rolls.)

Remember those special surprise toys you used to get as a kid inside the cereal boxes you bought? I used to spill the cereal out, retrieve the toy and put the cereal back in. No matter how cheap or crappy it was, I had to have it.

Now it’s time to revisit your childhood as a math-wielding  adult. Suppose a box of cereal costs 5$, and each box has a toy in it. There are 5 different toys for you to collect; by collecting all of them you can assemble them together and create a giant robot. If the toys have equal probabilities of turning up - that is, each toy is 1/5 likely to appear in a randomly chosen cereal box - how much will you have to spend, on average, before you can assemble the giant robot of your dreams?

EDITED, 4/June/2009: a solution has been posted here. Hats off, and a giant robot, to our reader Andrew Brandon for his solution.

]]> http://www.physicallyincorrect.com/2009/07/math-puzzle-the-cereal-box-surprise/feed/ http://www.physicallyincorrect.com/2009/07/more-blog-statistics/ http://www.physicallyincorrect.com/2009/07/more-blog-statistics/#comments Wed, 01 Jul 2009 14:13:45 +0000 Assaf http://www.physicallyincorrect.com/?p=792 At the beginning of last month I gave you a look into this blog’s statistics.  It was then that I decided to post on a daily basis and see whether that would bring more visitors to the blog. It’s time to face up to reality now and see if my strategy worked, so let’s see how well my traffic fared compared to previous months. Has the daily-posting routine achieved the desired effect? To be honest, posting daily is quite draining and time consuming, so I don’t think I’ll be able to keep up with it in the future. First, let’s take a look at the number of visits. Here’s a screenshot of the AWSTATS plugin I use to monitor traffic:

There was indeed a rise in the number of visits (excluding the March anomaly, which I still can’t explain). Why is it that the number of unique visitors increased by about 30% while the number of pages viewed increased by 100%? I’m not sure.  In terms of unique visitors 1768 amounts to about 60 people per day, which isn’t bad, but since I aim for world domination there’s still a ways to go.

Here are some of the popular  search key phrases that brought people to this site:

Interestingly the search terms follow a fat tails distribution, meaning that most of the visits come from the less common phrases. Put another way, there isn’t one phrase or keyword that brings people to this site; it’s lots of little different searches. If you add up all the contributions from the less frequent keywords you end up with something much more significant than, say, “guide to writing a physics engine”, which is the most popular phrase (but accounts for a mere 4.3%). Incidentally, the same thing is true for Amazon.com. They make most of their money not from bestsellers but from the more specialized, less glamorous publications.

However, all this is meaningless, since most visits to this site aren’t the result of using a search engine anyway! Even fewer end up via external links. Most people simply end up on the page by using a bookmark or typing in its name. Does this mean that most of my readers are return readers? I wish there was some way of finding this out:

If you’re a fellow blogger I’d love to hear about your stats. Drop a comment to this thread and tell us about it.

7 single men and 8 single women go together to see a movie. They’ve reserved in advance 15 seats. When they get to the cinema they seat themselves down randomly. What is the average number of pairs (man-woman, or woman-man) that are formed this way? For example, for the following configuration,

M M M M M M F M F F F F F F F

that number would be 3 (M stands for Male, F for Female).

Note: I’ve marked this problem as having ‘undergraduate’ difficulty since the solution I’m aware of is slightly formal (although elegant and short). If you end up with an elementary solution please share!

EDITED, 1/July/2009: a solution has been posted here.

1. RSS Feed: That orange, small square button on the top, to the right of the  “About this Blog” link, is the RSS feed. Every time this blog gets updated the RSS feed gets updated as well. There are lots of programs out there that allow you to follow RSS feeds, such as FeedReader3 (by the way, I don’t know those guys and I’m not affiliated with them). If you use FireFox as your browser, you have a reader built in already! Click on the orange RSS button and click “Subscribe now”. This will add a link to your  Bookmarks Toolbar. Clicking it will show you all the recent posts.
2. Email: On the right you have the option to get this blog’s posts delivered to you via email, just like the morning newspaper. Subscription is done via FeedBurner, owned by Google. It’s very simple: type in your email address and click “Subscribe”. You’ll then be asked to confirm by clicking on a link emailed to you. That’s it! It’s very simple and effective, and you have my personal word of honor that your email will not be used for anything else, nor given to anyone. In fact, I don’t even know my subscribers’ emails, it’s all handled automatically by FeedBurner.
3. Twitter: Forgive me father for I have tweeted. I can’t understand Twitter, but some people can. If you’re one of those people, my Twitter name is PhysicsBlog. I’ve installed a plugin that tweets whenever I post anything new, so you can keep yourself in the loop.

A prisoner is sentenced to death. Before being executed, the king announces that he will give the prisoner one last chance to save his skin. He gives him 100 balls, 50 of which are black and 50 of which are white, and two identical urns. He then asks the prisoner to distribute the balls as he sees fit between the two urns. “The idea,” explains the king, “is simple. Once you’ve distributed the balls, I’ll rearrange the urns at random so you won’t know which is which. Then I’ll let you pick an urn, and then I’ll have you reach in and take a ball out at random. If you’ll take out a white ball, I’ll spare your life.”

What would be the prisoner’s best strategy?

EDITED, 28/June/2009: just to make things clear, I’d like to reiterate that once the balls are distributed among the two urns, the urns are shaken. The choice of urn, as well as the choice of ball from within the selected urn, are uniformly random.

EDITED, 29/June/2009: hats off to our reader, Ricardo Cabral, for his solution. To read it (and my commentary), just read the comments to this post.

Okay, enough of these party problems. It’s back to physics!

A rope of uniform mass (say, x kg/meter) lies coiled on a smooth surface. You start pulling it up at a constant speed v (say, with your hand). What force will you have to apply? Hint: the force is a function of the length of rope that’s already up in the air. You may assume there is so much rope it never stops comin’!

Having answered that, calculate the work you put in, and compare that to the change in mechanical energy of the rope. Is there a discrepancy? If there is, can you explain it?

ADDED 28/June/2009: A detailed solution has been added here.

]]> http://www.physicallyincorrect.com/2009/06/physics-puzzle-pulling-up-the-rope/feed/ http://www.physicallyincorrect.com/2009/06/math-puzzle-how-many-people-were-at-the-party/ http://www.physicallyincorrect.com/2009/06/math-puzzle-how-many-people-were-at-the-party/#comments Fri, 26 Jun 2009 00:03:29 +0000 Assaf http://www.physicallyincorrect.com/?p=736 Here’s a puzzle I got asked over on physicsforums. My solution was intuitive, but later on a different forum member presented a much more concise and general approach, so there are at least two ways of solving this one. Here goes.

A party is thrown. Here’s what you know about it:

1. Every person at the party knows precisely 22 other people.
2. Now, pick two people, call them A and B. If A knows B, then they have no mutual friends.
3. If A doesn’t know B, they have precisely 6 mutual friends.

How many people are at the party?

EDITED, 27/June/2009: a solution has been posted here.

In billiards a stun shot is one in which the ball doesn’t roll, or, more correctly, doesn’t get a chance to start rolling before it hits its target ball. In such cases, and assuming the ball isn’t spinning at all (no English, Top Spin, Bottom Spin and all that jazz), a well known rule of thumb says that after impact, the angle between the cue ball (the one you hit with the stick) and target ball (the one that gets hit by the cue ball) is 90 degrees, as shown below. Can you prove this?

EDITED, 26/June/2009: hats off to our reader peter for his solution. A slightly more detailed one can be found here.

]]> http://www.physicallyincorrect.com/2009/06/physics-puzzle-stuns-shots-and-the-90-degrees-rule-in-billiard/feed/ http://www.physicallyincorrect.com/2009/06/physics-puzzle-the-climbers-free-lunch/ http://www.physicallyincorrect.com/2009/06/physics-puzzle-the-climbers-free-lunch/#comments Wed, 24 Jun 2009 10:40:18 +0000 Assaf http://www.physicallyincorrect.com/?p=710 Level of Difficulty: Highschool

A guy has to climb up a rope (left). He gets an idea: tie the rope around himself and put it through a pulley (right). Now by pulling on the rope, he reasons, he’ll get pulled up by two ropes instead of one - hence by twice the force, making it twice as easy to climb up.  Is he correct or is there a fallacy? If not, how can you explain this “free lunch” - doubling the force without “paying” for it? Treat the rope and pulley as massless. Neglect friction.

EDITED, 25/June/2009: hats off to our reader Mike for his solution in the comments section. I’ve added a more detailed solution, which you can view here.