Swing That Bowling Ball

Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.

Swing That Bowling Ball

September 7, 2017

Wait — why are all those bowling balls swinging on strings? They’re showing what happens when YOU ride a swing. Whether you push back a lot or a little to start, it will always take you the same amount of time to swing from back to front. Even a friend who’s bigger or smaller will take the same time as you. Try it! The only thing that changes the timing is the length of the chains. Longer swings take longer to go from front to back. In this amazing video, all those colorful bowling balls hang at slightly different heights. So even though they start swinging together, soon they separate in a ripple, clinking xylophone bars to prove it. Watch the video to see it — and hear it!

Wee ones: What shape is a bowling ball? Can you find that shape in your room?

Little kids: If 1 bowling ball swings 3 feet above the ground and another swings 4 feet above ground, which one is higher?  Bonus: If there are 14 balls between the first and last, how many are swinging in total?

Big kids: After every one of the 16 balls has made 4 full round trips (right to left and right again), how many notes have you heard them clink in total? (They clink at both the front and back.)  Bonus: If the balls play the musical notes C, D, E, F, G, A, B, then C to start the pattern again over, what note does the 16th ball play?

The sky’s the limit: If the slowest-period ball takes 5 seconds to swing from front to back, the one in the middle takes 4 seconds, and the fastest-period ball takes just 3 ½ seconds, how many seconds will it be before they all line up again on the right?

 

 

 

Answers:
Wee ones: A circle from the side, or in 3D, a “sphere.” Look for spheres of any kind in your room!

Little kids: The ball 4 feet above ground.  Bonus: 16 balls.

Big kids: 128 notes, since 4 full round trips means every ball clinks its bar 8 times.  Bonus: High D, since the first 14 cover the 1st 2 sets of 7 notes.

The sky’s the limit: At 280 seconds, or 4 minutes 40 seconds. The key here is that while 20 is a common multiple of 4 and 5, at 20 seconds the 5-second ball has done 2 full round trips and is on the right, while the 4-second ball has done only 2 ½ trips and is on the left. So they line up only every 40 seconds. For the 3 ½-second ball to meet them there, it needs to do an even number of 7-second round trips, and the smallest common multiple of 40 and 7 is 280.

Print Friendly, PDF & Email

About the Author

Laura Overdeck

Laura Overdeck

Laura Bilodeau Overdeck is founder and president of Bedtime Math Foundation. Her goal is to make math as playful for kids as it was for her when she was a child. Her mom had Laura baking while still in diapers, and her dad had her using power tools at a very unsafe age, measuring lengths, widths and angles in the process. Armed with this early love of numbers, Laura went on to get a BA in astrophysics from Princeton University, and an MBA from the Wharton School of Business; she continues to star-gaze today. Laura’s other interests include her three lively children, chocolate, extreme vehicles, and Lego Mindstorms.

More posts from this author

Comments (3)

  1. Theo Lengyel Reply

    September 11, 2017 at 12:49 pm

    Hi,

    First, I would like to say that my family and I really enjoy playing bedtime math!

    This particular issue was superb, and the video was spectacular. We all enjoyed the beauty of watching the pendula go in and out of phase in various configurations.

    I am commenting today to address an author’s blunder, in the hopes that it may help some confused souls out there.

    So, here goes …

    In the question and answer pair to “The sky’s the limit”, I believe there are two blunders (maybe three):

    1) The answer is not given in the terms stated in the question (i.e. the answer is given in total time, yet the question asks for “how many full rounds of swings must they all take …”.

    2) The answer, as given in total-time, is suffering from an off-by-a-factor-of-two error by failing to account for the fact that a “swing from front to back” is half a period, so the period is double the front-to-back time given.

    3) The question is a bit ambiguous “how many full rounds of swings must they all take …” … Does this ask for the sum of the full rounds for all three balls? Or, does it ask for the number of full rounds for each individual ball?

    Here is the full problem as stated, followed by my own solution:

    The sky’s the limit: If the slowest-period ball takes 5 seconds to swing from front to back, the one in the middle takes 4 seconds, and the fastest-period ball takes just 3 ½ seconds, how many full rounds of swings must they all take before they all line up again on the right?

    Solution:

    * the slowest-period ball takes 5 seconds to swing from front to back
    – The full-round period is thus 10 seconds

    * the one in the middle takes 4 seconds
    – The full-round period is thus 8 seconds

    * the faster-period ball takes 3.5 seconds
    – The full-round period is thus 7 seconds

    So, the least common multiple for the three periods gives us the total time it would take for them to all simultaneously realign on the front:

    10 * 8 * 7 = 560 seconds

    From here we can determine how many full rounds each ball takes in this time:

    * the slowest-period ball :

    560s / (10s/full-round) = 56 full-rounds

    * the one in the middle :

    560s / (8s/full-round) = 70 full-rounds

    * the faster-period ball :

    560s / (7s/full-round) = 80 full-rounds

    Therefore, the sum total of full rounds for all three balls is:

    56 + 70 + 80 = 206

    Well, I hope that helps, and I hope I haven’t made my own blunder(s)!

    Cheers,
    Theo

  2. Theo Lengyel Reply

    September 11, 2017 at 12:57 pm

    Oops, drat, I did make a blunder. The author is actually correct in that the total time is 280 seconds.

    So, my answer needs to be divided by two:

    28 + 35 + 40 = 103

    Doh!

    Cheers!
    Theo

    1. Dug Steen
      Dug Steen Reply

      September 11, 2017 at 3:34 pm

      Theo – No worries! This one was tricky; even more so because you were right: the wording was confusing. We’ve changed the Sky’s the Limit question here and on the app to read “How many seconds will it be before they all line up again on the right?” Hopefully, that’ll clear things up a bit. And thanks again for checking our math! It never hurts.

Leave a Reply

Your email address will not be published. Required fields are marked *