Batch 3 - Class 117 - Jenga Staircase (Center of Gravity)

• Petals around the roses - explain the puzzle to kids by rolling 5 dice, and answering them the code number in the roll of 5 dice. Take the first ten rolls, and ask them to note the rolls down along with answers. Ask them to come back next week with a solution to predict the answer for any given roll.
• Answer: Each roll of number 3 counts for 2 points and that of 5 for 4 points. Rest don't get any points. These can be seen as 2 and 4 "petals" around the central dot ("rose") for the roll of 3 and 5 respectively. All other numbers either do not have a "rose", i.e. central dot; or they dont have petals, i.e. dots around the center.

• Can you stack the 4 Jengas on top of each other in a staircase fashion near the edge of the table in such a way that the staircase is leading out over the edge of the table and such that no point on the topmost jenga is directly above the table?
• Let us start from smaller numbers - how far can the first block hang out - 1/2
• What if we had two blocks - how far can the edge of top block hang outside the edge? What determines when the block tips over

  What is required for the top block to tip over? Same as if it were the only block

  What is required for the next block to tip over? This can happen in the collective weight of two blocks is enough to tip over. Instructor Notes: Some kids may need to understand the notion of center of gravity

  So how far along can the blocks be so that the center of gravity still passes through the table and not outside it?This is 3/4, i.e. the bottom block may extend out 1/4 of the way
• What if we add a third block?

  Same argument as above. How far along can be move the third block so that the center of gravity of all three can lie within the table?Instructor Notes: This can be shown by seeing how much of the blocks lie outside and inside the tableAnswer: The third block can be pushed out 1/6th of the way, for total distance of 1/6+3/4=11/12
• What if we add the fourth block?

  Can you see a pattern?Can we derive the pattern - Assume that f(n) represents how far the first block can be pushed out in configuration of n blocks. Then f(n)-f(n-1) is the distance that the last block has been pushed out, and the prior (n-1) blocks center of gravity (which was earlier just at edge of the table) is now f(n)-f(n-1) distance away from the table. This torque of (n-1)(f(n)-f(n-1)) needs to be balanced by reverse torque of 1/2-(f(n)-f(n-1)). Hence, f(n)=1/2n + f(n-1)
• This also suggests that if we had infinite blocks, we could keep on increasing the hangover of the first block
• Instructor Notes: It might be an interesting aside to take kids through some simple balancing exercises, and try to derive the notion of center of gravity passing through point in touch with the ground for balance.

• In the jenga problem, what would happen if an ornament was to be hung 1/4 the distance from the extreme edge of first block, and the ornament weight was 1/5 the weight of the block
• This additional torque of 1/5(f(n)-1/4) now needs balancing, so the above equation can be adjusted accordingly