You discover that a certain bad Santa is scheming to build this bell so that he

Question

You discover that a certain bad Santa is scheming to build this bell so that he can wake you up at midnight on Christmas playing Jingle Bells ;) The bell is big and will be suspended from a strong support. It will also be heavy enough and swing at small oscillations which will ensure that the effects of friction are negligible (ignorable). Now you have figured out (from your 'sources') that the center of mass of the bell is 0.55 m below the point where the bell will be pivoted, that the mass of the bell is 40.0 kg , and that its moment of inertia about an axis passing through the pivot is 19.0 kgm2 . Now, one day, while sipping coffee with Santa, he tells you (without revealing his intentions) that he needs a uniform slender rod of mass 1.8 kg that he plans to use as a pendulum suspended by one end of the rod, and seeks your help in finding one such specific rod. You immediately realize (but don't let him know) that he is planning to use this rod as a clapper for his bell!! ... and will likely attach the other end of the rod to the inside of the bell about the same axis as the bell.

If you want to sleep in peace, you can do something clever:

Get him a rod (for his clapper) of a specific length L so that the bell rings silently and you can sleep peaceful :-) [i.e. a clapper that will oscilate with the same timing as that of the bell]. How long of a uniform rod, of the given mass (see above), will you get Santa so that the bell rings silently?   

FIVE sig figs

sorry my professor is weird...

Explanation / Answer

Time period of bell, Tb = 2(Ib/mbgLb)1/2

Time period of rod, Tr = 2(2Ir/mrgL)1/2

Now, moment of inertia of rod, Ir = mrL2/3

So, Tr = 2(2L/3g)1/2

Now, Tb = Tr

=> 2(Ib/mbgLb)1/2 = 2(2L/3g)1/2

=> Ib/mbLb = 2L/3

=> Length of rod, L = 3Ib / 2mbLb = 3 * 19 / (2 * 40 * 0.55) = 1.2954 m

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