In an isolated system, a red ball of mass m moves to the right with speed v. It strikes a green ball, of
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The length to which the pendulum will be adjusted to keep perfect time is 29.59 inches. See the explanation below.
<h3>What is the justification for the above answer?</h3>T1 = 2πR√(L1/GM)
and
T2 = 2πR√(L1/GM)
T1/T2 = √(L1/L2).
If the pendulum has an efficient period, that means it executes with perfect frequency.
Thus,
T2 = (24 * 60)/x
= 1440/x
This means that in one day, there are perfect cycles of represented by "x". Note that there are 1440 minutes in one day.
If the other Pendulum is slower by 10 minutes, that means
T1 = 1450/x
Hence
(1450/x)/(1440/x) = √(L1/L2).
⇒ 1450/1440 = √(L1/L2).
Thus,
(1450/1440)² = 30/L
L = 30/(1450/1440)²
L = 30/(1.00694444444)²
L = 30/1.01393711419
L = 29.5876337695
L 29.59 inches.
Hence, the pendulum will need to be adjusted by 29.59 inches to ensure that the clock keeps perfect time.
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