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according to the guinness book of world records (1990 edition, p. 169), the highest rotary speed ever attained was 2010 m/s (4500 mph). the rotating rod was 15.6 cm (6.1 in) long. assume the speed quoted is that of the end of the rod. what is the centripetal acceleration of the end of the rod? answer in units of m/s 2 .

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Final answer:

The centripetal acceleration of the end of a rod rotating at 2010 m/s with a length of 0.156 m is calculated to be 51834615.38 m/s².

Step-by-step explanation:

The question asks for the centripetal acceleration of the end of a rotating rod which attained a speed of 2010 m/s.

Given that the length of the rod is 15.6 cm, we first convert the length to meters to make our units consistent:

15.6 cm = 0.156 m.

The formula for calculating centripetal acceleration (ac) is ac = v2/r, where v is the linear speed and r is the radius of the circular path.

In this case, the linear speed is 2010 m/s and the radius (r) is half the length of the rod, which is 0.156 m / 2 = 0.078 m.

Using these values:

ac = (2010 m/s)2 / 0.078 m
ac = 4040100 m2/s2 / 0.078 m
ac = 51834615.38 m/s2

Therefore, the centripetal acceleration of the end of the rod is 51834615.38 m/s2.

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