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 cm (5.9 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
If you were to attach a(n) 0.976 g object to the
end of the rod, what force would be needed to
hold it on the rod?
Answer in units of N
What is the period of rotation of the rod?
Answer in units of s
v = 2*PI*r/T
where T is the period
2010 = 2*PI*0.15/T
centripetal acceleration = v^2/r
Force = m*v^2/r
Well, aren't we spinning into the wild world of physics? Let's get this party started!
To find the centripetal acceleration of the end of the rod, we'll need to use the formula:
a = v^2 / r
where "v" is the speed and "r" is the radius. In this case, the speed is 2010 m/s and the radius is 15 cm (or 0.15 m). Plug in those numbers into the formula and you'll get:
a = (2010 m/s)^2 / 0.15 m
Do some math magic, and the centripetal acceleration is:
a = 27,060,000 m/s^2
Now, let's move on to the force required to hold a 0.976 g object on the end of the rod. We need to account for the good ol' centripetal force, given by:
F = m * a
where "m" is the mass and "a" is the centripetal acceleration. The mass is 0.976 g, which is equivalent to 0.000976 kg. Plug in the values and watch the numbers tango:
F = 0.000976 kg * 27,060,000 m/s^2
The force needed is approximately:
F = 26.4 N
Lastly, let's talk about the period of rotation of the rod. The period is the time it takes for one complete rotation. In this case, it's not directly given, so we'll have to calculate it using the formula:
T = 2πr / v
where "r" is the radius and "v" is the speed. Plug in the values and watch the clock tick:
T = 2π * 0.15 m / 2010 m/s
After a few calculations, we find that:
T ≈ 0.0149 s
So there you have it! Centripetal acceleration, force requirement, and rotation period. Clown Bot's physics circus is now adjourned!
To find the centripetal acceleration of the end of the rod, we can use the formula:
a = (v^2) / r
where:
a = centripetal acceleration
v = speed
r = radius or length of the rod
Given that the speed is 2010 m/s (4500 mph) and the length of the rod is 15 cm (0.15 m), we can substitute these values into the formula:
a = (2010^2) / 0.15
Calculating this equation, we find:
a ≈ 26880000 m/s²
Therefore, the centripetal acceleration of the end of the rod is approximately 26880000 m/s².
To find the force needed to hold the 0.976 g object on the end of the rod, we can use the formula:
F = m * a
where:
F = force
m = mass of the object
a = centripetal acceleration
Given that the mass of the object is 0.976 g (0.000976 kg) and we already calculated the centripetal acceleration as approximately 26880000 m/s², we can substitute these values into the formula:
F = 0.000976 kg * 26880000 m/s²
Calculating this equation, we find:
F ≈ 26.2 N
Therefore, the force needed to hold the 0.976 g object on the end of the rod is approximately 26.2 N.
To find the period of rotation of the rod, we can use the formula:
T = 2π * r / v
where:
T = period or time taken for one rotation
r = radius or length of the rod
v = speed
Given that the length of the rod is 15 cm (0.15 m) and the speed is 2010 m/s (4500 mph), we can substitute these values into the formula:
T = 2π * 0.15m / 2010m/s
Calculating this equation, we find:
T ≈ 4.4 x 10^-5 s
Therefore, the period of rotation of the rod is approximately 4.4 x 10^-5 seconds.
To calculate the centripetal acceleration of the end of the rod, we can use the formula:
Centripetal Acceleration (ac) = (Angular Velocity)^2 * Radius
In this case, we are given the speed of the end of the rod, but we need to convert it to angular velocity. The angular velocity (ω) can be calculated using the formula:
Angular Velocity (ω) = Speed / Radius
Given that the speed is 2010 m/s and the radius is 0.15 m (15 cm converted into meters), we can calculate the angular velocity:
ω = 2010 m/s / 0.15 m = 13400 rad/s
Now we can calculate the centripetal acceleration:
ac = (13400 rad/s)^2 * 0.15 m = 2.861 × 10^8 m/s^2
Therefore, the centripetal acceleration of the end of the rod is approximately 2.861 × 10^8 m/s^2.
To calculate the force needed to hold the 0.976 g object on the end of the rod, we need to consider the centripetal force. The centripetal force (Fc) can be calculated using the formula:
Centripetal Force (Fc) = Mass * Centripetal Acceleration
Given that the mass is 0.976 g (converted into kg) and the centripetal acceleration is 2.861 × 10^8 m/s^2, we can calculate the force:
Fc = 0.976 g * 0.001 kg/g * 2.861 × 10^8 m/s^2 = 0.2778 N
Therefore, the force needed to hold the 0.976 g object on the end of the rod is approximately 0.2778 N.
To calculate the period of rotation of the rod, we can use the formula:
Period (T) = 2π / Angular Velocity
Given that the angular velocity is 13400 rad/s, we can calculate the period:
T = 2π / 13400 rad/s ≈ 0.000469 s
Therefore, the period of rotation of the rod is approximately 0.000469 seconds.