what is the acceleration of a point on the time of a flywheel 0.90 m in diameter, turning at the rate of 1200 rev/min?
1200 rev/min = 20 rev/s = 12.57 rad/s
Call that the angular speed, w.
The centripetal acceleration is R*w^2.
R is the 0.45 m radius.
a = 0.45*(12.57)^2 = ___ m/s^2
To find the acceleration of a point on the rim of a rotating object like a flywheel, you can use the following formula:
a = (4π²r)/T²
where:
a is the acceleration of the point,
r is the radius of the flywheel, which is half of its diameter,
π is a mathematical constant approximately equal to 3.14159,
T is the period of rotation, which is the time taken for one complete revolution.
First, let's find the radius of the flywheel:
The diameter of the flywheel is given as 0.90 m, so the radius is half of that:
r = 0.90 m / 2 = 0.45 m
Next, we need to find the period of rotation (T):
The flywheel is turning at a rate of 1200 rev/min. To convert this to seconds, divide by 60:
T = 1 min / (1200 rev/min) = 1/1200 min/rev = 1/1200 min/rev × (1/60 min/s) = 1/7200 s/rev
Now we have all the values we need to calculate the acceleration (a):
a = (4π²r)/T²
= (4 × 3.14159² × 0.45) / (1/7200)²
≈ 4 × 3.14159² × 0.45 × (7200/1)²
≈ 4 × 3.14159² × 0.45 × 7200²
Using a calculator, we can evaluate this expression to find the acceleration of the point on the rim of the flywheel.