A helicoptor bladehas an angular speed of 6.50 rev/s and has an angular acceleration of 1.30 rev/s. point 1 on the blade is 3.00m and point 2 is 6.7m. Can you find the speeds of A) the tangential speeds and B) the tangential accelerations?
To find the tangential speed, we need to multiply the given angular velocity by the radius of the circular path. Similarly, to find the tangential acceleration, we need to multiply the given angular acceleration by the radius. Let's solve it step by step:
A) Tangential speed:
The formula for tangential speed is given by:
V = ω * r
where
V = tangential speed,
ω = angular speed,
r = radius.
Given:
ω = 6.50 rev/s (angular speed),
r = 3.00 m (radius at point 1).
Substituting the values in the formula:
V = 6.50 rev/s * 2π rad/rev * 3.00 m
(Note: 2π rad/rev is the conversion factor between revolutions and radians. One revolution is equal to 2π radians.)
Calculating:
V = 6.50 * 2π * 3.00 ≈ 122.52 m/s
Therefore, the tangential speed at point 1 is approximately 122.52 m/s.
To find the tangential speed at point 2, we use the same formula but with a different radius. Given:
r = 6.7 m (radius at point 2).
V = 6.50 rev/s * 2π rad/rev * 6.7 m
Calculating:
V = 6.50 * 2π * 6.7 ≈ 272.52 m/s
Therefore, the tangential speed at point 2 is approximately 272.52 m/s.
B) Tangential acceleration:
The formula for tangential acceleration is given by:
a_t = α * r
where
a_t = tangential acceleration,
α = angular acceleration,
r = radius.
Given:
α = 1.30 rev/s (angular acceleration),
r = 3.00 m (radius at point 1).
Substituting the values in the formula:
a_t = 1.30 rev/s * 2π rad/rev * 3.00 m
Calculating:
a_t = 1.30 * 2π * 3.00 ≈ 24.66 m/s^2
Therefore, the tangential acceleration at point 1 is approximately 24.66 m/s^2.
For point 2:
r = 6.7 m (radius at point 2).
a_t = 1.30 rev/s * 2π rad/rev * 6.7 m
Calculating:
a_t = 1.30 * 2π * 6.7 ≈ 55.77 m/s^2
Therefore, the tangential acceleration at point 2 is approximately 55.77 m/s^2.
To summarize:
A) The tangential speeds are approximately 122.52 m/s at point 1 and 272.52 m/s at point 2.
B) The tangential accelerations are approximately 24.66 m/s^2 at point 1 and 55.77 m/s^2 at point 2.