What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius 3220 km at a speed of 29000 km/h?
change speed to m/second, radius to meters
tangential speed=angular velociyt*radius
you are given tangential speed, radius, solve for angular velocity w
radial acceleration= w^2*radius
tangential acceleration: zero
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To find the magnitudes of the angular velocity, radial acceleration, and tangential acceleration of the spaceship, we can use the following equations:
(a) Angular velocity (ω) is given by the formula:
ω = v / r,
where v is the linear velocity and r is the radius of the circular path.
(b) Radial acceleration (ar) is given by the formula:
ar = v^2 / r.
(c) Tangential acceleration (at) is given by the formula:
at = α * r,
where α is the angular acceleration and r is the radius of the circular path.
Given:
Linear velocity (v) = 29000 km/h,
Radius (r) = 3220 km.
Let's calculate each magnitude step-by-step:
First, let's convert the linear velocity from km/h to km/s:
v (km/s) = v (km/h) / 3600.
v (km/s) = 29000 / 3600 = 8.0556 km/s.
(a) Angular velocity:
ω = v / r = 8.0556 / 3220 = 0.0025 rad/s (rounded to 4 decimal places).
(b) Radial acceleration:
ar = v^2 / r = (8.0556^2) / 3220 = 0.0201 km/s^2 (rounded to 4 decimal places).
(c) Tangential acceleration:
The problem does not provide information about the angular acceleration (α). Without that, we cannot directly calculate the tangential acceleration.
To find the magnitudes of the angular velocity, radial acceleration, and tangential acceleration of the spaceship, we can use the following equations:
(a) Angular velocity (ω) is the rate at which the spaceship is changing its angle as it moves along the circular path. It is given by the formula:
ω = v / r, where
v is the linear velocity (speed) of the spaceship, and
r is the radius of the circular path.
To find the angular velocity, plug in the given values:
v = 29000 km/h, and
r = 3220 km.
Convert the linear velocity to m/s for consistency:
v = 29000 km/h * (1000 m/km) / (3600 s/h) ≈ 805.6 m/s.
Convert the radius to meters:
r = 3220 km * 1000 m/km ≈ 3,220,000 m.
Now, substitute the values into the equation to calculate the angular velocity:
ω = 805.6 m/s / 3,220,000 m ≈ 0.00025 rad/s.
(b) Radial acceleration (ar) is the component of acceleration directed towards the center of the circular path. It is given by the formula:
ar = (v^2) / r, where
v is the linear velocity (speed) of the spaceship, and
r is the radius of the circular path.
Substitute the given values into the equation to calculate the radial acceleration:
ar = (805.6 m/s)^2 / 3,220,000 m ≈ 0.202 m/s^2.
(c) Tangential acceleration (at) is the component of acceleration tangent to the circular path. It can be calculated by multiplying the angular velocity (ω) by the linear velocity (v):
at = ω * v, where
ω is the angular velocity, and
v is the linear velocity (speed) of the spaceship.
Substitute the values to calculate the tangential acceleration:
at = 0.00025 rad/s * 805.6 m/s ≈ 0.202 m/s^2.
Therefore, the magnitudes of the angular velocity, radial acceleration, and tangential acceleration are approximately:
(a) Angular velocity: 0.00025 rad/s.
(b) Radial acceleration: 0.202 m/s^2.
(c) Tangential acceleration: 0.202 m/s^2.