1)Astronauts use a centrifuge to simulate the acceleration of a rocket launch. T
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Question
1)Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 20.0 s to speed up from rest to its top speed of 1 rotation every 1.20 s . The astronaut is strapped into a seat 5.30 m from the axis.
What is the astronaut's tangential acceleration during the first 20.0 s ?
How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s2 of acceleration is 1 g.
2)A particle's trajectory is described by x=(12t32t2)m and y =(12t22t)m, where t is in s.
What is the particle's speed at t=4.0s ?
What is the particle's direction of motion, measured as an angle from the x-axis, at t=4.0s ?
3)A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 200 m above the glacier at a speed of 190 m/s .
How far short of the target should it drop the package?
4)A ball thrown horizontally at 24 m/s travels a horizontal distance of 40 m before hitting the ground.
From what height was the ball thrown?
5) Abicycle wheel is rotating at 48 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration of 0.47 rad/s2 .
What is the wheel's angular velocity, in rpm, 9.0 s later?
How many revolutions does the wheel make during this time?
6)Object A has a position as a function of time given by r A(t) = (3.00 m/s)t i^ + (1.00 m/s2)t2 j^. Object B has a position as a function of time given by r B(t) = (4.00 m/s)t i^ + (1.00 m/s2)t2 j^. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?
Explanation / Answer
1) Angular vekicity at the end of 20 s is = [2 / 1.2] = 5.24 radian /s
Linear velocity at the end of 20 second is r = 5.3 x 5.24 = 27.772 m/s
Tangential acceleration is change in velocity / time = 27.772 / 20 = 1.3886 m/s^2.
The centripetal acceleration is r ^2 = r * = 27.772 x 5.24 = 145.53 m/s^2
= 145.53/ 9.8 = 14.85 g force.
2) the displacement vector is, r(t) = <1/2t^32t^2, 1/2t^22t>.
Hence v(t) = <3 / 2 t^2 - 4t, t - 2 > and speed = s = |v(t)| = sqrt([3/ 2 t^2 - 4t]^2 + [t - 2]^2).
The speed at t = 4 is <8, 2> = 8.25 m/s
direction = arctan(2/8) = 14.04 degree
3) H = 200 m
Vo = 190 m/s
g = 9.8 m/s^2
R = Vo * SQRT [ 2H / g }
R = (190 m/s) * SQRT { [ 2 * (190 m/s) ] / (9.8 m/s^2) }
R = (190 m/s) * SQRT { [ 380 m/s ] / (9.8 m/s^2) }
R = (190 m/s) * SQRT { 38.78 s^2 }
R = (190 m/s) * (6.23 s)
R = 1183.7 m
4) The time it takes for the arrow to reach the ground is just the distance divided by the horizontal speed.
time = t = 40/24 = 1.67 seconds
You can then use the formula for distance traveled under constant acceleration.
h = (1/2)g*t2
h = (1/2)(9.8)(1.67)^2
h = 13.67 m
So the arrow was fired from a height of 13.67 meters
5) Given: i (initial angular velocity)=(48 rev/min)(2/1rev)(1min/60sec)
i = 5.03 rad/sec
f = final angular velocity
f = i + t
f = 5.03 rad/sec + (0.47rad/sec^2)(9sec)
f = 9.26 rad/sec or
(9.26 rad/sec)(1rev/2rad)(60sec/1min) = 88.43 rev/min
b) = (avg)(t) = [(i + f)/2](9sec)
= [(5.03+9.26)/2](9)
= 64.305 rad or 64.305rad (1rev/2rad) = 10.23 rev.
.
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