An apparatus is designed to study insects at an accelera�tion of magnitude 980 m/s 2 ( = 100 g). The apparatus
consists of a 2.0-m rod with insect containers at either
end. The rod rotates about an axis perpendicular to the
rod and at its center. (a) How fast does an insect move
when it experiences a radial acceleration of 980 m/s 2
?
(b) What is the angular speed of the insect?
Ac = v^2/R
980 = v^2/2
v = 44.3 meters /second
v = omega * R
omega = 44.3/2 = 22.2 radians / second
(a) When an insect experiences a radial acceleration of 980 m/s², you can imagine it being quite "bugged" out. It's like going from standing still to being tossed around at a whopping 100 times the acceleration due to gravity! That's quite a thrill ride for our little insect friend.
(b) As for the angular speed of the insect, well, let's just say it's doing some serious "spin bug-gy". With the rod rotating about its center, the rapid spinning motion might make our insect feel a bit "dizzy"! So, the angular speed is dependent on the rotational motion of the rod.
(a) To determine how fast the insect moves when it experiences a radial acceleration of 980 m/s^2, we can use the formula for radial acceleration:
ar = v^2 / r
where ar is the radial acceleration, v is the velocity of the insect, and r is the radius of rotation.
In this case, the radial acceleration is given as 980 m/s^2. Assuming the insect is located at a distance r from the axis of rotation (which is the radius of the rod), we can rearrange the formula to solve for v:
v = sqrt(ar * r)
Plugging in the values ar = 980 m/s^2 and r = 2.0 m, we can calculate v:
v = sqrt(980 m/s^2 * 2.0 m) = sqrt(1960 m^2/s^2) = 44.2 m/s
Therefore, the insect moves at a speed of 44.2 m/s when it experiences a radial acceleration of 980 m/s^2.
(b) The angular speed (ω) of the insect can be determined using the following formula:
v = ω * r
where v is the linear speed of the insect and r is the radius of rotation.
From part (a), we know that the speed of the insect (v) is 44.2 m/s and the radius of the rod (r) is 2.0 m. Plugging in these values, we can solve for ω:
ω = v / r = 44.2 m/s / 2.0 m = 22.1 rad/s
Therefore, the angular speed of the insect is 22.1 rad/s.
To answer these questions, we can use the concept of centripetal acceleration and the relationship between linear and angular velocity.
(a) How fast does an insect move when it experiences a radial acceleration of 980 m/s^2?
The radial acceleration of the insect is equal to the centripetal acceleration, which is given by the formula:
a = rω^2
where a is the centripetal acceleration, r is the radius of the circular path, and ω is the angular velocity.
In this case, the radial acceleration is given as 980 m/s^2 and we need to find the linear velocity (v) of the insect.
To find the linear velocity, we first need to determine the radius of the circular path. The rod has a length of 2.0 m, and since the insect is located at one of the container ends, the radius (r) is equal to half the length of the rod.
r = 2.0 m / 2
r = 1.0 m
Now, we can rewrite the formula for radial acceleration as:
a = rω^2
980 m/s^2 = (1.0 m)ω^2
To find ω, we rearrange the equation:
ω^2 = 980 m/s^2 / (1.0 m)
ω^2 = 980 s^-2
Taking the square root of both sides, we find:
ω = sqrt(980 s^-2)
ω ≈ 31.30 s^-1
Therefore, the angular velocity of the insect is approximately 31.30 s^-1.
(b) What is the angular speed of the insect?
The angular speed (ω) is the magnitude of the angular velocity and is measured in radians per second (rad/s).
In this case, we already found the value of the angular velocity ω in part (a), which is approximately 31.30 s^-1. Therefore, the angular speed of the insect is also approximately 31.30 rad/s.