A typical laboratory centrifuge rotates at 4100 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations.
(a) What is the acceleration at the end of a test tube that is 11 cm from the axis of rotation?
dropped from a height of 1.2 m and stopped in a 1.0 ms long encounter with a hard floor?
To calculate the acceleration at the end of a test tube in a centrifuge, we can use the centripetal acceleration formula:
a = ω²r
Where:
a = acceleration
ω (omega) = angular velocity in radians per second
r = radius or distance from the axis of rotation
First, we need to convert the rotational speed from rpm to radians per second. We know that 1 revolution is equal to 2π radians, and 1 minute is equal to 60 seconds. Therefore, the conversion factor is:
(4100 rpm) * (2π radians / 1 revolution) * (1 revolution / 60 seconds) = ω
Let's calculate ω:
ω = 4100 * 2π / 60 = 215.66 radians per second
Now we can calculate the acceleration. Given that the distance from the axis of rotation (r) is 11 cm, we need to convert it to meters:
r = 11 cm = 0.11 m
Now we can substitute the values into the formula:
a = (215.66 radians per second)² * 0.11 m
Calculating the above expression will give us the acceleration at the end of the test tube.
To find the acceleration of the test tube, we need to convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). Here's how to do it:
Step 1: Convert rpm to rad/s
1 revolution = 2π radians
1 minute = 60 seconds
So, to convert rpm to rad/s, we can use the formula:
Angular velocity (ω) in rad/s = (2π * rpm) / 60
In this case, the rotational speed is 4100 rpm. So, we can calculate the angular velocity as follows:
ω = (2π * 4100) / 60
Step 2: Calculate the linear speed
The linear speed of a point on the circumference of a rotating object can be determined using the formula:
v = ω * r
Where:
v = linear speed
ω = angular velocity
r = radius of rotation
In this case, the test tube is 11 cm (or 0.11 m) away from the axis of rotation. So, substituting the values into the formula, we get:
v = ω * r
v = ((2π * 4100) / 60) * 0.11
Step 3: Calculate acceleration using kinematic equations
To find the acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0, as the test tube stops)
u = initial velocity (unknown)
a = acceleration (what we need to find)
s = displacement (in this case, the distance the test tube falls)
In this case, the test tube falls from a height of 1.2 m, so s = 1.2 m.
Substituting the known values into the equation, we get:
0^2 = u^2 + 2as
0 = u^2 + 2(a)(1.2)
Solving for u^2, we have:
u^2 = -2(a)(1.2)
Now, let's find the acceleration using the given information:
u = 0 (as the test tube starts from rest)
s = 1.2 m
We can rearrange the equation to solve for acceleration:
a = -u^2 / (2s)
Substituting the known values, we get:
a = -0^2 / (2 * 1.2)
a = 0 m/s^2
Therefore, the acceleration at the end of the test tube is 0 m/s^2.
acceleration in the centifruge
w^2r=(4100rpm*1min/60 sec * 2PIrad/sec)^2*.11m = appx 275 g's. Work it out. check my math.
Now dropping it,
1/2 mv^2=mgh
v=sqrt(2gh)=sqrt(2*9.8*1.2)=4.95m/s
a=changevelocity/time=4.95/.001=4950m/s^2=about 500 g's