A computer is reading data from a rotating CD-ROM. At a point that is 0.0302 m from the center of the disk, the centripetal acceleration is 284 m/s2. What is the centripetal acceleration at a point that is 0.0803 m from the center of the disc?
omega = v/R is the same for each
Ac = omega^2 R = v^2/R
284 = omega^2 (.0302)
SO
Ac = [284/.0302](.0803)
Well, isn't this a "disk-y" question? Let's spin right into it!
To solve this problem, we can use the formula for centripetal acceleration, which is given by:
a = ω²r
Here, "a" is the centripetal acceleration, "ω" is the angular velocity, and "r" is the distance from the center of rotation.
Now, the angular velocity "ω" is the same throughout the entire CD-ROM since it's rotating uniformly. So, we can say that:
a₁ / r₁ = a₂ / r₂
Given that a₁ = 284 m/s² and r₁ = 0.0302 m, we can substitute these values into the equation and solve for a₂:
284 / 0.0302 = a₂ / 0.0803
Cross-multiplying, we get:
(284 * 0.0803) / 0.0302 = a₂
So, the centripetal acceleration at a point that is 0.0803 m from the center of the disc would be... *drumroll*... approximately 755 m/s²!
And there you have it! Remember, centrifugal force is just a "spin-off" of centripetal force, so keep rotating those questions my way!
To solve this problem, we can use the formula for centripetal acceleration:
a = (v^2) / r
where:
a is the centripetal acceleration
v is the linear velocity
r is the distance from the center of the disk
Given information:
At point 0.0302 m from the center (r1), the centripetal acceleration is 284 m/s^2.
Let's calculate the linear velocity (v1) at this point first.
Since the disk is rotating, the linear velocity is given by:
v = ω * r
where:
v is the linear velocity
ω is the angular velocity
r is the distance from the center of the disk
Now, we can derive the formula for the angular velocity:
ω = 2πf
where:
ω is the angular velocity
f is the frequency of rotation
Assuming a constant frequency of rotation, we can rewrite the formula for linear velocity as:
v = 2πf * r
Given that r1 = 0.0302 m and a1 = 284 m/s^2, we can solve for f and v1:
284 = v1^2 / 0.0302 (equation 1)
v = 2πf * 0.0302 (equation 2)
Now, let's solve equation 2 for f:
f1 = v1 / (2π * r1)
Substituting the given values, we can calculate f1.
Next, we can use equation 1 to calculate v1:
v1 = sqrt(a1 * r1)
Now that we know f1 and v1, we can calculate the centripetal acceleration at a point that is 0.0803 m from the center (r2) using the same formula:
a2 = (v2^2) / r2
Let's calculate the values:
f1 = v1 / (2π * r1)
v1 = sqrt(a1 * r1)
a2 = (v2^2) / r2
Note: To solve the equation, we need to know the value of v2 or f2. Please provide additional information to proceed further.
To find the centripetal acceleration at a point 0.0803 m from the center of the disk, we can use the formula for centripetal acceleration:
a = ω²r
Where:
a = centripetal acceleration
ω = angular velocity
r = distance from the center of the disk
In this case, we are given that the centripetal acceleration at a point 0.0302 m from the center is 284 m/s². We can find the angular velocity using this information.
First, we need to convert the distance from the center of the disk to the radius. Since the radius is half the diameter, we have:
r₁ = 0.0302 m
Next, we can find the angular velocity using the formula:
a = ω²r
Rearranging the formula, we can solve for ω:
ω = √(a/r)
Plugging in the values:
ω = √(284 m/s² / 0.0302 m)
Calculating this, we find:
ω ≈ 324.67 rad/s
Now that we have the angular velocity, we can use it to find the centripetal acceleration at a different distance from the center of the disk.
r₂ = 0.0803 m
Using the formula:
a = ω²r
Plugging in the values:
a = (324.67 rad/s)² * 0.0803 m
Calculating this, we get:
a ≈ 831.06 m/s²
Therefore, the centripetal acceleration at a point 0.0803 m from the center of the disk is approximately 831.06 m/s².