A student on a stool rotates freely at 180 rounds per minute. The student holds a 1.00 kg mass in each outstretched arm, 0.8 m from the axis of rotation. The combined moment of inertia of the student and the stool is 6.00 kgm2 which remains constant. How far should the student pull his arms inward so that the rotation becomes 4.00 Hz?
180 rounds per minute = 3.00 Hz
Angular momentum of each mass is:
L = (m r^2)ω
Mass and Angular momentum are conserved, so:
r0^2 ω0 = r1^2 ω1
or
r1 = r0 √(ω0/ω1)
r1 = 0.800 √(3.00/4.00)
r1 ≈ 6.93m
how u calculate i calculate get 0.693m .
can u explain to me ??how u did this ~ why the 3Hz /4Hz?
My apology for the typo. I must have miskeyed during calculation, but I should have realised that seven meter was a bit of an arm stretch.
So yes, r1 ≈ 0.693m
Because angular momentum and mass are conserved, the angular velocity is inversely proportional to the square of the radius. Thus to increase the frequency from 3.00Hz to 4.00Hz requires decreasing the radius from 0.800m to 0.693m.
(r1/r0) = √(ω0/ω1)
(6.93m/0.800m) = √(3.00Hz/4.00Hz)
To solve this problem, we need to use the principles of rotational dynamics. The formula we will use is:
I₁ω₁ = I₂ω₂
where I₁ and ω₁ are the initial moment of inertia and angular velocity respectively, and I₂ and ω₂ are the final moment of inertia and angular velocity respectively.
Let's begin by calculating the initial angular velocity (ω₁) in radians per second. We are given that the student rotates at 180 rounds per minute. To convert this to radians per second, we use the formula:
ω₁ = (2π * f₁)
where f₁ is the frequency in Hertz (cycles per second).
So, ω₁ = (2π * 180/60) = 2π * 3 = 6π rad/s
Next, we need to calculate the initial moment of inertia (I₁). The combined moment of inertia of the student and the stool is given as 6.00 kgm². Since this value remains constant, we can use it as the initial moment of inertia.
Now, let's calculate the final angular velocity (ω₂) in radians per second. We are given that the rotation becomes 4.00 Hz. So, ω₂ = 2π * 4 = 8π rad/s.
We can now substitute the values into the formula I₁ω₁ = I₂ω₂ to find the final moment of inertia (I₂).
I₁ * ω₁ = I₂ * ω₂
6.00 kgm² * (6π rad/s) = I₂ * (8π rad/s)
Simplifying the equation gives:
6 * 6π = I₂ * 8π
36π = 8π * I₂
Dividing both sides by 8π gives:
I₂ = 36π/(8π) = 4.5 kgm²
Now that we know the final moment of inertia (I₂), we can calculate the new distance the student should pull his arms inward to achieve this. We will use the principle of conservation of angular momentum:
I₁r₁ = I₂r₂
where r₁ and r₂ are the initial and final distances from the axis of rotation respectively.
Initially, the 1.00 kg masses are 0.8 m from the axis of rotation. So, r₁ = 0.8 m.
We can now rearrange the formula to solve for r₂:
r₂ = (I₁r₁)/I₂
Substituting the known values into the formula gives:
r₂ = (6.00 kgm² * 0.8 m) / 4.5 kgm²
Simplifying the equation gives:
r₂ = 1.0667 m
Therefore, the student should pull his arms inward by approximately 1.0667 meters to achieve a rotation of 4.00 Hz.