A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 24.5 kg. The child grabs and clings to a bar that is 1.60 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 55.0 rpm to 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?
initial angular momentum of system = I wi
where I is out unknown and wi is omega i, the initial angular velocity
wi= = 55 * 2pi radians/min * 1 min/60 s = 5.76 rad/s
so initial ang momentum = 5.76 I
final angular velocity = wf = 19 * 2 pi /60 = 1.99 If
If = I + m r^2 = I + 24.5 (1.6)^2 = I + 62.7
BUT, final angular momentum = initial angular momentum
so
1.99 ( I + 62.7) = 5.76 I
solve for I
Well, well, well. Seems like we got ourselves a physics problem here. Hold on tight, because I'm about to spin some knowledge your way!
To solve this, we need to figure out the moment of inertia of the merry-go-round. Moment of inertia is basically a measure of how an object resists changes in its rotational motion.
Now, we can use the principle of conservation of angular momentum. The initial angular momentum of the merry-go-round is equal to the final angular momentum when the child grabs on. So, let's crunch some numbers!
The initial angular momentum is given by L₁ = I₁ * ω₁, where I₁ is the initial moment of inertia and ω₁ is the initial angular velocity. The final angular momentum is given by L₂ = I₂ * ω₂, where I₂ is the final moment of inertia and ω₂ is the final angular velocity.
We know the initial and final angular velocities, so we can plug in those values. But we need to find the final moment of inertia, so we rearrange the equation to solve for I₂:
I₂ = (L₂ * ω₁) / (L₁ * ω₂)
Now, let's plug in the numbers. L₁ = I₁ * ω₁ = I₁ * (55.0 rev/min) = I₁ * (55.0 * 2π rad/min), and L₂ = I₂ * ω₂ = I₂ * (19.0 rev/min) = I₂ * (19.0 * 2π rad/min). Let's keep the units in radians for now to avoid any confusion.
Substituting these values into the equation, we get:
I₂ = (I₁ * (55.0 * 2π rad/min)) / (19.0 * 2π rad/min)
Wait a minute... look at that! The 2π's cancel out! So we're left with:
I₂ = (I₁ * 55.0) / 19.0
Now carefully rearranging the equation, we find:
I₂ = I₁ * (55.0 / 19.0)
So my final answer is that the moment of inertia of the merry-go-round with respect to its central axis is equal to 55/19 times the initial moment of inertia.
I hope I didn't spin you in circles with my explanation. Remember, if there's ever a physics problem you can't solve, call in the clowns!
To solve this problem, we can use the principle of conservation of angular momentum. The angular momentum before the child grabs onto the merry-go-round is equal to the angular momentum after the child grabs onto it.
The formula for angular momentum is given by:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Let's denote the initial angular velocity as ω1 and the final angular velocity as ω2.
Using the conservation of angular momentum, we have:
I1 * ω1 = I2 * ω2
where I1 is the moment of inertia before the child grabs onto the merry-go-round (which we need to find) and I2 is the moment of inertia after the child grabs onto it.
We are given:
ω1 = 55.0 rpm
ω2 = 19.0 rpm
r = 1.60 m (distance from the center of the merry-go-round)
To convert the angular velocities from rpm to rad/s, we use the following conversion factor:
1 rpm = (2π / 60) rad/s
So we have:
ω1 = (55.0 rpm) * (2π / 60) rad/s
ω2 = (19.0 rpm) * (2π / 60) rad/s
Now we can rearrange the equation to solve for I1:
I1 = (I2 * ω2) / ω1
Substituting the given values, we get:
I1 = (I2 * (19.0 rpm) * (2π / 60) rad/s) / ((55.0 rpm) * (2π / 60) rad/s)
Simplifying the equation, the units of rpm and (2π / 60) rad/s cancel out:
I1 = (I2 * 19.0) / 55.0
To solve for I2, we need to use the radius of the bar from the center of the merry-go-round and the mass of the child.
The formula for the moment of inertia of a point mass rotating about an axis is given by:
I = m * r^2
where m is the mass and r is the distance from the rotation axis.
Substituting the given values, we get:
I2 = (24.5 kg) * (1.60 m)^2
Calculating this, we find:
I2 = 62.4 kg * m^2
Now we can substitute this value back into the equation for I1:
I1 = (62.4 kg * m^2 * 19.0) / 55.0
Calculating this, we find:
I1 ≈ 21.51 kg * m^2
Therefore, the moment of inertia of the merry-go-round with respect to its central axis is approximately 21.51 kg * m^2.
To find the moment of inertia of the merry-go-round, we can use the conservation of angular momentum principle. The angular momentum of an object is given by the formula:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
At the initial state, the angular momentum of the child can be calculated as:
L_initial = I_initial * ω_initial
At the final state, the angular momentum of the child and the merry-go-round together can be calculated as:
L_final = (I_initial + I_child) * ω_final
Since the child grabs onto the merry-go-round and becomes a part of the system, we can add the moment of inertia of the child to the moment of inertia of the merry-go-round.
Assuming the child is a point mass, the moment of inertia of the child can be calculated as:
I_child = m_child * r_child^2
Where m_child is the mass of the child and r_child is the distance of the child from the center of the merry-go-round.
Now, let's plug in the given values:
Mass of the child, m_child = 24.5 kg
Distance of the child from the center, r_child = 1.60 m
Initial angular velocity, ω_initial = 55.0 rpm
Final angular velocity, ω_final = 19.0 rpm
First, convert the angular velocities from rpm to radians per second:
ω_initial = (55.0 rpm * 2π rad/rev) / (60 s/min) = 5.75 rad/s
ω_final = (19.0 rpm * 2π rad/rev) / (60 s/min) = 1.99 rad/s
Substituting the values into the equations:
L_initial = I_initial * ω_initial
L_final = (I_initial + I_child) * ω_final
Since angular momentum is conserved:
L_initial = L_final
So, we can write:
I_initial * ω_initial = (I_initial + I_child) * ω_final
Now, rearrange the equation to solve for I_initial:
I_initial * ω_initial - I_initial * ω_final = I_child * ω_final
I_initial * (ω_initial - ω_final) = I_child * ω_final
I_initial = (I_child * ω_final) / (ω_initial - ω_final)
Substituting the values:
I_initial = (24.5 kg * (1.60 m)^2 * 1.99 rad/s) / (5.75 rad/s - 1.99 rad/s)
Calculating this expression will give you the moment of inertia of the merry-go-round.