A horizontal disc rotating freely about a vertical axis makes 90 revolutions per minute. A small piece of wax of mass “m” falls vertically on the disk & sticks to it at a distance “r” from the axis. If the number of revolutions reduces to 60, then find the moment of inertia of the disc?
1. mr2
2.3/2mr2
3.2mr2
4.3mr2
(the last ‘2’in each option should be considered as square, please)
To find the moment of inertia of the disc, we can use the concept of conservation of angular momentum.
Before the wax falls onto the disc, the initial angular momentum of the disc is given by:
L1 = I1 * ω1
Where:
L1 = initial angular momentum
I1 = initial moment of inertia of the disc
ω1 = initial angular velocity of the disc
After the wax falls onto the disc and sticks to it, the final angular momentum of the disc is given by:
L2 = I2 * ω2
Where:
L2 = final angular momentum after the wax falls
I2 = final moment of inertia of the disc (which we need to find)
ω2 = final angular velocity of the disc
According to the law of conservation of angular momentum, the initial and final angular momenta are equal. Therefore, we can write:
L1 = L2
I1 * ω1 = I2 * ω2
Given that the initial angular velocity, ω1, is 90 revolutions per minute, and the final angular velocity, ω2, is 60 revolutions per minute, we need to convert these values to radians per second for consistency.
1 revolution = 2π radians
So, ω1 = (90 revolutions/minute) * (2π radians/revolution) * (1 minute / 60 seconds) = 3π radians/second
And, ω2 = (60 revolutions/minute) * (2π radians/revolution) * (1 minute / 60 seconds) = 2π radians/second
Substituting these values into the conservation equation, we get:
I1 * (3π) = I2 * (2π)
Simplifying, we can cancel out π from both sides:
3I1 = 2I2
Dividing both sides by 2, we get:
I2 = (3/2) * I1
Therefore, the moment of inertia of the disc after the wax falls for a horizontal disc rotating freely about a vertical axis is given by option 2:
3/2mr²
Option 2 is the correct answer.
To solve this problem, we need to consider the principle of conservation of angular momentum.
The initial angular momentum of the disc before the wax falls on it can be given as L1 = I1 * ω1, where I1 is the moment of inertia of the disc at that time and ω1 is the initial angular velocity.
After the wax falls on the disc, the combined system (disc + wax) will have a new moment of inertia I2 and a new angular velocity ω2. The angular momentum of the combined system can be written as L2 = I2 * ω2.
According to the conservation of angular momentum, L1 = L2.
Given that the initial number of revolutions is 90, we can calculate the initial angular velocity ω1 as follows:
ω1 = (90 revolutions/minute) * (2π radians/revolution) * (1 minute/60 seconds) = 3π radians/second.
After the wax falls on the disc, the number of revolutions reduces to 60, so the new angular velocity ω2 can be calculated as follows:
ω2 = (60 revolutions/minute) * (2π radians/revolution) * (1 minute/60 seconds) = 2π radians/second.
Now, let's substitute the values into the conservation equation and solve for the moment of inertia I2:
I1 * ω1 = I2 * ω2.
Since the mass of the wax (m) falls vertically on the disk and sticks to it at a distance r from the axis, the moment of inertia of the wax can be written as Iwax = m * r^2.
The moment of inertia of the disc can be denoted as Idisc.
Since the wax sticks to the disk, we can treat the combined system as a single rigid body and write the moment of inertia of the combined system as I2 = Idisc + Iwax.
Substituting the values into the conservation equation, we have:
Idisc * ω1 = (Idisc + Iwax) * ω2.
Simplifying the equation:
Idisc * 3π = (Idisc + m * r^2) * 2π.
2Idisc * π = m * r^2 * π.
2Idisc = m * r^2.
Dividing both sides by 2:
Idisc = m * r^2 / 2.
Comparing the equation obtained with the options given:
The moment of inertia of the disc is 1/2 * m * r^2, which matches with option 2.
Therefore, the correct answer is option 2: 3/2 * mr^2.
Conservation of momentum is applied.
Momentum before=momentum after
I disk*(90*2PI/60)=Idisk(60*2PI/60)+mr^2*60(2PI/60)
solve for I disk.