A centrifuge rotor rotating at 11500 is shut off and is eventually brought uniformly to rest by a frictional torque of 1.52 .
a.If the mass of the rotor is 6.07 and it can be approximated as a solid cylinder of radius 0.0750 , through how many revolutions will the rotor turn before coming to rest?
b.How long will it take?
Units????????
SOrry! the units are
11500rmp
1.52 m*N
6.07kg
0.0750m
To find the number of revolutions the rotor will turn before coming to rest, we can use the principle of conservation of angular momentum.
The angular momentum (L) of the rotor is given by the formula: L = I * ω
where I is the moment of inertia of the rotor and ω is the angular velocity.
The moment of inertia for a solid cylinder rotating about its axis is given by the formula: I = (1/2) * m * r^2
where m is the mass of the rotor and r is the radius of the rotor.
Therefore, the moment of inertia (I) of the rotor is: I = (1/2) * 6.07 * (0.0750)^2
Next, we need to find the initial angular velocity (ω) of the rotor. The angular velocity is given in terms of revolutions per minute (rpm), so we need to convert it to radians per second (rad/s).
To do this, we use the conversion factor: 1 rev/min = 2π rad/min = (2π/60) rad/s. Therefore, the initial angular velocity (ω) is: ω = (11500 * 2π/60) rad/s.
Now, we can calculate the initial angular momentum (L) of the rotor: L = I * ω.
To find the time it takes for the rotor to come to rest, we can use the formula: τ = I * α
where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
In this case, the torque applied is given as 1.52 N·m. The angular acceleration (α) can be calculated using the formula: α = τ/I.
Finally, the time it takes for the rotor to come to rest can be calculated using the formula: t = ω/α.
Let's plug in the values and calculate:
I = (1/2) * 6.07 * (0.0750)^2
ω = (11500 * 2π/60)
L = I * ω
α = τ/I
t = ω/α
After plugging in the values and performing the calculations, you will obtain the answers to both parts of the question.