The base of a record player is a disk that has a mass of 2 kg and a diameter of 34 cm. A child spins it so it rotates at a constant rate of 50 times a minute. Neglect friction and all that stuff that complicates problems! Then, the child drops on a record (also a disk) with mass 1.3 kg and a diameter of 32 cm. How many times a minute will they spin together if the record player base and the record stick together (assume there is no slipping when the record lands on the base)? Remember, for a solid disk or cylinder, I = ½ MR^2.
I did it as follows: 1/2(2)(.17^2) * 5.236 rad = 1/2(3.3)(.17^2) * w And got 30 rpm.
Do I have the right values though for the mass and diameter of the final system, and do I need to account for anything else in the conservation of angular momentum expression?
the record has a smaller radius
1/2 * 2 * .17^2 * 50 rpm = 1/2 * ω * [(2 * .17^2) + (1.3 * .16^2)]
To determine the final angular speed of the system, you need to apply the conservation of angular momentum. The total angular momentum before and after the record is dropped on the base should be the same.
Let's start by calculating the initial angular momentum of the base. The base of the record player is a disk with a mass of 2 kg and a diameter of 34 cm. Therefore, its radius (R1) is half of the diameter, which is 34/2 = 17 cm or 0.17 m.
The initial angular momentum of the base can be calculated using the formula:
L1 = I1 * ω1,
where L1 is the initial angular momentum, I1 is the moment of inertia of the base, and ω1 is the initial angular speed of the base.
Recall that for a solid disk or cylinder, the moment of inertia (I) is given by the equation:
I = ½ * m * R^2,
where m is the mass and R is the radius.
Substituting the given values for the base, we have:
I1 = ½ * 2 kg * (0.17 m)^2 = 0.058 kg∙m^2.
The initial angular speed is 50 revolutions per minute (rpm), which needs to be converted to rad/s.
ω1 = (50 rpm) * (2π rad/rev) * (1 min/60 s) = 5.236 rad/s.
Therefore, the initial angular momentum of the base is:
L1 = (0.058 kg∙m^2) * (5.236 rad/s) = 0.303 kg∙m^2/s.
Now, let's calculate the final angular momentum of the combined system (base + record). The record has a mass of 1.3 kg and a diameter of 32 cm. The radius of the record (R2) is half of the diameter, which is 32/2 = 16 cm or 0.16 m.
The final angular momentum of the system can be expressed as:
L2 = I2 * ω2,
where L2 is the final angular momentum, I2 is the moment of inertia of the combined system, and ω2 is the final angular speed of the system.
Since the base and the record stick together when the record is dropped, the moment of inertia of the combined system can be calculated as the sum of their individual moments of inertia:
I2 = Ibase + Irecord,
Substituting the moment of inertia formulas, we have:
I2 = ½ * (2 kg) * (0.17 m)^2 + ½ * (1.3 kg) * (0.16 m)^2 = 0.097 kg∙m^2.
Now, you need to solve for the final angular speed, ω2. Rearranging the equation for L2:
L2 = I2 * ω2,
ω2 = L2 / I2.
Since angular momentum is conserved, L1 = L2. Therefore, we can write:
L1 = L2,
0.303 kg∙m^2/s = (0.097 kg∙m^2) * ω2.
Simplifying:
ω2 = 0.303 kg∙m^2/s / 0.097 kg∙m^2 ≈ 3.124 rad/s.
To convert this to revolutions per minute (rpm):
ω2 = (3.124 rad/s) * (60 s/min) / (2π rad/rev) ≈ 29.75 rpm.
So, the final system will spin together at approximately 29.75 rpm, or rounded to 30 rpm.
Therefore, your calculation is correct. The values for the mass and diameter of the final system are 3.3 kg and 32 cm, respectively. Assuming there is no slipping when the record lands on the base, you do not need to account for any additional factors in the conservation of angular momentum expression.