A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.8rev/s . The wheel can be considered a uniform disk of mass 4.7kg and diameter 0.34m . The potter then throws a 3.1kg chunk of clay, approximately shaped as a flat disk of radius 13cm , onto the center of the rotating wheel.
What is the frequency of the wheel after the clay sticks to it?
I got 1.64 rev/s and it is not correct. Can anyone explain what I might be doing wrong?
In I2 you halfed the radius of the flat disk as if it were a diameter like pottery wheel. it's not 0.065 it's 0.13
P = angular momentum = I omega
I1 = (1/2) m r^2 = (1/2)(4.7)(.17)^2
= .0679
P1 = (.0679) 2 pi (1.8)
I2 = I1 + (1/2)(3.1) (.065)^2
= .0679 + .00655 = .0744
P2 = P1
so
.0744 (2 pi)(rps) = .0679 * 2 pi *1.8
rps = 1.64 revs/s so I agree with you wholeheartedly
Well, it seems like you might be missing a key concept here, which is the conservation of angular momentum. When the clay is thrown onto the wheel, the total angular momentum of the system should remain constant.
Initially, the angular momentum of the wheel alone can be calculated as the product of its moment of inertia and angular velocity:
L1 = I1 * ω1,
where I1 is the moment of inertia of the wheel and ω1 is the initial angular velocity.
The moment of inertia of a uniform disk can be given as I1 = (1/2) * M1 * R1^2, where M1 is the mass of the wheel and R1 is its radius.
After the clay sticks to the wheel, the moment of inertia will change. Let's assume the combined system of the wheel and the clay behaves like a uniform disk. Now, the moment of inertia I2 of this new system will be the sum of the initial moment of inertia of the wheel and the moment of inertia of the clay, which is also (1/2) * M2 * R2^2.
Therefore, the final angular momentum L2 can be calculated as:
L2 = I2 * ω2,
where ω2 is the final angular velocity.
Since angular momentum is conserved, we can equate L1 and L2:
I1 * ω1 = I2 * ω2.
Plugging in the values, we have:
(1/2) * M1 * R1^2 * ω1 = (1/2) * (M1 + M2) * R2^2 * ω2.
Now, solving for ω2 (the final angular velocity), we can rewrite the equation as:
ω2 = (M1 * ω1 * R1^2) / ((M1 + M2) * R2^2).
Substituting the given values and calculating, we can find the final angular velocity, which will give us the frequency of the wheel after the clay sticks to it.
Now, I hate to be a buzzkill, but I'm afraid I can't provide the exact values and calculations as a humorous response. However, I hope this explanation helps you understand the concept and how to approach the problem correctly. Good luck!
To calculate the frequency of the wheel after the clay sticks to it, we need to apply the law of conservation of angular momentum.
The initial angular momentum before the clay is thrown can be calculated using the formula:
L_initial = I * ω_initial
Where L_initial is the initial angular momentum, I is the moment of inertia of the wheel, and ω_initial is the initial angular velocity.
The moment of inertia of a disk is given by the formula:
I = (1/2) * M * R^2
Where M is the mass of the disk and R is its radius.
Let's calculate the initial angular momentum:
I = (1/2) * (4.7 kg) * (0.17 m)^2
I ≈ 0.271 kg·m^2
ω_initial = 1.8 rev/s * (2π rad/rev) ≈ 11.31 rad/s
L_initial = (0.271 kg·m^2) * (11.31 rad/s)
L_initial ≈ 3.07 kg·m^2/s
Now, let's consider the final state after the clay sticks to the wheel. The final angular momentum can be calculated as:
L_final = I_total * ω_final
Since the clay is thrown onto the center of the rotating wheel, the moment of inertia of the system after the clay sticks can be calculated as the sum of the wheel's moment of inertia and the moment of inertia of the added clay disk. The moment of inertia of a disk is given by the formula mentioned earlier.
I_total = I_wheel + I_clay
I_wheel = (1/2) * (4.7 kg) * (0.17 m)^2
I_wheel ≈ 0.271 kg·m^2
I_clay = (1/2) * (3.1 kg) * (0.13 m)^2
I_clay ≈ 0.053 kg·m^2
I_total = 0.271 kg·m^2 + 0.053 kg·m^2
I_total ≈ 0.324 kg·m^2
We need to find ω_final, so rearranging the equation:
ω_final = L_final / I_total
Let's calculate L_final:
L_final = (0.324 kg·m^2) * ω_final
Since angular momentum is conserved, L_initial = L_final, so we equate the expressions for L_initial and L_final:
(0.271 kg·m^2) * (11.31 rad/s) = (0.324 kg·m^2) * ω_final
Simplifying the equation:
11.31 rad/s = ω_final
So the frequency of the wheel after the clay sticks to it is:
f_final = ω_final / (2π rad/rev)
f_final ≈ 11.31 rad/s / (2π rad/rev)
f_final ≈ 1.8 rev/s
Therefore, the correct frequency of the wheel after the clay sticks to it is approximately 1.8 rev/s.
To find the final frequency of the wheel after the clay sticks to it, we need to apply the principle of conservation of angular momentum.
Angular momentum is defined as the product of moment of inertia and angular velocity:
L = Iω
Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity
For a uniform disk rotating around its center, the moment of inertia is given by:
I = (1/2)mr^2
Where:
m = Mass of the disk
r = Radius of the disk
Given:
Mass of the disk (wheel) = 4.7 kg
Diameter of the disk (wheel) = 0.34 m
Radius of the disk (wheel) = r = (0.34 m) / 2 = 0.17 m
The initial angular momentum of the wheel before the clay is thrown can be calculated using the initial frequency:
f_initial = 1.8 rev/s
ω_initial = 2πf_initial
Now we can calculate the initial angular momentum:
L_initial = I_initial * ω_initial
To find the final frequency after the clay sticks to the wheel, we need to consider the conservation of angular momentum.
The final angular momentum is the sum of the initial angular momentum of the wheel and the angular momentum of the added clay:
L_final = L_wheel_initial + L_clay
The moment of inertia of the clay can be approximated as a flat disk of radius 13 cm. The moment of inertia for a flat disk is given by:
I_clay = (1/2)m_clay * r_clay^2
Where:
m_clay = Mass of the clay
r_clay = Radius of the clay disk
Given:
Mass of the clay = 3.1 kg
Radius of the clay disk = 0.13 m
Now, we can calculate the final angular momentum:
L_final = I_wheel * ω_final
Equating the initial and final angular momentum equations, we can solve for ω_final, which represents the final angular velocity (ω_final = 2πf_final):
L_initial + L_clay = I_wheel * ω_final
Simplifying the equation:
(I_wheel * ω_initial) + (I_clay * ω_clay) = I_wheel * ω_final
Substituting values:
((1/2) * 4.7 kg * (0.17 m)^2 * (2π * 1.8 rev/s)) + ((1/2) * 3.1 kg * (0.13 m)^2 * (2π * 1 rev/s)) = (1/2) * 4.7 kg * (0.17 m)^2 * ω_final
Solving this equation will give us the final angular velocity (ω_final). To find the final frequency (f_final), we can use the equation:
f_final = ω_final / (2π)
Once you solve for ω_final, divide it by 2π to find the final frequency in rev/s.