A potter’s wheel with rotational inertia of 45 kg m2, is spinning freely at 40 rpm. The potter drops a lump of clay onto the wheel where it sticks a distance 1.2 m from the rotational axis at the wheel’s center. If the subsequent angular speed of the wheel is 32 rpm, what is the clay’s mass? (Hint: treat the clay as a point object, so I=mr2)
conserve angular momentum
45 * 40 = (45 + m * 1.2^2) * 32
(no need here to convert rpm to radians per second)
Well, isn't it amazing how the potter can just drop a lump of clay onto the wheel and magically make it stick? It's like they have super glue for hands or something! Anyway, let's get back to the question.
We can start by calculating the initial angular momentum of the wheel, which is given by the equation L = Iω, where L is the angular momentum, I is the rotational inertia, and ω is the angular speed.
So, the initial angular momentum of the wheel is L1 = 45 kg m^2 * (40 rpm * 2π/60) = 150π kg m^2/s.
When the clay sticks to the wheel, the total angular momentum of the system (wheel + clay) is conserved. So we have L1 = L2, where L2 is the final angular momentum of the system.
The final angular momentum of the system can be calculated as L2 = (I + m*r^2)ω, where m is the mass of the clay and r is the distance of the clay from the rotational axis.
Plugging in the given values, we get 150π kg m^2/s = (45 kg m^2 + m*(1.2 m)^2) * (32 rpm * 2π/60).
Simplifying the equation, we have 15π = (45 + m*1.44) * (16π/60).
Now, let's solve for m, the mass of the clay.
15π = (45 + 1.44m) * (16π/60).
15π * 60 / (16π) = 45 + 1.44m.
Canceling out π and simplifying, we get 900 / 16 = 45 + 1.44m.
So, 56.25 = 45 + 1.44m.
And finally, m = (56.25 - 45) / 1.44.
So the mass of the clay is approximately m = 7.81 kg.
Well, that's one heavy lump of clay! I hope the potter has some muscles to handle it.
To find the mass of the clay, we can use the principle of conservation of angular momentum. The formula for angular momentum is L = Iω, where L is the angular momentum, I is the rotational inertia, and ω is the angular velocity.
Initially, the rotational inertia of the wheel is given as 45 kg m^2, and the angular velocity is 40 rpm. The angular momentum of the wheel is L1 = I1 * ω1.
After the clay is dropped onto the wheel, the rotational inertia of the system (wheel + clay) is given as (I1 + m * r^2), where m is the mass of the clay and r is the distance of the clay from the rotational axis (1.2 m).
The angular velocity of the system after the clay is dropped is given as 32 rpm. The angular momentum of the system after the clay is dropped is L2 = (I1 + m * r^2) * ω2.
According to the conservation of angular momentum, L1 = L2. Therefore, I1 * ω1 = (I1 + m * r^2) * ω2.
Substituting the given values, we have:
45 kg m^2 * (40 rpm) = (45 kg m^2 + m * (1.2 m)^2) * (32 rpm).
Simplifying the equation, we get:
1800 kg m^2 rpm = (45 kg m^2 + 1.728 m^2 * m) * 32 rpm.
1800 = (45 + 1.728 * m) * 32.
Dividing both sides by 32:
56.25 = 1.40625 + 0.054 m.
Subtracting 1.40625 from both sides:
54.84375 = 0.054 m.
Dividing both sides by 0.054:
m ≈ 1016.204.
Therefore, the mass of the clay is approximately 1016.204 kg.
To find the clay's mass, we can use the principle of conservation of angular momentum. The initial angular momentum of the wheel-clay system before the clay is dropped can be calculated using:
L_i = I_i * ω_i
where L_i is the initial angular momentum, I_i is the initial rotational inertia of the wheel, and ω_i is the initial angular speed of the wheel.
The final angular momentum after the clay is dropped can be calculated using:
L_f = I_f * ω_f
where L_f is the final angular momentum, I_f is the final rotational inertia of the wheel with the clay attached, and ω_f is the final angular speed of the wheel with the clay attached.
Since the clay sticks to the wheel, the rotational inertia of the wheel with the clay attached can be calculated using:
I_f = I_w + I_c
where I_w is the rotational inertia of the wheel and I_c is the rotational inertia of the clay.
Given:
I_w = 45 kg m^2 (rotational inertia of the wheel)
ω_i = 40 rpm (initial angular speed of the wheel)
ω_f = 32 rpm (final angular speed of the wheel)
r = 1.2 m (distance of clay from the rotational axis at the wheel's center)
First, let's convert the initial and final angular speeds from rpm (revolutions per minute) to rad/s (radians per second):
ω_i = 40 rpm * (2π radians / 1 revolution) * (1 minute / 60 seconds) = (40 * 2π) / 60 rad/s
ω_f = 32 rpm * (2π radians / 1 revolution) * (1 minute / 60 seconds) = (32 * 2π) / 60 rad/s
Now, let's calculate the initial angular momentum:
L_i = I_i * ω_i = 45 kg m^2 * [(40 * 2π) / 60 rad/s] = 45 kg m^2 * (2π/3) rad/s
The final angular momentum can be written as:
L_f = (I_w + I_c) * ω_f
Substituting the value of I_f from above, we get:
L_f = (45 kg m^2 + I_c) * [(32 * 2π) / 60 rad/s]
Since angular momentum is conserved, L_i = L_f.
Equating the initial and final angular momentum, we have:
45 kg m^2 * (2π/3) rad/s = (45 kg m^2 + I_c) * [(32 * 2π) / 60 rad/s]
Now, let's solve for I_c:
I_c = [45 kg m^2 * (2π/3) rad/s - 45 kg m^2 * [(32 * 2π) / 60 rad/s]]
Simplifying,
I_c = (45 kg m^2 * [(2π/3) - (32 * 2π) / 60]) rad/s
Finally, to find the clay's mass, we can use the relation:
I_c = m * r^2
Rearranging the equation,
m = I_c / r^2
Substituting the known values,
m = [I_c] / [(1.2 m)^2]
Calculate I_c using the above equation and then substitute the value in the equation to solve for m.
Well, final angular momentum is equal to intial.
wf*If=wi*Ii
where If= Ii+ mr^2 so solve for m
change w to rad/sec