A thin uniform cylindrical turntable of radius 2.1 m and mass 18 kg rotates in a horizontal plane with an initial angular speed of 8 rad/s.
The turntable bearing is frictionless. A clump of clay of mass 5.5 kg is dropped onto the
turntable and sticks at a point 0.9 m from the point of rotation.
Find the angular speed of the clay and turntable.
Answer in units of rad/s
The rotational
inertia of a disc is 1/2(MR-squared).
OK... so i am "supposedly" doing everything right, by using an elastic equation and substituting mass with inertia and velocity with angular velocity... but i keep getting the same answer... 7.575 rad/s and its wrong...
It's an inelastic conservation of angular question.
I omegai = I omegaf (turntable)+ I omegaf (Clay, I = mr^2)
And obviously omegaf is the same for both.
To solve this problem, we can use the principle of conservation of angular momentum. Initially, the turntable is rotating with an angular speed of 8 rad/s, and after the clay is dropped onto it, the system (turntable + clay) will continue to rotate with a new angular speed.
The principle of conservation of angular momentum states that the initial angular momentum must be equal to the final angular momentum.
The initial angular momentum (L_initial) of the system is given by the product of the initial moment of inertia (I_initial) and the initial angular speed (ω_initial).
L_initial = I_initial * ω_initial
The moment of inertia (I_initial) of the turntable alone can be calculated using the given formula for the rotational inertia of a disk:
I_initial = (1/2) * M * R^2
where M is the mass of the turntable (18 kg) and R is the radius (2.1 m).
Substituting the values, we get:
I_initial = (1/2) * 18 kg * (2.1 m)^2
= 21.42 kg·m²
Substituting this into the equation for initial angular momentum, we have:
L_initial = 21.42 kg·m² * 8 rad/s
= 171.36 kg·m²/s
Now, when the clay is dropped onto the turntable and sticks at a point 0.9 m from the point of rotation, the moment of inertia of the system changes.
The final moment of inertia (I_final) of the system can be calculated as the sum of the initial moment of inertia of the turntable (I_initial) and the moment of inertia of the clay (I_clay).
The moment of inertia of the clay can be calculated using the formula for a point mass:
I_clay = m_clay * r^2
where m_clay is the mass of the clay (5.5 kg) and r is the distance of the clay from the point of rotation (0.9 m).
Substituting the values, we get:
I_clay = 5.5 kg * (0.9 m)^2
= 4.955 kg·m²
Now, the final moment of inertia (I_final) of the system is:
I_final = I_initial + I_clay
= 21.42 kg·m² + 4.955 kg·m²
= 26.375 kg·m²
Since the angular momentum is conserved, we can equate the initial and final angular momenta:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Solving for ω_final, we have:
ω_final = (I_initial * ω_initial) / I_final
Substituting the values, we get:
ω_final = (21.42 kg·m² * 8 rad/s) / 26.375 kg·m²
= 6.549 rad/s
Therefore, the angular speed of the clay and turntable after the clay is dropped onto it is approximately 6.549 rad/s.
To find the final angular speed of the clay and turntable, we can use the principle of conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum of the system.
The initial angular momentum can be calculated using the formula:
L_initial = I_initial * ω_initial
Where:
I_initial is the initial moment of inertia of the turntable (1/2 * mass * radius^2)
ω_initial is the initial angular speed of the turntable (given as 8 rad/s)
The final angular momentum can be calculated using the formula:
L_final = I_final * ω_final
Where:
I_final is the final moment of inertia of the system, considering both the turntable and the clay together
ω_final is the final angular speed of the system
Given that the clay sticks at a point 0.9 m from the point of rotation, we can calculate the final moment of inertia as follows:
I_final = I_turntable + I_clay
Where:
I_turntable = 1/2 * mass_turntable * radius_turntable^2
I_clay = mass_clay * distance_from_rotation^2
Now we have all the necessary formulas to solve the problem. Let's substitute the values into these formulas and calculate the final angular speed.
I_turntable = 1/2 * 18 kg * (2.1 m)^2 = 41.58 kg*m^2
I_clay = 5.5 kg * (0.9 m)^2 = 4.455 kg*m^2
I_final = 41.58 kg*m^2 + 4.455 kg*m^2 = 46.035 kg*m^2
Using the conservation of angular momentum:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
(1/2 * 18 kg * (2.1 m)^2) * 8 rad/s = 46.035 kg*m^2 * ω_final
Simplifying:
159.12 kg*m^2/s = 46.035 kg*m^2 * ω_final
ω_final = 159.12 kg*m^2/s / 46.035 kg*m^2 ≈ 3.457 rad/s
Therefore, the final angular speed of the clay and turntable is approximately 3.457 rad/s.