1 point possible)
A uniform disc of radius R and mass m1 is rotating with an angular speed ωi about an axis passing through its center and perpendicular to the disc's plane. A small box of mass m2, initially at the center of the disc, moves away from the center and stops moving relative to the disc when it is at a distance 12R from the disc's center.
What is the angular speed of the disc--block system at the instant when the block stops moving relative to the disc?
Symbol m1 m2 R ωi
You could enter m1 m2 R omega_i
ωf=
To solve this problem, we need to apply the principle of conservation of angular momentum. The angular momentum of a rotating object is given by the product of its moment of inertia and angular velocity.
Let's break down the problem step by step to find the angular speed (ωf) of the disc-block system at the instant when the block stops moving relative to the disc.
Step 1: Determine the initial angular momentum (Li) of the system.
The initial angular momentum of the system is the sum of the angular momentum of the disc and the angular momentum of the block. Since the block is initially at rest, its angular momentum is zero.
Therefore, Li = Angular momentum of the disc = moment of inertia of the disc (Idisc) × initial angular velocity of the disc (ωi).
Step 2: Determine the final angular momentum (Lf) of the system.
The final angular momentum of the system is the sum of the angular momentum of the disc and the angular momentum of the block. Since the block comes to rest, its final angular momentum is zero.
Therefore, Lf = Angular momentum of the disc = moment of inertia of the disc (Idisc) × final angular velocity of the disc (ωf).
Step 3: Apply the principle of conservation of angular momentum.
According to the principle of conservation of angular momentum, the total angular momentum before the block moves (Li) is equal to the total angular momentum after the block stops (Lf).
Li = Lf
Step 4: Substitute the expressions for initial and final angular momentum.
Idisc × ωi = Idisc × ωf
Step 5: Solve for ωf.
Divide both sides of the equation by Idisc to isolate ωf:
ωf = (Idisc × ωi) / Idisc
Step 6: Simplify.
The moment of inertia of a disc is given by I = (1/2) × m × R^2, where m is the mass of the disc and R is its radius. Substituting this expression into the equation, we get:
ωf = ( (1/2) × m1 × R^2 × ωi) / ( (1/2) × m1 × R^2)
Step 7: Simplify further.
The R^2 terms will cancel out, leaving us with:
ωf = ωi
Therefore, the angular speed (ωf) of the disc-block system at the instant when the block stops moving relative to the disc is equal to the initial angular speed (ωi).