Rotational Motion

Step 1: Volume conservation:



Step 2: Disc moment of inertia:


Step 3: Sphere moment of inertia:


Step 4: Hence:

Step 1: The peg moves horizontally with speed v₀ while remaining in contact with the semicircular frame.

Step 2: The instantaneous velocity of the contact point on the frame is equal to the velocity of the peg.

Step 3: At the given position, the perpendicular distance of point P from the hinge O is: OP_⊥ = 2r sin 60^∘

Step 4: Using the relation for rotational motion: v₀ = ω × OP_⊥ v₀ = ω × (2r sin 60^∘) ⟹ ω = (v₀)/(2r)

To solve the problem, we need to apply the conservation of angular momentum. Initially, only the disc is rotating with angular velocity . The total initial angular momentum of the system is:

When the ring is placed on the disc, it reaches the same angular velocity as the disc due to the frictional force, and they rotate together. Let be the moment of inertia of the ring. If both the disc and the ring have the same mass and radius , the moment of inertia of the ring is:

Thus, the total final moment of inertia of the system is:

The final angular velocity is . According to the conservation of angular momentum, the initial angular momentum equals the final angular momentum:

Substitute into the equation:

Rearrange to solve for :

Since the ring has the same mass and radius as the disc, if the disc's moment of inertia is , then:

Simplify the numerator and denominator:

Cancel from the numerator and denominator:

There seems to be a mistake in the initial computation regarding moments of inertia. Let's correct it assuming mass distribution:
,
replace I_r similarly:

Thus:

Simplifying gives:

Therefore, the final angular velocity of the system is .