As you said since the disc is moving linearly but the particles perform circular motion so the particle at r/2 would also do the same and hence its angular velocity will be 2u/R please correct me if i am wrong...... Hope it helps

Problem Statement:
A disc of radius R is rolling (without slipping) on a horizontal surface with a linear velocity u. We need to find the speed of a particle located at a distance R/2 just above the center of the disc.

Solution:

1. Understanding the Problem:
To solve this problem, we need to understand the concept of rolling motion and the relationship between linear velocity and angular velocity.

2. Relationship between Linear Velocity and Angular Velocity:
In rolling motion, the linear velocity of any point on the disc can be related to its angular velocity using the formula:

v = ω * r

where,
v = linear velocity
ω = angular velocity
r = distance of the point from the axis of rotation

3. Linear Velocity of the Center of the Disc:
Since the disc is rolling without slipping, the linear velocity of the center of the disc is given by:

v_center = u

where u is the given linear velocity.

4. Angular Velocity of the Center of the Disc:
The angular velocity of the center of the disc can be calculated using the formula:

ω_center = u / R

where R is the radius of the disc.

5. Linear Velocity of the Particle at R/2:
To find the speed of the particle at a distance R/2 above the center of the disc, we need to calculate the linear velocity at that point. Since the particle is located on the disc, its distance from the axis of rotation is R/2.

Using the formula mentioned in step 2, we can calculate the linear velocity of the particle as:

v_particle = ω_center * (R/2)

Substituting the value of ω_center from step 4, we get:

v_particle = (u / R) * (R/2)
v_particle = u/2

6. Final Answer:
Therefore, the speed of the particle located at a distance R/2 just above the center of the disc is equal to u/2.