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Step-by-Step Solution
Step 1: Identify the Translational Velocity of the Centre
The wheel rolls on a plane surface without slipping, and its center moves with a constant speed v_0 . This speed is the translational speed of the entire wheel.
Step 2: Express the Angular Velocity of the Wheel
Let the radius of the wheel be R . Since there is no slipping, the relation between translational velocity v_0 and angular velocity \omega is:
\omega = \dfrac{v_0}{R} .
Step 3: Determine the Rotational Velocity of the Rim Particle
A point on the rim of the wheel also has rotational velocity around the center. The linear speed of any point on the rim, due to rotation, is given by v_{\text{rot}} = \omega R . Substituting \omega = \dfrac{v_0}{R} , we get:
v_{\text{rot}} = \omega R = \dfrac{v_0}{R} \times R = v_0.
Step 4: Combine the Velocities for the Point at the Same Level as the Center
The particle on the rim, at the same horizontal level as the center, will have its translational velocity v_0 (horizontally to the right, say) and its rotational velocity v_0 (vertically upward or downward, depending on the wheel’s sense of rotation) at right angles to each other.
Step 5: Calculate the Resultant Velocity
The total velocity of that point is the vector sum of the two perpendicular components v_0 (horizontal) and v_0 (vertical). Thus, the magnitude v_{\text{total}} is:
v_{\text{total}} \;=\; \sqrt{v_0^2 + v_0^2} \;=\; \sqrt{2}\,v_0.
Step 6: Conclude the Value of x
Since the question states this speed as \sqrt{x}\,v_0 , we have:
\sqrt{x}\,v_0 = \sqrt{2}\,v_0.
By comparing both sides, we get:
\sqrt{x} = \sqrt{2} \quad \Longrightarrow \quad x = 2.
