JEE Main 2025 Apr 4 Shift 2, Physics Q12: Rolling Motion

For pure rolling without slipping, the contact point is the instantaneous centre of rotation. The speed of any point on the rim equals $\omega$ times its distance from the contact point.

Top of the rim is at a distance $2R$ from the contact, so its speed is $2R\omega =8$ m/s, giving the centre speed $v=R\omega =4$ m/s.

A point at the same horizontal level as the centre (e.g. the side of the rim) lies at a distance $\sqrt{R^2+R^2}=R\sqrt{2}$ from the contact point. Its speed is

$v_P=\omega \cdot R\sqrt{2}=v\sqrt{2}=4\sqrt{2}$ m/s.