The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity 1. A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor. What is the angular velocity 2 of the bar PQ?

GATE 2023 ME – Question 38

Engineering Mechanics · Kinematics and dynamics of rigid bodies in plane motion · 2 marks · Multiple choice

The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity $\omega_1$. A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor.

What is the angular velocity $\omega_2$ of the bar PQ?

a wheel of radius 3 m with centre O. The point P is on the horizontal line through O, 2 m from O. The bar PQ goes from P down to the point Q on the floor, which is 8 m to the right of P horizontally. The contact point of the wheel with the floor is R.

$\omega_2 = 2\omega_1$
$\omega_2 = \omega_1$
$\omega_2 = 0.5\omega_1$
$\omega_2 = 0.25\omega_1$

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Correct answer: (D) $\omega_2 = 0.25\omega_1$

Explanation

Take the origin at the contact point R of the wheel. P is at $(2, 3)$ relative to R (2 m from the centre horizontally, and the centre is 3 m above the floor). The wheel rotates about R, so $\mathbf{v}_P = \omega_1\hat{k} \times (2, 3) = \omega_1(-3, 2)$. The end Q is at $(8, -3)$ relative to P, and its velocity has no vertical component because it slides on the floor. With $\mathbf{v}_Q = \mathbf{v}_P + \omega_2\hat{k} \times (8, -3) = \mathbf{v}_P + \omega_2(3, 8)$, the vertical component is $2\omega_1 + 8\omega_2 = 0$, so $|\omega_2| = \frac{\omega_1}{4} = 0.25\omega_1$ (the bar turns in the sense opposite to the wheel).