Question 1 of 6: Disk Rolling on a Free-Rolling Wedge

Question 1: Disk Rolling on a Free-Rolling Wedge (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A disk of mass $m_d=2\text{ kg}$ (1 m diameter, radius $R=0.5\text{ m}$) starts at rest on the inclined face of a wedge of mass $M_s=5\text{ kg}$ inclined at $\theta=30^\circ$; the wedge sits on massless, frictionless wheels so it is free to translate horizontally; the disk rolls without slipping on the incline as it descends.

Find. (a) the type of kinematic constraint governing the disk's rolling; (b) the number of generalized coordinates (degrees of freedom) needed for the disk's motion; (c) the conservation-of-momentum relation linking the wedge's and disk's motions.

Approach. Classify the rolling condition, count the independent coordinates left after the contact and rolling constraints, then apply $\Sigma F_{ext,x}=0$ to the combined disk+wedge system.

Part (a) — type of constraint. Rolling without slipping means the disk's contact-point velocity relative to the incline surface is zero; because this incline is a fixed STRAIGHT line embedded in the wedge, that condition integrates directly to a linear relation between the disk's spin angle and its sliding distance along the surface, $s=R\,\phi+\text{const}$. A constraint like this — expressible as an algebraic relation among the coordinates once integrated — is, by definition, holonomic, even though it is stated as a velocity condition. (Rolling only becomes genuinely nonholonomic when the contact path is not a fixed straight line, e.g. a sphere rolling freely over a 2-D surface.) The disk's staying in contact with the incline is a second, purely geometric, holonomic constraint.

Part (b) — degrees of freedom. A free disk in the plane has 3 raw coordinates $(x_d,y_d,\phi_d)$. The contact constraint removes 1 (fixing the disk's offset perpendicular to the incline); the rolling constraint removes a second (slaving $\phi_d$ to the sliding distance). That leaves exactly one coordinate for the disk's motion relative to the wedge — call it $s$, the distance travelled along the incline. But the wedge itself is free to translate horizontally (coordinate $X_s$), and the disk's absolute position depends on both $X_s$ and $s$ together. So describing the disk's actual (absolute) motion needs two generalized coordinates, $(X_s,\,s)$: one for the wedge's own position, one for the disk's position along the moving incline; the disk's spin angle is not independent (Part a).

1. Identify the external forces on {disk + wedge}. Gravity acts vertically on both bodies; the ground reacts on the (frictionless, massless) wheels with a purely vertical normal force. No external force has a horizontal component, so the horizontal linear momentum of the combined system is conserved: $$\frac{d}{dt}\big(m_d\dot x_d + M_s\dot X_s\big) = 0$$ Since the system starts from rest, the constant is zero for all time: $$\boxed{m_d\dot x_d + M_s \dot X_s = 0}$$
2. Relate the disk's absolute velocity to the two chosen coordinates. The disk's centre sits a fixed height above the incline, so its horizontal position is $x_d = X_s + s\cos\theta+\text{const}$, giving $\dot x_d = \dot X_s + \dot s\cos\theta$. Substituting into the boxed relation and solving for $\dot X_s$: $$m_d(\dot X_s+\dot s\cos\theta)+M_s\dot X_s=0 \ \Rightarrow\ \boxed{\dot X_s = -\frac{m_d\cos\theta}{m_d+M_s}\,\dot s}$$ This is the reduced, one-equation kinematic link between the wedge's recoil and the disk's slide — it lets the whole problem be solved with the single coordinate $s$ once $\dot X_s$ is eliminated using this relation.
3. Evaluate the coefficient numerically. With $m_d=2\text{ kg}$, $M_s=5\text{ kg}$, $\theta=30^\circ$ ($\cos30^\circ=0.8660$): $$\frac{m_d\cos\theta}{m_d+M_s} = \frac{2(0.8660)}{7} = 0.2474$$ so $\dot X_s \approx -0.2474\,\dot s$: the wedge recoils in the direction opposite the disk's downslope slide, at about one quarter of the slide's horizontal-projection speed — the heavier wedge moves the less of the two, as expected.