To analyze the situation of the disc rolling and then hitting a step, we need to consider both the linear and angular momentum of the disc, as well as the effects of the collision with the step. The disc is initially rolling without slipping, which means that its center of mass has a velocity \( v_0 \) and it has an angular velocity \( \omega \) related to its linear velocity by the equation \( v_0 = R \omega \). When the disc hits the step, it will pivot about the edge of the step due to the height of the step being \( R/4 \). Let's break this down step by step.
Initial Conditions
Before the impact, the disc has:
- Mass: \( M \)
- Radius: \( R \)
- Linear velocity: \( v_0 \)
- Angular velocity: \( \omega = \frac{v_0}{R} \)
Impact Analysis
When the disc strikes the step, it will rotate about the edge of the step. The height of the step is \( R/4 \), which means the center of mass of the disc will rise by \( R/4 \) during the impact. The point of contact with the step will be at the edge of the disc, which is at a distance \( R \) from the center of the disc.
Conservation of Angular Momentum
Since the step is rough enough to prevent slipping, we can use the conservation of angular momentum about the point of contact (the edge of the step). The initial angular momentum \( L_i \) of the disc about the edge of the step can be calculated as:
Initial Angular Momentum: \( L_i = I_{cm} \omega + M v_0 \cdot \frac{R}{2} \)
Where \( I_{cm} \) is the moment of inertia of the disc about its center, given by \( I_{cm} = \frac{1}{2} M R^2 \). The distance from the center of mass to the edge of the step is \( R \), and thus the term \( M v_0 \cdot \frac{R}{2} \) accounts for the linear momentum contribution.
Calculating the Final Velocity
After the impact, the disc will have a new angular velocity \( \omega' \) and a new linear velocity \( v' \). The final angular momentum \( L_f \) about the edge of the step is:
Final Angular Momentum: \( L_f = I_{step} \omega' \)
Where \( I_{step} \) is the moment of inertia about the edge of the step, which can be calculated using the parallel axis theorem:
Moment of Inertia about the edge: \( I_{step} = I_{cm} + M \left( R \right)^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2 \)
Setting Up the Equation
By conservation of angular momentum, we have:
Equation: \( L_i = L_f \)
Substituting the expressions we derived:
\( \left( \frac{1}{2} M R^2 \cdot \frac{v_0}{R} + M v_0 \cdot \frac{R}{2} \right) = \left( \frac{3}{2} M R^2 \right) \omega' \)
Solving for Final Velocity
After simplifying and solving for \( \omega' \), we can find the new linear velocity \( v' \) using the relationship \( v' = R \omega' \). The calculations will yield:
\( v' = \frac{2}{3} v_0 \)
Final Result
Thus, the velocity of the disc just after the impact with the step is:
Final Velocity: \( v' = \frac{2}{3} v_0 \)
This result shows how the impact with the step reduces the linear velocity of the disc due to the change in motion dynamics as it pivots about the edge of the step.
