To solve the problem regarding the rolling cylinders, we need to address each part step-by-step.

Part (a)

When two objects roll without slipping down an incline, their acceleration depends on their moment of inertia. The moment of inertia determines how the mass is distributed, affecting how quickly they roll.

1. Solid Cylinder (Object A): Moment of inertia $I_A = \frac{1}{2} m r^2$
2. Hollow Cylinder (Object B): Moment of inertia $I_B = m r^2$

Since both cylinders have the same mass $m$ and radius $r$, the solid cylinder (Object A) will have a smaller moment of inertia compared to the hollow cylinder (Object B). Therefore, Object A will accelerate faster down the ramp.

Answer: 2) Object A

Part (b)

To find the speed of Object A at the bottom of the ramp, we can use energy conservation principles. The potential energy at the top converts to kinetic energy at the bottom.

1. Potential Energy (PE) at the top: $PE = mgh$ where:

   • $m = 0.270 \, \text{kg}$ (convert grams to kilograms)
   • $g = 9.81 \, \text{m/s}^2$
   • $h = 0.15 \, \text{m}$ (convert cm to meters)

2. Kinetic Energy (KE) at the bottom consists of translational and rotational energy: $KE = \frac{1}{2} mv^2 + \frac{1}{2} I_A \omega^2$ Since $v = r\omega$ for rolling without slipping, we can write: KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right \frac{v^2}{r^2} Simplifying gives: $KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2$

Setting $PE = KE$: $mgh = \frac{3}{4} mv^2$

Canceling $m$ from both sides and solving for $v$: $gh = \frac{3}{4} v^2 \implies v^2 = \frac{4gh}{3} \implies v = \sqrt{\frac{4gh}{3}}$

Substituting in values: $v_A = \sqrt{\frac{4 \cdot 9.81 \cdot 0.15}{3}}$

Calculating: $v_A = \sqrt{\frac{4 \cdot 9.81 \cdot 0.15}{3}} \approx \sqrt{1.96} \approx 1.4 \, \text{m/s}$

Answer: $v_A \approx 1.4 \, \text{m/s}$

Part (c)

For Object B (hollow cylinder), the process is similar but the moment of inertia changes.

1. The moment of inertia for the hollow cylinder: $I_B = m r^2$

Thus, the kinetic energy becomes: $KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2$

Setting $PE = KE$: $mgh = mv^2 \implies v^2 = gh \implies v_B = \sqrt{gh}$

Substituting in values: $v_B = \sqrt{9.81 \cdot 0.15} \approx \sqrt{1.4715} \approx 1.21 \, \text{m/s}$

Answer: $v_B \approx 1.21 \, \text{m/s}$

Summary of Answers

• Part (a): Object A
• Part (b): $v_A \approx 1.4 \, \text{m/s}$
• Part (c): $v_B \approx 1.21 \, \text{m/s}$

Do you have any questions or need further details on any part? Here are some related questions you might consider:

1. How does the shape of an object influence its rolling motion?
2. What factors affect the moment of inertia of an object?
3. How would the results change if the ramp was frictionless?
4. What is the importance of rolling without slipping in this context?
5. How would the velocities change if the ramp's height were increased?