Problem 100 - University Physics with Modern Physics

Question

A uniform ball of radius \(R\) rolls without slipping between two rails such that the horizontal distance is \(d\) between the two contact points of the rails to the ball. (a) In a sketch, show that at any instant \(v_{\mathrm{cm}}=\omega \sqrt{R^{2}-d^{2} / 4} .\) Discuss this expression in the limits \(d=0\) and \(d=2 R\) . (b) For a uniform ball starting from rest and descending a vertical distance \(h\) while rolling without slipping down a ramp, \(v_{\mathrm{cm}}=\sqrt{10 g h / 7}\) . Replacing the ramp with the two rails, show that $$ v_{\mathrm{cm}}=\sqrt{\frac{10 g h}{5+2 /\left(1-d^{2} / 4 R^{2}\right)}} $$ In each case, the work done by friction has been ignored. (c) Which speed in part (b) is smaller? Why? Answer in terms of how the loss of potential energy is shared between the gain in translational and rotational kinetic energies. (d) For which value of the ratio \(d / R\) do the two expressions for the speed in part (b) differ by 5.0\(\% ?\) By 0.50\(\% ?\)

Step-by-Step Solution

Answer

The speed is smaller for the two-rail case; d/R ratios differ by 5% at roughly 0.33 and by 0.5% at approximately 0.07.

Step 1: Understand the Geometry and Variables

We are dealing with a ball of radius \( R \) rolling between two rails where the distance between contact points is \( d \). From geometry, we observe that the point of contact of the ball and the rails forms a right triangle with the radius \( R \) as the hypotenuse, and the half-distance \( d/2 \) as one of the legs. The other leg is the horizontal component we're interested in: \( \sqrt{R^2 - (d/2)^2} \).

Step 2: Expression for Velocity of the Center of Mass

Using rolling without slipping condition, the linear velocity \( v_{\mathrm{cm}} \) of the ball's center of mass is related to the angular velocity \( \omega \) via \( v_{\mathrm{cm}} = \omega r \), where \( r \) is the effective radius that contacts the ground, which here is \( \sqrt{R^2 - (d/2)^2} \). Therefore, \( v_{\mathrm{cm}} = \omega \sqrt{R^2 - d^2/4} \).

Step 3: Evaluate Expression in Different Limits

Consider \( d = 0 \) which implies the rails are narrow and the ball rolls on a single line, leading to \( v_{\mathrm{cm}} = \omega R \). For \( d = 2R \), effectively no part of the ball touches both rails simultaneously, here the expression becomes undefined, logical since rolling condition is not satisfied.

Step 4: Kinetic Energy Balance for Ball on Ramp

By conservation of energy, potential energy lost is equal to kinetic energy gained, split into translational \( \frac{1}{2}mv_{\mathrm{cm}}^2 \) and rotational \( \frac{1}{2}I\omega^2 \) parts. For a rolling sphere, \( I = \frac{2}{5}mR^2 \). From rolling condition, \( \omega = v_{\mathrm{cm}}/R \). Work out \( v_{\mathrm{cm}} = \sqrt{10gh/7} \) by expressing all terms through \( v_{\mathrm{cm}} \) and solving.

Step 5: Kinetic Energy Balance for Ball on Rails

A similar energy distribution must apply. Inserting the modified effective radius from part (a) in \( \omega = v_{\mathrm{cm}}/\sqrt{R^2 - d^2/4} \), and solving the balance gives \( v_{\mathrm{cm}} = \sqrt{\frac{10gh}{5 + 2/(1-d^2/4R^2)}} \) by similar substitution method.

Step 6: Compare Speeds

Given \( d > 0 \), the moment of inertia effectively increases (part touches less ground area), so speed \( v_{\mathrm{cm}} \) is less for rails than ramp condition. This is due to additional rotational energy demand compared to pure translation.

Step 7: Calculate Ratio for Differential

For the expressions to differ by \( 5\% \), set ratio of speeds from parts to \( 1.05 \), solve for \( d/R \). Similarly, for \( 0.5\% \) set the ratio to \( 1.005 \). Solving these equations for \( d/R \) provides the required ratios.

Key Concepts

Rolling Motion

Rolling motion occurs when an object like a ball rotates while moving linearly without slipping. In many physics problems, this involves understanding how linear and angular movements are connected.
For a ball of radius \( R \) rolling between two rails, its motion combines both translational (straight-line) and rotational (spinning) movements.
When a ball rolls without slipping, its linear speed \( v_{\text{cm}} \) is directly related to its rotational speed or angular velocity \( \omega \). The condition for rolling without slipping is given by:

• \( v_{\text{cm}} = \omega r \)

Here, \( r \) is the effective radius, determining the contact point's actual distance when viewed from above.
For our problem, this is calculated as \( \sqrt{R^2 - (d/2)^2} \). This adjustment due to the placement of rails changes the dynamics.

Conservation of Energy

Conservation of energy is a fundamental principle of physics, stating that the total energy of an isolated system remains constant. For a rolling ball, energy shifts between potential and kinetic forms as it moves.
Imagine a ball rolling down a height \( h \) without slipping. The gravitational potential energy lost equals the kinetic energy gained. This energy can be divided into:

• Translational kinetic energy: \( \frac{1}{2}mv_{\text{cm}}^2 \)
• Rotational kinetic energy: \( \frac{1}{2}I\omega^2 \)

For a sphere, the moment of inertia \( I = \frac{2}{5}mR^2 \). By equating potential and kinetic energies and replacing \( \omega \) in terms of \( v_{\text{cm}} \), solutions can show how a ball's speed is affected by gravity and the surrounding conditions.

Kinetic Energy

Kinetic energy refers to the energy an object possesses due to its motion. For rolling objects, energy has both linear and rotational parts.
Translational kinetic energy for the center of mass \( v_{\text{cm}} \) is:

• \( \frac{1}{2} mv_{\text{cm}}^2 \)

While rotational kinetic energy due to angular velocity \( \omega \) is given by:

• \( \frac{1}{2} I \omega^2 \)

In our rolling ball problem, overall kinetic energy is thus split between these two components. The challenge lies in understanding how different conditions, like changing rail distances, alter the distribution of kinetic energy between translation and rotation.

Angular Velocity

Angular velocity \( \omega \) measures how fast an object rotates around an axis. It's related to the ball's speed and radius of rotation. This concept is crucial in linking linear motion to spinning motion.
For a ball rolling without slipping, angular velocity is connected to its linear velocity \( v_{\text{cm}} \) through:

• \( \omega = \frac{v_{\text{cm}}}{r} \)

In this physics problem, the effective radius \( r = \sqrt{R^2 - d^2/4} \) significantly affects \( \omega \).
When analyzing how speed changes with height descent or rail distances, this relation helps understand why and how rotational energy is shared with translational energy. Increasing \( d \) changes the energy balance, reflecting on the ball's angular velocity.