Problem 85 - University Physics with Modern Physics

Question

In a lab experiment you let a uniform ball roll down a curved track. The ball starts from rest and rolls without slipping.While on the track, the ball descends a vertical distance \(h .\) The lower end of the track is horizontal and extends over the edge of the lab table; the ball leaves the track traveling horizontally. While free-falling after leaving the track, the ball moves a horizontal distance \(x\) and a vertical distance \(y\) . (a) Calculate \(x\) in terms of \(h\) and \(y\) ignoring the work done by friction. (b) Would the answer to part (a) be any different on the moon? (c) Although you do the experiment very carefully, your measured value of \(x\) is consistently a bit smaller than the value calculated in part (a). Why? (d) What would \(x\) be for the same \(h\) and \(y\) as in part (a) if you let a silver dollar roll down the track? You can ignore the work done by friction.

Step-by-Step Solution

Answer

(a) \( x = \sqrt{\frac{20hy}{7}} \). (b) No, same on the moon. (c) Likely air resistance. (d) \( x = \sqrt{2hy} \) for a silver dollar.

Step 1: Understand the Scenario

We are considering a ball rolling down a curved track and then falling freely from a horizontal plane. We need to express the horizontal distance \( x \) (traveled by the ball) in terms of vertical fall \( y \) and initial drop height \( h \).

Step 2: Use Energy Conservation for Velocity Calculation

Since the ball rolls without slipping, we apply conservation of mechanical energy. The potential energy lost when descending height \( h \) is converted into translational and rotational kinetic energy. The equation is \( mgh = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 \). For a solid sphere, \( I = \frac{2}{5}mr^2 \) and \( v = r\omega \). Solving gives \( v = \sqrt{\frac{10}{7}gh} \).

Step 3: Determine Time of Free Fall

The time \( t \) it takes to fall a vertical distance \( y \) can be calculated using \( y = \frac{1}{2}gt^2 \). Solving for \( t \), we find \( t = \sqrt{\frac{2y}{g}} \).

Step 4: Calculate Horizontal Distance \( x \)

The horizontal velocity \( v_x = v = \sqrt{\frac{10}{7}gh} \). The horizontal distance \( x \) the ball travels in time \( t \) is \( x = v_x \cdot t = \sqrt{\frac{10}{7}gh} \cdot \sqrt{\frac{2y}{g}} = \sqrt{\frac{20hy}{7}} \).

Step 5: Consider the Effect on the Moon

On the Moon, acceleration due to gravity \( g' = \frac{1}{6}g \). Using \( x = \sqrt{\frac{20hy}{7}} \), the formula for \( x \) remains the same because the factor \( g \) cancels out in the expression for \( x \). Hence, \( x \) would be the same as on Earth.

Step 6: Address the Measurement Discrepancy

If measured \( x \) is consistently smaller, air resistance or slight energy losses (due to rolling friction or deformation) can be causes, even though friction is overlooked in the theoretical calculation.

Step 7: Consider a Silver Dollar Rolling Down

For a silver dollar, the moment of inertia is different. As a coin (disk), \( I = \frac{1}{2}mr^2 \). Solving \( mgh = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 \) with \( I = \frac{1}{2}mr^2 \) gives \( v = \sqrt{gh} \). Then \( x = v \cdot t = \sqrt{gh} \cdot \sqrt{\frac{2y}{g}} = \sqrt{2hy} \).

Key Concepts

Rolling Motion

Rolling motion is a fascinating concept in physics where an object moves forward while rotating around its axis. This dual motion combines translation and rotation, typical for objects like wheels, balls, and cylinders. In our scenario, a uniform ball rolls without slipping down a curved track.

When a ball rolls without slipping, there is a relationship between its linear velocity (\(v\)) and its angular velocity (\(\omega\)). This relationship can be expressed as \(v = r\omega\), where \(r\) is the radius of the ball. The ball's kinetic energy is split into translational kinetic energy and rotational kinetic energy, due to its rotational motion.

For a solid sphere like our ball, the moment of inertia \(I\) is given by \(I = \frac{2}{5}mr^2\). As the ball descends the height \(h\), its potential energy converts to these forms of kinetic energy. The total kinetic energy can be computed using: \(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\).

This energy conversion helps us determine that the speed \(v = \sqrt{\frac{10}{7}gh}\) as it leaves the track, ready to enter free fall.

Free Fall

Free fall describes the motion of an object under the influence of gravity alone, without any air resistance. In our experiment, once the ball leaves the horizontal end of the track, it enters free fall and descends vertically while also moving horizontally.

The key component of free fall is the acceleration due to gravity, denoted as \(g\). On Earth, \(g\) is approximately \(9.81 \,ms^{-2}\). In free fall, the only force acting on the ball is this gravitational force. The time it takes to fall a vertical distance \(y\) can be calculated using the equation:
\[y = \frac{1}{2}gt^2 \]
Solving for \(t\) yields \(t = \sqrt{\frac{2y}{g}}\), giving the duration of the fall.

This falling time, combined with the horizontal velocity from rolling motion, determines the horizontal distance \(x\) that the ball travels before hitting the ground. Importantly, whether on the Earth or moon, \(g\) cancels out when calculating \(x\), so \(x\) would be the same in both places.

Kinematics

Kinematics involves the study of motion without considering the forces that cause it. It's all about how objects move. In our experiment, we analyze the ball’s path from the track and during its fall.

During its fall, the ball follows a parabolic trajectory due to its initial horizontal velocity and the downward gravitational pull. The horizontal distance travelled by the ball is influenced by its initial velocity \(v_x\), which we found to be \(\sqrt{\frac{10}{7}gh}\).

To find how far the ball travels horizontally, we use the equation \(x = v_x \cdot t\), where \(t\) is the time derived from the free fall exploration. The resulting formula is \(x = \sqrt{\frac{20hy}{7}}\).

Any deviation in the measured value of \(x\) from the calculated one might be due to factors like air resistance or small energy losses from rolling friction, which are usually ignored in theoretical calculations.

Rotational Dynamics

Rotational dynamics studies the motion of objects that rotate and helps us understand the complexities of this type of movement. For the ball in our experiment, rotational dynamics plays a crucial role.

The primary aspect involved here is the moment of inertia \(I\), a measure of an object's resistance to changes in its rotation. For different shapes, \(I\) varies significantly. For a solid sphere, \(I = \frac{2}{5}mr^2\), while for a disk, \(I = \frac{1}{2}mr^2\).

When we apply conservation of energy to the rotating ball, we account for its rotational kinetic energy as it involves both the sphere’s mass and how that mass is distributed around its rotational axis.

If instead of a ball we used a silver dollar (a disk), its different moment of inertia would give a final velocity calculated as \(v = \sqrt{gh}\), leading to a different expression for the distance \(x\) compared to the sphere's own calculation.