Problem 7 - College Physics

Question

A bocce ball with a diameter of \(6.00 \mathrm{~cm}\) rolls without slipping on a level lawn. It has an initial angular speed of \(2.35 \mathrm{rad} / \mathrm{s}\) and comes to rest after \(2.50 \mathrm{~m}\). Assuming constant deceleration, determine (a) the magnitude of its angular deceleration and (b) the magnitude of the maximum tangential acceleration of the ball's surface (tell where that part is located)

Step-by-Step Solution

Answer

(a) \(-0.033\, \mathrm{rad/s^2}\), (b) \(0.00099\, \mathrm{m/s^2}\) at the point of contact with the ground.

Step 1: Identify the Given Quantities

We are given the diameter of the bocce ball, which is \(6.00\, \mathrm{cm}\) and converted to radius \(r = 3.00\, \mathrm{cm} = 0.030\, \mathrm{m}\). The initial angular speed is \(\omega_i = 2.35\, \mathrm{rad/s}\) and the distance traveled is \(s = 2.50\, \mathrm{m}\). We need to find the angular deceleration (\(\alpha\)) and the maximum tangential acceleration of the ball's surface.

Step 2: Relate Linear and Angular Quantities

The relationship between linear distance \(s\) and angular displacement \(\theta\) is given by \(s = r \theta\). Thus, \(\theta = \frac{s}{r}\). Substitute the known values, \(\theta = \frac{2.50}{0.030}\, \mathrm{rad}\).

Step 3: Calculate Angular Displacement

Calculate \(\theta\):\[ \theta = \frac{2.50}{0.030} = 83.33\, \mathrm{rad} \]

Step 4: Use Angular Motion Equation

Using the equation \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\), where \(\omega_f = 0\, \mathrm{rad/s}\) (since the ball comes to rest) to find \(\alpha\):\[ 0 = (2.35)^2 + 2 \alpha \times 83.33 \]

Step 5: Solve for Angular Deceleration

Solving the equation:\[ 0 = 5.5225 + 166.66 \alpha \] gives\[ \alpha = -\frac{5.5225}{166.66} = -0.033\, \mathrm{rad/s^2} \]The negative sign indicates deceleration.

Step 6: Find Maximum Tangential Acceleration

The maximum tangential acceleration \(a_t\) at the surface is given by \(a_t = r\alpha\). Substituting the known values:\[ a_t = 0.030 \times 0.033 = 0.00099\, \mathrm{m/s^2} \]

Step 7: Describe Location of Maximum Tangential Acceleration

The maximum tangential acceleration occurs at the point on the ball's surface that is instantaneously in contact with the ground, where the linear velocity is due solely to the angular deceleration.

Key Concepts

Understanding Angular Deceleration

Angular deceleration is the rate at which an object slows down its rotation. It's essentially the opposite of angular acceleration. When a bocce ball rolls to a stop, its angular deceleration indicates how quickly it reduces its rotational speed to zero. In our exercise, the bocce ball had an initial angular speed of 2.35 rad/s and came to a halt over a distance of 2.50 meters. The angular deceleration can be found using the angular motion equation:\[0 = \omega_i^2 + 2\alpha\theta\]where:

• \(\omega_i\) is the initial angular speed (2.35 rad/s)
• \(\alpha\) is the angular deceleration
• \(\theta\) is the angular displacement (83.33 rad)

Through solving, we find that the angular deceleration \(\alpha\) is \(-0.033\, \mathrm{rad/s^2}\). The negative sign indicates that the ball is slowing down, not speeding up. This deceleration is constant throughout the rolling process, making the ball gradually slow.Understanding this concept is crucial for interpreting how rotating objects behave when they decelerate. It forms a basis for broader physics concepts applicable in areas like automotive wheel movement and even in planetary rotations.

Introduction to Tangential Acceleration

Tangential acceleration refers to how quickly the points on a rotating object's surface speed up or slow down as it rotates, measured along the tangent to the path. Simply put, it tells us how the object accelerates along a curved path. For the bocce ball in our exercise, the tangential acceleration was found where the surface contacts the ground.Here, we use the formula:\[a_t = r\alpha\]where:

• \(r\) is the radius of the ball (0.030 m)
• \(\alpha\) is the previously calculated angular deceleration (\(-0.033\, \mathrm{rad/s^2}\)).

Solving gives us a maximum tangential acceleration \(a_t\) of approximately \(0.00099\, \mathrm{m/s^2}\). This acceleration occurs at the edge of the ball, where the linear and angular motions combine. Tangential acceleration plays a critical role in applications such as roller coasters, where it governs the speed changes as cars navigate curved tracks, and in vehicle dynamics on curved roads.

Angular Displacement Simplified

Angular displacement represents the angle through which an object rotates during its motion. It's a crucial concept that links linear and rotational motions. In our scenario, it's determined by how far the bocce ball rolls, which is connected to the ball's angular movement over the lawn.To find the angular displacement \(\theta\), you use the relationship between the linear distance covered \(s\) and radius \(r\):\[\theta = \frac{s}{r}\]Given in the exercise:

• \(s = 2.50\, \mathrm{m}\) (distance travelled)
• \(r = 0.030\, \mathrm{m}\) (radius of the ball)

Solving gives \(\theta = 83.33\, \mathrm{rad}\). This translates the linear motion into angular terms, showing how much the ball has rotated. Angular displacement is integral in fields such as engineering and robotics, where the conversion between linear and rotational measurements is common and crucial for designing functional systems.