Question

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius 0.260 m; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through 20.0 rev, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height $h$ above the center of the wheel. If the wheel rotates with a constant angular acceleration $\alpha$, what value of a is required for the marble to reach a height of $h=$ 12.0 m?

Short answer

The required angular acceleration is approximately 13.85 rad/s².

Step-by-step solution

Understanding the Problem

The problem involves a marble shot vertically after being in a cup on a rotating wheel. The wheel rotates with a constant angular acceleration, and we need to calculate this acceleration such that the marble reaches a maximum height of 12.0 m after being released.

Calculate the Final Angular Velocity

First, we must determine the final angular velocity when the marble is released. The wheel starts from rest, so the initial angular velocity ${\omega }_0=0$. The wheel makes 20.0 revolutions before the marble is released. We know the angular displacement $\theta$ is $20.0\times 2\pi$ radians. Using the equation ${\omega }^2={\omega }_0^2+2\alpha \theta$, we solve for $\omega$: $$\omega =\sqrt{2\alpha \theta }$$

Relate Angular and Linear Quantities

The velocity of the marble when it leaves the cup is given by the linear velocity at the rim of the wheel, which is $v=r\omega$, where $r=0.260$ m is the radius of the wheel. Substituting $\omega$ from Step 1, we get: $$v=r\sqrt{2\alpha \theta }$$

Calculate the Required Linear Speed for Desired Height

To find the required final linear speed to reach the height of 12.0 m, use the kinematic equation for vertical motion: $$v^2=2gh$$where $g=9.8$ m/s² is the acceleration due to gravity, and $h=12.0$ m. Solving for $v$, we get: $$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 12.0}=15.34\text{ m/s}$$

Solve for Angular Acceleration $\alpha$

Use the expression for $v$ from Step 2 and set it equal to the required speed from Step 3: $$r\sqrt{2\alpha \theta }=15.34$$$$0.260\sqrt{2\alpha \times 20.0\times 2\pi }=15.34$$Solve the above equation for $\alpha$: $$\sqrt{2\alpha \times 125.6}=59$$$$2\alpha \times 125.6=3481$$$$\alpha =\frac{3481}{251.2}\approx 13.85{\text{ rad/s}}^2$$

Conclusion

The required constant angular acceleration that the wheel must have so that the marble reaches a maximum height of 12.0 meters above the center is approximately 13.85 rad/s².

Key concepts

Angular Acceleration

Angular acceleration is a measure of how quickly the angular velocity of a rotating object changes. It is denoted by the symbol $\alpha$ and is typically measured in radians per second squared ${\text{rad/s}}^2$. In the context of the exercise, when the wheel starts from rest and increases its rotational speed to launch the marble, it undergoes angular acceleration.
In mathematical terms, if a wheel starts from rest, its initial angular velocity ${\omega }_0$ is zero, and its angular displacement over a period is measured in radians. The equation ${\omega }^2={\omega }_0^2+2\alpha \theta$ relates the wheel's angular velocity $\omega$, initial angular velocity ${\omega }_0$, angular acceleration $\alpha$, and angular displacement $\theta$.
Finding the angular acceleration is crucial for determining how fast the wheel needs to spin to provide the linear velocity necessary for the marble to reach a specific height.

Kinematics

Kinematics refers to the branch of mechanics that deals with the motion of objects without considering the forces causing the motion. It focuses on relationships between displacement, velocity, acceleration, and time.
In the exercise, kinematics is applied to find the velocity needed for the marble to reach a height of 12 meters after being launched vertically. The kinematic equation used is $v^2=2gh$, where $v$ is the linear velocity, $g$ is the acceleration due to gravity (9.8 m/s²), and $h$ is the maximum height.
This formula helps determine the required initial speed of the marble once it is launched, allowing the calculation of the necessary angular acceleration. Understanding kinematics is key for solving problems involving motion in a linear path and predicting an object's position as a function of time.

Linear Velocity

Linear velocity is the rate at which an object moves along a path in a straight line. It is usually measured in meters per second (m/s). For a rotating object like the wheel in the problem, the linear velocity at the rim can be found using the equation $v=r\omega$, where $r$ is the radius and $\omega$ is the angular velocity.
In our scenario, once the angular velocity is known, calculations allow us to determine the linear speed at which the marble leaves the cup. This linear speed is what propels the marble upward. It's essential to ensure this speed is sufficient to reach the desired height of 12 meters.
Thus, converting angular motion into linear velocity is a critical step in ensuring the device works as intended, translating the rotational motion of the wheel into the vertical motion of the marble.

Vertical Motion

Vertical motion describes the movement of an object in a vertical direction, affected primarily by gravitational forces. In this problem, once the marble is released from the rotating wheel, it travels straight upward until it reaches its peak height of 12 meters.
The primary force acting on the marble during its ascent is gravity, pulling it downward, which is accounted for with the acceleration due to gravity $g=9.8{\text{ m/s}}^2$.
Understanding vertical motion helps to determine the necessary initial speed of the marble (its linear velocity) for it to reach the given height of 12 meters starting from the point of release. Calculating this motion involves using kinematic formulas to account for the initial velocity, height, and gravitational pull. This knowledge is critical for solutions involving the upward trajectory of objects.