91Ó°ÊÓ

When we talk about tangential acceleration, we're referring to the linear acceleration experienced by a point on a rotating object, like a gyroscope wheel, along the direction of rotation. Tangential acceleration is a key concept in understanding how quickly an object speeds up its rotation. The formula for calculating tangential acceleration is

• \(a_t = r \cdot \alpha\)

Here, \(a_t\) is the tangential acceleration, \(r\) is the radius, and \(\alpha\) is the angular acceleration.
In this exercise, the gyroscope flywheel has a radius \(r\) of 2.83 cm (or 0.0283 m in meters), and it's given an angular acceleration of 14.2 rad/s². Thus, each moment it spins faster, every point on the edge of the flywheel becomes quicker too.
Calculating using these values:

• \(a_t = 0.0283 \times 14.2 = 0.40186 \; \text{m/s}^2\)

This tells us that every second, the point on the rim speeds up by 0.40186 meters per second squared, which might seem modest, but it's crucial for the efficient functioning of devices like gyroscopes.

Radial acceleration, also known as centripetal acceleration, keeps the points on the gyroscope's rim moving in a curved path rather than shooting off in a straight line.
This acceleration always points towards the center of the object's rotation.
The formula to calculate it is:

• \(a_r = r \cdot \omega^2\)

where \(a_r\) is the radial acceleration, \(r\) is the radius, and \(\omega\) is the angular velocity.
In our case, at the full angular velocity of 289 rad/s (after converting from the original 2760 rpm), and with the radius as 0.0283 m, the radial acceleration is:

• \(a_r = 0.0283 \times (289)^2 = 2360.61 \; \text{m/s}^2\)

This large value indicates the immense force needed to keep the rim of the spinning gyroscope on its circular path. It's essential for preventing objects from flying outward due to inertia during high-speed rotation.

Angular displacement describes the total angle through which a point on the gyroscope rim has traveled over the course of its spinning up from rest. It's a measure of how much rotation has happened.
Angular displacement is crucial in determining the distance a point on the rim has traveled. To find it, you'll typically use this formula:

• \(\omega_f^2 = \omega_0^2 + 2\alpha\theta\)

where \(\omega_f\) is the final angular speed, \(\omega_0\) is the initial angular speed (which is zero if starting from rest), \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement.
From the exercise:

• With an initial angular speed of 0 and a final speed of 289 rad/s, the angular displacement is:
• \(\theta = \frac{289^2}{2 \times 14.2} = 2945.14 \; \text{rad}\)

To find the linear distance traveled by the rim's point, the formula used is:

• \(s = r\theta\)

where \(s\) is the distance.
Plugging in the numbers:

• \(s = 0.0283 \times 2945.14 = 83.3345 \; \text{m}\)

This means, during its speed increase, a point on the rim travels approximately 83.33 meters, illustrating a considerable motion for such a compact device.