91Ó°ÊÓ

Angular position is a key concept in rotational motion, which describes the orientation of an object as it rotates around a fixed axis. It is similar to "distance" in linear motion but in a circular path.
In this exercise, the angular position \( \theta \) is given as a function of time \( t \), specifically \( \theta = 0.40 e^{2t} \). This format signifies that as time goes on, the object’s position along its circular path changes at an exponential rate.

• \( \theta \) is commonly measured in radians, an angle unit where one full revolution is \( 2\pi \) radians.
• The formula tells us how fast or slow the object rotates; for instance, an increasing value of \( \theta \) indicates the object is rotating and its position on the circle is changing.

Understanding angular position is vital because it lays the foundation for calculating further properties like angular velocity \( \omega \) and angular acceleration \( \alpha \). Both are directly derived from \( \theta \) by differentiating it with respect to time.

Tangential acceleration refers to how quickly a point on a rotating object speeds up or slows down along its circular path. It is analogous to linear acceleration but occurs on a circular path. In this problem, our job is to determine the tangential acceleration at time \( t = 0 \) for a point located \( 0.04 \) meters from the axis.
The formula for tangential acceleration \( a_t \) is:
\[ a_t = r \cdot \alpha \]
where \( r \) is the radius, the distance from the axis, and \( \alpha \) is the angular acceleration.

• Angular acceleration \( \alpha \) was found by differentiating the angular velocity \( \omega = 0.80 e^{2t} \), giving \( 1.60 e^{2t} \).
• At \( t = 0 \), \( \alpha(0) = 1.6 \, \text{rad/s}^2 \).
• Substitute \( r = 0.04 \) meters and \( \alpha(0) = 1.60 \text{ rad/s}^2 \) into the formula, resulting in \( a_t = 0.064 \text{ m/s}^2 \).

This value represents how the speed of the point in its path changes as the object rotates.

Radial acceleration, commonly known as centripetal acceleration, describes the acceleration that keeps an object moving in a circular path by constantly changing its direction. This acceleration is always directed towards the center of the rotation. In this exercise, we find it for a point at the same 0.04 meters from the rotation axis at time \( t = 0 \).
The formula to compute radial acceleration \( a_r \) is:
\[ a_r = r \cdot \omega^2 \]
where \( \omega \) is the angular velocity, in radians per second.

• We previously calculated the angular velocity \( \omega = 0.80 e^{2t} \) and found that at \( t = 0 \), \( \omega(0) = 0.80 \, \text{rad/s} \).
• Substituting \( r = 0.04 \text{ m} \) and \( \omega^2 = (0.80 \text{ rad/s})^2 = 0.64 \text{ rad/s}^2 \) into the formula results in \( a_r = 0.0256 \text{ m/s}^2 \).

This radial acceleration value means that there is a constant pull directed inward to maintain circular motion.