A car has wheels each with a radius of 30 cm . It starts from rest and (without slipping) accel

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A car has wheels each with a radius of 30 cm . It starts...

A car has wheels each with a radius of $30 \mathrm{~cm} .$ It starts from rest and (without slipping) accelerates uniformly to a speed of $15 \mathrm{~m} / \mathrm{s}$ in a time of $8.0 \mathrm{~s}$. Find the angular acceleration of its wheels and the number of rotations one wheel makes in this time. Remember that the center of the rolling wheel accelerates tangentially at the same rate as does a point on its circumference. We know that $a_{T}=\left(v_{f}-v_{i}\right) / t$, and so $$a_{T}=\frac{15 \mathrm{~m} / \mathrm{s}}{8.0 \mathrm{~s}}=1.875 \mathrm{~m} / \mathrm{s}^{2}$$ Then $a_{T}=r \alpha$ yields $$\alpha=\frac{a_{T}}{r}=\frac{1.875 \mathrm{~m} / \mathrm{s}^{2}}{0.30 \mathrm{~m}}=6.2 \mathrm{rad} / \mathrm{s}^{2}$$ Notice that we must introduce the proper angular measure, radians. Now use $\theta=\omega_{i} t+\frac{1}{2} \alpha t^{2}$ to find $$\theta=0+\frac{1}{2}\left(6.2 \mathrm{rad} / \mathrm{s}^{2}\right)(8.0 \mathrm{~s})^{2}=200 \mathrm{rad}$$ and to get the corresponding number of turns divide by $2 \pi$, $$ (200 \mathrm{rad})\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right)=32 \mathrm{rev} $$

A car has wheels each with a radius of $30 \mathrm{~cm} .$ It starts from rest and (without slipping) accelerates uniformly to a speed of $15 \mathrm{~m} / \mathrm{s}$ in a time of $8.0 \mathrm{~s}$. Find the angular acceleration of its wheels and the number of rotations one wheel makes in this time.
Remember that the center of the rolling wheel accelerates tangentially at the same rate as does a point on its circumference. We know that $a_{T}=\left(v_{f}-v_{i}\right) / t$, and so
$$a_{T}=\frac{15 \mathrm{~m} / \mathrm{s}}{8.0 \mathrm{~s}}=1.875 \mathrm{~m} / \mathrm{s}^{2}$$
Then $a_{T}=r \alpha$ yields
$$\alpha=\frac{a_{T}}{r}=\frac{1.875 \mathrm{~m} / \mathrm{s}^{2}}{0.30 \mathrm{~m}}=6.2 \mathrm{rad} / \mathrm{s}^{2}$$
Notice that we must introduce the proper angular measure, radians.
Now use $\theta=\omega_{i} t+\frac{1}{2} \alpha t^{2}$ to find
$$\theta=0+\frac{1}{2}\left(6.2 \mathrm{rad} / \mathrm{s}^{2}\right)(8.0 \mathrm{~s})^{2}=200 \mathrm{rad}$$
and to get the corresponding number of turns divide by $2 \pi$,
$$
(200 \mathrm{rad})\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right)=32 \mathrm{rev}
$$

Key Concepts

Uniform Acceleration

Uniform acceleration refers to a constant rate of change of velocity over time. In both linear and rotational contexts, this concept allows for the use of standard kinematic equations to determine the change in position, velocity, or angle over time, assuming the acceleration remains constant.

Angular Acceleration

Angular acceleration is the rate at which an object’s angular velocity changes with time. It is a fundamental concept in rotational dynamics, playing a similar role to linear acceleration in translational motion, and is typically measured in radians per second squared.

Relationship Between Linear and Angular Quantities

In rotational motion, linear (tangential) acceleration and angular acceleration are linked by the radius of the rotation. This relationship, expressed by the formula a = r × ?, connects the rate of change of linear speed along the circumference of a circle to the rate of change of angular speed, under the no-slip condition.

Kinematic Equations for Rotational Motion

Kinematic equations for rotational motion serve to relate angular displacement, angular velocity, angular acceleration, and time, analogous to the equations used for linear motion. These equations are essential for solving problems where an object is experiencing constant angular acceleration.

Unit Conversion in Angular Measure

Unit conversion in angular measure involves switching between different units such as radians and revolutions. Recognizing that one full rotation is equal to 2? radians is crucial for correctly interpreting and converting angular displacements in various problems in rotational dynamics.

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