Question

An automobile traveling at \(97 \mathrm{~km} / \mathrm{h}\) has wheels of diameter \(76 \mathrm{~cm} .(a)\) Find the angular speed of the wheels about the axle. \((b)\) The car is brought to a stop uniformly in 30 turns of the wheels. Calculate the angular acceleration. \((c)\) How far does the car advance during this braking period?

Short answer

The angular speed of the wheels is \(709.06 \, \text{rad/s}\). The angular acceleration is \(-3.76 \, \text{rad/s}^2\). The car travels about \(71.12 \, \text{m}\) during the stopping period.

Step-by-step solution

Find the angular speed

The linear speed \(v\) can be converted to angular speed \(\omega\) through the relation: \(\omega = \frac{v}{r}\), where \(r\) is the radius of the wheel. Convert speed to cm/s: \(v = 97 \, \text{km/h} * \frac{1000 \, \text{m}}{1 \, \text{km}} * \frac{100 \, \text{cm}}{1 \, \text{m}} * \frac{1 \, \text{h}}{3600 \, \text{s}} = 26944.44 \, \text{cm/s}\). Afterward, take \(r = \frac{76 \, \text{cm}}{2} = 38 \, \text{cm}\). Substituting the values, we have: \(\omega = \frac{26944.44 \, \text{cm/s}}{38 \, \text{cm}} = 709.06 \, \text{rad/s}\).

Calculate the angular acceleration

Given that the car came to a stop uniformly in 30 turns of the wheels, we can use \(\alpha = \frac{-\omega}{\theta}\) to find the angular acceleration \(\alpha\), where \(\theta\) is the total angle in radians (which is \(30\) turns \(* 2\pi\)). Hence, \(\theta = 30 * 2\pi = 60\pi \, \text{rad}\). Substituting the values, we have: \(\alpha = \frac{-709.06 \, \text{rad/s}}{60\pi \, \text{rad}} = -3.76 \, \text{rad/s}^2\).

Find the distance covered during braking

Finally, the distance \(d\) covered during this period can be calculated from the relation: \(d = r \times \theta\). Substituting the given radius and calculated theta: \(d = 38 \, \text{cm} * 60\pi \, \text{rad} = 7111.8 \, \text{cm or} \, 71.12 \, \text{m}\).

Key concepts

Angular Speed

In circular motion, angular speed, denoted by \(\omega\), describes how fast an object rotates around a fixed point. It's analogous to linear speed, but instead of covering distance over time, it measures how much of an angle is covered per unit of time. This is expressed in radians per second (rad/s). To find angular speed, we utilize the relationship between linear speed (\(v\)) and radius (\(r\)):
\[ \omega = \frac{v}{r} \]
In our exercise, the car travels at 97 km/h. We convert this to 26,944.44 cm/s to make calculations consistent with the wheel's diameter of 76 cm. With a radius of 38 cm (half the diameter), we calculate:
\[ \omega = \frac{26944.44 \, \text{cm/s}}{38 \, \text{cm}} = 709.06 \, \text{rad/s} \]
This means the wheels complete approximately 709 radians in one second as the car travels.

Angular Acceleration

Angular acceleration (\(\alpha\)) measures how the angular speed changes with time. It's critical for understanding how quickly an object speeds up or slows down its rotation. The unit of angular acceleration is radians per second squared (\(\text{rad/s}^2\)). When a vehicle slows down uniformly, as in our exercise, it can be captured by the formula:
\[ \alpha = \frac{-\omega}{\theta} \]
where \(\theta\) is the total angular displacement in radians. The car stops in 30 rotations, or \(30 \times 2\pi\), which is equal to \(60\pi\) radians. With an initial angular speed of 709.06 rad/s, we calculate:
\[ \alpha = \frac{-709.06 \, \text{rad/s}}{60\pi \, \text{rad}} = -3.76 \, \text{rad/s}^2 \]
The negative sign indicates deceleration, as expected when braking.

Linear and Angular Relationship

The relationship between linear and angular quantities is foundational in rotational kinematics. Linear distance, speed, and acceleration have their angular counterparts (angle, angular speed, angular acceleration). This allows us to translate motion descriptions in linear terms into rotational terms and vice versa.

• Linear distance \(d\) is related to angular displacement \(\theta\) by the formula:
  \[ d = r \times \theta \]
• For our problem, the car's wheels cover a distance during the braking period. Using the calculated \(\theta = 60\pi\) radians, and \(r = 38\) cm, we have:
  \[ d = 38 \, \text{cm} \times 60\pi \, \text{rad} = 7111.8 \, \text{cm} \]
• Converting to meters: 71.12 m.
  This showcases how angular motion translates into linear advance, helping to understand the car's travel during wheel rotations.