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An automobile travels to the left at a constant speed of \(72 \mathrm{km} / \mathrm{h}\). Knowing that the diameter of the wheel is \(560 \mathrm{mm}\), determine the acceleration (a) of point \(B,(b)\) of point \(C,(c)\) of point \(D .\)

Short Answer

Expert verified
(a) 0 m/s², (b) 0 m/s², (c) 1428.57 m/s².

Step by step solution

01

Convert the Speed to Meters Per Second

The car's speed is given as 72 km/h. To convert this to meters per second (m/s), use the conversion factor: 1 km/h = 0.27778 m/s.So, the speed in m/s is:\[72 \times 0.27778 = 20 \, \text{m/s}\]
02

Determine Angular Velocity

To find the angular velocity \( \omega \) of the wheel, use the formula:\[\omega = \frac{v}{r}\]where \(v = 20 \, \text{m/s}\) is the speed of the car and \(r\) is the radius of the wheel. The radius is half the diameter:\[ r = \frac{560 \, \text{mm}}{2} = 280 \, \text{mm} = 0.28 \, \text{m}\]Therefore, the angular velocity is:\[\omega = \frac{20}{0.28} \approx 71.43 \, \text{rad/s}\]
03

Determine Acceleration of Point B

Point B is the point of contact between the wheel and the ground. For a wheel rolling without slipping, point B has zero velocity relative to the ground. Thus, its acceleration is zero:\[a_B = 0 \, \text{m/s}^2\]
04

Determine Acceleration of Point C (Center of the Wheel)

The acceleration of point C, the center of the wheel, is directed horizontally. Since it is moving at a constant speed, its linear acceleration is zero:\[a_C = 0 \, \text{m/s}^2\]
05

Determine Acceleration of Point D (Top of the Wheel)

Point D, located at the top of the wheel, experiences both the tangential acceleration (which is zero as it moves at constant speed) and the centripetal acceleration directed towards the center:\[a = r\omega^2\]Substitute known values:\[a_D = 0.28 \times (71.43)^2 \approx 1428.57 \, \text{m/s}^2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Acceleration
Centripetal acceleration is a crucial concept when dealing with objects moving in a circular path. It's that invisible force keeping an object moving in a curved trajectory. Think of it as the glue that holds the object's path together rather than letting it fly off in a straight line.
The formula for centripetal acceleration is given by:
  • \[ a_c = r\omega^2 \]
  • Where \( r \) is the radius of the circle and \( \omega \) is the angular velocity.
In the context of the exercise, when calculating the acceleration of point D at the top of the wheel, we use the centripetal acceleration to account for the change in direction. Despite moving at a constant speed, a change in direction necessitates an acceleration inwards, towards the center of the wheel's rotation.
This leads to high values, such as the noted \( 1428.57 \, \text{m/s}^2 \), illustrating how something moving rapidly in a circle can experience significant accelerative forces without a change in speed. Understanding centripetal acceleration is vital in analyzing circular motions such as the dynamics of wheels and planetary orbits.
Wheel Dynamics
Wheel dynamics involves understanding the forces and motions acting on a rolling wheel. This concept explains how a wheel behaves as it travels and encounters various conditions.
  • Rotation: Important to note is the wheel's rotation around its axis, where each point on the wheel follows a circular path.
  • Contact points: For a rolling wheel, the point of contact with the ground (point B) is briefly stationary relative to the ground.
In this exercise, two significant implications arise:
  • First, point B has zero velocity and acceleration in context to the road's surface.
  • Secondly, the wheel moves forward without slipping, maintaining a harmony between its rotational speed and linear speed.
Wheel dynamics involves complex interactions between linear and angular components, crucial for explaining behaviors like traction, braking, and turning. Students must grasp these principles to analyze practical situations where wheel behavior directly impacts performance and efficiency.
Rolling Motion
Rolling motion combines both translational (straight line) and rotational motion. When a wheel rolls on a surface, every point on its rim follows a trochoid path, which consists of loops or cusps at the point of contact.
  • A critical feature is no relative motion between the wheel and the surface at the point of contact, allowing for grip and traction without slipping.
  • The wheel's center moves forward with a linear velocity equal to the wheel's outer surface speed.
This problem demonstrates how, even in constant-speed motion, there's complex interrelation between different points on the wheel. Point C (center) experiences zero acceleration since its speed is constant.
Understanding rolling motion helps in analyzing vehicles, where the interaction between tires and the road surface affects movement. This knowledge aids in designing systems, from basic bicycles to advanced vehicles, ensuring stability, efficiency, and safety. Rolling motion is foundational to not just physics problems but mechanical engineering and robotics as well.

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