Problem 63 - College Physics

Question

\(\bullet\) A vacuum cleaner belt is looped over a shaft of radius 0.45 \(\mathrm{cm}\) and a wheel of radius 2.00 \(\mathrm{cm} .\) The motor turns the shaft at 60.0 \(\mathrm{rev} / \mathrm{s}\) and the moving belt turns the wheel, which in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed (see Figure 9.32 ). Assume that the belt doesn't slip on either the shaft or the wheel. (a) What is the speed of a point on the belt? (b) What is the angular velocity of the wheel, in rad/s?

Step-by-Step Solution

Answer

(a) 1.7 m/s, (b) 85 rad/s.

Step 1: Find the Linear Speed of the Shaft

The linear speed \( v \) of a point on the belt is the same as the linear speed of the shaft because the belt does not slip. The linear speed can be calculated using the formula \( v = r \cdot \omega \), where \( r \) is the radius of the shaft and \( \omega \) is the angular velocity of the shaft. Here, the radius \( r = 0.45 \) cm or 0.0045 m, and \( \omega = 60.0 \) revolutions per second or \( 60 \times 2\pi \) rad/s. Therefore:\[v = 0.0045 \times 60 \times 2\pi = 1.7 \space \text{m/s}\]

Step 2: Calculate the Angular Velocity of the Wheel

Since the belt doesn't slip, the linear speed of the belt at the wheel is the same as the linear speed of the shaft. Therefore, the linear speed \( v = 1.7 \) m/s is also the speed for the point on the wheel. Using the same formula \( v = r \cdot \omega' \), solve for \( \omega' \), the angular velocity of the wheel, where \( r = 2 \) cm or 0.02 m:\[1.7 = 0.02 \times \omega'\]Solving for \( \omega' \):\[\omega' = \frac{1.7}{0.02} = 85 \space \text{rad/s}\]

Key Concepts

Angular Velocity

In the context of rotational dynamics, angular velocity is a measure of how fast an object is rotating. Essentially, angular velocity describes the rate of change of angular displacement and is typically measured in radians per second (rad/s). It is comparable to linear velocity, but instead of describing straight-line motion, it describes rotational motion.
When calculating angular velocity, we often use the formula:

• \[ \omega = \frac{\theta}{t} \]

where \( \omega \) represents angular velocity, \( \theta \) is the angle of rotation in radians, and \( t \) is time in seconds.
In systems involving components like belts and pulleys, calculating angular velocity is crucial for understanding how motion is transferred.
The angular velocity of a shaft or a wheel depends on its rotational speed. In the exercise above, the shaft rotates at \( 60 \space \text{rev/s} \) which converts to an angular velocity of \( 60 \times 2\pi \space \text{rad/s} \). The belt moves at the same linear speed as the edge of the shaft, enabling us to tie its motion to the wheel's angular velocity through the relation \( v = r \cdot \omega \).
This understanding allows us to find the speed of rotating parts even in complex systems.

Linear Speed

Linear speed, also known as tangential speed, refers to how fast a point on the edge of a rotating object is moving along its circular path. Linear speed is particularly important in systems where rotational motion is translated into linear motion, such as belts and pulleys.
The formula to calculate linear speed is given by:

• \[ v = r \cdot \omega \]

where \( v \) is the linear speed, \( r \) is the radius of the circular path, and \( \omega \) is the angular velocity.
In the exercise, we determine the linear speed of a vacuum cleaner belt by considering the shaft's radius and angular velocity. The shaft has a radius of \( 0.45 \space \text{cm} \) (or \( 0.0045 \space \text{m} \)) and an angular velocity of \( 60 \times 2\pi \space \text{rad/s} \). Therefore, the linear speed of the shaft, and thus the belt which doesn’t slip over it, can be calculated as:

• \[ v = 0.0045 \times 60 \times 2\pi = 1.7 \space \text{m/s} \]

This linear speed is key to understanding how fast the belt moves and how it can transfer motion efficiently to the wheel. Mastering linear speed calculations ensures a firm grasp on transitioning between rotational and linear movements in mechanical systems.

Belt and Pulley Systems

Belt and pulley systems are a fundamental part of mechanics, commonly used to transfer motion and power between shafts in various machines. This setup allows for the efficient distribution of rotational motion across different distances and angles.
In the exercise situation, a belt looped over a small shaft and a larger wheel helps translate the rotational motion of the motor to the wheel without slipping. This characteristic means that the linear speed of the belt remains constant across both the shaft and the wheel, simplifying calculations.
Pulleys, when used in conjunction with belts, can effectively convert rotational motion to linear motion and vice versa. Advantages include:

• Flexibility in the arrangement of components.
• Reduced wear and tear since the components don't need to be in direct contact.
• Ability to change the direction and speed of motion with different pulley sizes.

To solve problems involving belt and pulley systems, remember to:

• Check that the belt is not slipping so that linear speeds at contact points are the same,
• Use the formula \( v = r \cdot \omega \) to find speeds and velocities,
• Be mindful of the radii to calculate each component’s individual characteristics effectively.

Grasping the mechanics of belts and pulleys is vital for understanding and managing rotational dynamics in practical and complex machinery settings.