Problem 72 - University Physics with Modern Physics

Question

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 \(\mathrm{m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.

Step-by-Step Solution

Answer

(a) 75.14 m/s, (b) 54,180 m/s²

Step 1: Determine Linear Speed Formula

The linear speed of an object moving in a circle is given by the formula \(v = r \cdot \omega\), where \(r\) is the radius of the circle and \(\omega\) is the angular velocity.

Step 2: Calculate Angular Velocity

The angular velocity \(\omega\) is the speed in radians per second. First, convert the motor's speed from revolutions per minute to radians per second. Since there are \(2\pi\) radians in one revolution and 60 seconds in a minute, we have:\[ \omega_{motor} = 3450 \text{ rev/min} \times \frac{2\pi \text{ radians}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \approx 361.28 \text{ rad/s} \]

Step 3: Equate Pulley Angular Velocities

Since the second pulley is half the diameter of the first, it spins at twice the angular speed of the motor pulley because the linear speed must be the same for both connected by the V-belt:\[ \omega_{blade} = 2 \cdot \omega_{motor} = 2 \times 361.28 \text{ rad/s} \approx 722.56 \text{ rad/s} \]

Step 4: Calculate the Blade's Linear Speed

Now, apply the linear speed formula. Substitute \(r = \frac{0.208}{2} = 0.104\) meters for the blade's radius:\[ v = 0.104 \text{ m} \cdot 722.56 \text{ rad/s} \approx 75.14 \text{ m/s} \]

Step 5: Determine Radial Acceleration

Radial (or centripetal) acceleration is given by \(a = r \cdot \omega^2\). Using the blade's radius and its angular velocity:\[ a = 0.104 \text{ m} \cdot (722.56 \text{ rad/s})^2 \approx 54180 \text{ m/s}^2 \]

Key Concepts

Understanding Linear Speed

Linear speed is essentially how fast a point on the edge of a rotating object, like a saw blade, is moving. It combines the rotation of the object with the distance from the center to that edge.

The formula for linear speed is given by:

• \( v = r \cdot \omega \)

where:

• \( v \) is the linear speed,
• \( r \) is the radius of the circle, and
• \( \omega \) is the angular velocity in radians per second.

To find the linear speed, we first need to know the radius of the rotating object and its angular velocity. From the provided exercise, we find that the motor's pulley angular velocity is converted from revolutions per minute into radians per second allowing us to calculate the blade's angular velocity.

The linear speed is crucial because it tells us how fast the edge of the blade moves – this could literally be the speed that a thrown object would travel if it came off the edge, like a piece of wood as mentioned in the problem.

The Role of Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is the acceleration that occurs when an object moves in a circular path, directed towards the center of the circle. It is what keeps the object moving around, rather than shooting off in a straight line.

The formula to compute radial acceleration is:

• \( a = r \cdot \omega^2 \)

where:

• \( a \) is the radial acceleration,
• \( r \) is the radius, and
• \( \omega \) is the angular velocity.

In the context of the exercise, the radial acceleration helps explain why particles such as sawdust do not stick to the blade. This high acceleration pushes them off the edge rapidly. The higher the angular velocity, the greater this radial acceleration becomes.

So, when you see sawdust flying away from a spinning saw blade, radial acceleration is doing the job of flinging it into the air.

Exploring Circular Motion

Circular motion refers to the movement of an object along the circumference of a circle or rotation along a circular path. Key elements of circular motion include the radius, angular velocity, and centripetal (radial) acceleration.

In circular motion, the linear speed at any point is perpendicular to the radius of the motion. This is why objects in circular motion can change direction constantly yet maintain a steady speed along the edge. This perpendicular aspect ensures that, theoretically, any object breaking away from the path will move tangentially off.

The exercise shows a practical example of circular motion using a saw blade. The pulley system changes the speed of rotation, which then translates into the high linear speeds discussed in problems like this. Observing the blade, all points on the edge move at the same linear speed due to the consistent radius, evidencing uniform circular motion.

• This shows the fundamental relationship between the motion’s radius and speed.
• Understanding this helps foresee the effects of different factors like diameter and speed changes.

The continuous direction change in circular motion necessitates constant inward acceleration, maintaining the path of motion.