Question

What power is developed by a grinding machine whose wheel has a radius of \(20.7 \mathrm{~cm}\) and runs at \(2.53\) rev/s when the tool to be sharpened is held against the wheel with a force of \(180 \mathrm{~N}\) ? The coefficient of friction between the tool and the wheel is \(0.32\)

Short answer

The power developed by the grinding machine is approximately 1.89 watts.

Step-by-step solution

Calculate the circumference of the wheel

The first step is to calculate the circumference of the wheel using the formula for the circumference of a circle, which is \(2\pi r\). Here, \(r\) is the radius of the wheel. So, the circumference \(C\) would be: \(C = 2\pi r = 2\pi (20.7 \mathrm{~cm}) = 130.1 \mathrm{~cm}\).

Convert Revolutions per second to Velocity

The Second step is to convert the speed of the wheel from revolutions per second into cm/sec. The speed \(V\) of the wheel can be calculated by multiplying the number of revolutions per second by the circumference of the wheel. So, the speed would be: \(V = 2.53 \text{ rev/s} \times 130.1 \mathrm{~cm} = 328.95 \mathrm{~cm/sec}\)

Calculate the frictional force

Next, calculate the frictional force \(F_f\). This can be calculated by multiplying the coefficient of friction (\(0.32\)) by the force that's being applied (\(180 N\)). Hence the frictional force would be: \(F_f = 180 N \times 0.32 = 57.6 N\).

Compute the Power

Finally, compute the power \(P\) developed by the grinding machine using the formula \(P=Fv\), where \(F\) is the frictional force and \(v\) is the velocity. Substituting the given values, we get: \(P = 57.6 N \times 328.95 \mathrm{~cm/sec} = 18941.82 \mathrm{~N.cm/sec}\). However, the standard SI unit of power is the watt, so we convert it from \(N.cm/sec\) to \(W\) by multiplying by \(10^{-7}\). Hence, \(P = 18941.82 \mathrm{~N.cm/sec} \times 10^{-7} = 1.89 \mathrm{~W}\).

Key concepts

Circumference of a Circle

Understanding the circumference of a circle is essential for a variety of physical problems, including those involving rotational motion. The circumference is the total length around the circle and is calculated using the formula \( C = 2\pi r \) where \( r \) is the radius of the circle. In the context of the grinding wheel problem, knowing the circumference is crucial since it tells us how far the wheel travels in one full revolution. By using the radius provided, \(20.7 \text{cm}\), we can determine the circumference of the wheel, which is key to further calculations.

When a wheel spins, it covers a distance equal to its circumference with each revolution, thus becoming the foundation for solving the next steps of velocity and power in a rotational system.

Converting Revolutions to Velocity

Another fundamental concept in physics is to convert between different units of motion, such as revolutions to linear velocity. In our exercise, we need to understand how fast the wheel is moving in terms of \(\text{cm/sec}\). Linear velocity \(V\) is the distance traveled per unit of time. For a wheel, this can be found by multiplying the number of revolutions per second by the wheel's circumference.

By using the formula \(V = \text{revolutions per second} \times \text{circumference}\), we can find that the wheel, spinning at \(2.53\) rev/s, moves forward \(328.95 \text{cm/sec}\). Comprehending this step is important as it sets the stage for calculating the frictional force exerted by the tool against the wheel.

Frictional Force Calculation

Frictional force is a key factor in many physics problems and calculating it correctly is vital for accurate results. Frictional force \(F_f\) is generated when two surfaces interact, and it resists motion. In the grinding wheel example, this force is what enables the scraping and shaping action on a tool. The amount of frictional force can be found using the formula \(F_f = \mu F_n\), where \(\mu\) is the coefficient of friction and \(F_n\) is the normal force.

In the given problem, multiplying the coefficient of friction (\(0.32\)) by the force applied (\(180\) N) gives us the frictional force, \(57.6\) N. This value is crucial as it represents the force with which the grinding wheel and the tool interact.

Power Computation in Physics

Power in physics is the rate at which work is done, or energy is transferred, often calculated as force times velocity. It is a concept that frequently arises in various applications, including machinery and rotational movements. In our exercise, we're looking at the power developed by a machine, defined as the product of the frictional force and the velocity of the grinding wheel.

By the formula \(P = F_f \times v\), we multiply the frictional force (\(57.6\) N) by the linear velocity of the wheel (\(328.95 \text{cm/sec}\)) to find the machine's power output in \(N\cdot cm/sec\). Afterward, to convert to the standard unit of power, the watt (W), we multiply our result by \(10^{-7}\), bringing us to \(1.89\) W. Understanding this power computation is crucial in determining the efficiency and capability of machines in real-world applications.