Question

A 2.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during that time? (c) Use Eq. (10.21) to calculate the work done by the torque. (d) What is the grinding wheel's kinetic energy when it is rotating at 1200 rev/min ? Compare your answer to the result in part (c).

Short answer

(a) 0.224\(\pi\) Nm; (b) 50\(\pi\) radians; (c) 11.2\(\pi^2\) J; (d) 11.2\(\pi^2\) J, same as (c).

Step-by-step solution

Convert Units

First, convert the angular speed from revolutions per minute to radians per second. Since 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds, we have:\[\omega = 1200 \, \text{rev/min} = 1200 \times \frac{2\pi}{60} \, \text{rad/s} = 40\pi \, \text{rad/s}\]

Calculate Moment of Inertia

The moment of inertia \(I\) for a solid cylinder is given by \(I = \frac{1}{2} m r^2\), where \(m = 2.80 \, \text{kg}\) and \(r = 0.100 \, \text{m}\).\[I = \frac{1}{2} \times 2.80 \times (0.100)^2 = 0.014 \, \text{kg} \, \text{m}^2\]

Calculate Angular Acceleration

To find the angular acceleration \(\alpha\), use the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega = 40\pi \, \text{rad/s}\) and \(\Delta t = 2.5 \, \text{s}\):\[\alpha = \frac{40\pi}{2.5} = 16\pi \, \text{rad/s}^2\]

Calculate Torque

Use the relationship between torque \(\tau\) and angular acceleration \(\alpha\), given by \(\tau = I \alpha\).\[\tau = 0.014 \times 16\pi = 0.224\pi \, \text{Nm}\]

Calculate Angle Turned

The angle \(\theta\) through which the wheel turns is given by \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\). Since \(\omega_0 = 0\), it simplifies to:\[\theta = \frac{1}{2} \times 16\pi \times (2.5)^2 = 50\pi \, \text{radians}\]

Calculate the Work Done by the Torque

Work done by the torque \(W\) is given by \(W = \tau \theta\).\[W = 0.224\pi \times 50\pi = 11.2\pi^2 \, \text{J}\]

Calculate Kinetic Energy

The kinetic energy \(K\) when the wheel is rotating is \(K = \frac{1}{2}I \omega^2\).\[K = \frac{1}{2} \times 0.014 \times (40\pi)^2 = 11.2\pi^2 \, \text{J}\]The kinetic energy is the same as the work done by the torque.

Key concepts

Torque

In the realm of angular mechanics, torque is a fundamental concept. It is essentially the rotational counterpart of force and dictates how effectively an object can be set into rotational motion. Torque (\( \tau \)) is calculated using the formula \( \tau = I \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. It is measured in Newton-meters (Nm).
In our example, to calculate the torque required to spin the grinding wheel, we first found the moment of inertia and the angular acceleration. Then, by multiplying these two, we determined the torque. This technique is used frequently in machinery design, where precise levels of torque must be applied to achieve desired rotation rates.
Torque reflects not just the force applied but also the distance from the pivot point. When the distance from the axis of rotation increases, the potential for greater torque is present. This relationship is akin to using a wrench, where a longer handle allows for easier turning due to increased torque.

Moment of Inertia

The moment of inertia (\( I \)) is crucial in understanding how an object resists changes in its rotational motion. For different shapes and mass distributions, the moment of inertia varies. For a solid cylinder like our grinding wheel, the formula is \( I = \frac{1}{2}mr^2 \), where \( m \) is the mass and \( r \) is the radius.
In essence, it's the measure of how spread out an object's mass is in relation to the axis of rotation. As the mass moves further away, the moment of inertia increases, requiring more torque to achieve the same angular acceleration. Imagine spinning an empty wheel versus one filled with a heavy liquid; the second one spins more reluctantly because of the higher moment of inertia.
Understanding the moment of inertia is vital in applications ranging from designing car engines to understanding how to optimize the spins of athletic moves.

Angular Acceleration

Angular acceleration (\( \alpha \)) describes how quickly an object changes its rotational speed. Mathematically, it is defined as the change in angular velocity over time: \( \alpha = \frac{\Delta \omega}{\Delta t} \).
In our example, we calculated that the angular acceleration required to bring the grinding wheel to the desired speed was \( 16\pi \) radians per second squared. This tells us how fast the wheel reaches its peak speed from rest within the timeframe of 2.5 seconds.
Angular acceleration plays a critical role in many engineering systems, such as conveyor belts, where a controlled increase in speed is necessary for proper operation. It is also crucial in robotics, where precise movement and adjustments are essential for the efficient running of tasks.

Kinetic Energy

Once an object is rotating, it possesses kinetic energy due to its motion. The rotational kinetic energy (\( K \)) can be expressed using the formula \( K = \frac{1}{2}I \omega^2 \).
For our rotating grinding wheel, we calculated the kinetic energy to verify that it matches the work done by the torque used to accelerate the wheel. Both calculations resulted in \( 11.2\pi^2 \) Joules, confirming the conservation of energy principle.
Kinetic energy is an essential concept in physics and engineering. It allows us to quantify how much work an object in motion can do. Understanding kinetic energy is necessary for everything from designing efficient industrial machines to calculating the impacts during vehicular accidents.