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Chapter 10: Problem 5

An electric motor rotating a grinding wheel at \(100 \mathrm{rev} / \mathrm{min}\) is switched off. With constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2},\) (a) how long does it take the wheel to stop? (b) Through how many radians does it turn while it is slowing down?

Short Answer

Expert verified
The wheel takes 5.235s to stop and turns through 27.57 radians while it is slowing down.

Step by step solution

01

Translate Given Data

First, remember that the velocity given is in revolutions per minute. So, to proceed, it needs to be converted to rad/s using the conversion factor \(2蟺 \, rad/rev\). Therefore, the initial angular velocity \(蠅_i = 100 rev/min * (2蟺 rad/rev) * (1 min/60s) = 10.47 rad/s \) . The angular acceleration is \(伪 = -2.00 rad/s^2\), this is negative, indicating that it is a deceleration.
02

Time Calculation

Using the equation of motion \(蠅_f = 蠅_i + 伪t\), where \(蠅_f\) is the final angular velocity (which is 0 when the wheel stops), \(蠅_i\) is the initial angular velocity, \(伪\) is the angular acceleration, and t is the time, we can solve for t which gives \(t = (蠅_f - 蠅_i)/伪 = (-10.47 rad/s) / (-2.00 rad/s^2) = 5.235 s\). This is the time it takes for the wheel to stop.
03

Angle Calculation

To find the number of radians the wheel has turned while slowing down, use the equation of motion \(胃 = 蠅_i*t + 0.5*伪*t^2\). Note that even if the acceleration is negative, the wheel still turns an additional amount in the original direction. Substituting provided values, we get \(胃 = 10.47 rad/s * 5.235 s + (0.5 * -2.00 rad/s^2 * (5.235 s)^2) = 27.57 rad\). This is the total angle in radians that the wheel has turned while slowing down.

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Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how quickly the angular velocity of an object changes. It is analogous to linear acceleration, but instead of affecting the straight-line speed of an object, it influences the rate of rotation. In our example of the electric motor, the angular acceleration is constant and has a negative value of \( -2.00 \mathrm{rad} / \mathrm{s}^{2} \). The negative sign indicates that the acceleration is acting to decrease the angular velocity, leading to a deceleration or slowing down of the rotating grinding wheel.

Understanding angular acceleration is crucial, as it is a key factor in the equations of rotational motion. In terms of units, it is expressed in radians per second squared (\(\mathrm{rad}/\mathrm{s}^2\)). It's important to note that while the deceleration is negative, the wheel still covers a certain angular distance in the direction it was spinning before it comes to a halt.
Angular Velocity
Angular velocity measures how fast an object rotates or revolves relative to another point, usually the center of rotation. It is expressed in radians per second (\(\mathrm{rad}/\mathrm{s}\)). In the step by step solution of our problem, the initial angular velocity is determined by converting the given speed from revolutions per minute to radians per second, resulting in an initial velocity of \(蠅_i = 10.47 \, \mathrm{rad}/\mathrm{s}\). This value reflects the wheel's rotational speed when the motor is first shut off.

While linear velocity gives us the speed of an object moving along a path, angular velocity tells us how quickly the angle (or position) changes on a circular path. Therefore, when an object's angular velocity changes, it is undergoing angular acceleration or deceleration. In the example, the final angular velocity is zero, indicating that the wheel has come to a stop after being subjected to a negative angular acceleration.
Equations of Rotational Motion
Equations of rotational motion allow us to predict the future state of rotating systems. They're similar to the equations for linear motion, but they apply to rotation about an axis. For our motor and grinding wheel, we used two primary equations:
  • \(蠅_f = 蠅_i + 伪t\) to calculate the time taken for the wheel to stop.
  • \(胃 = 蠅_i*t + 0.5*伪*t^2\) to determine the angle through which the wheel turns while slowing down.

Here, \(蠅_f\) is the final angular velocity, \(蠅_i\) is the initial angular velocity, \(伪\) is the angular acceleration, \(t\) is the time, and \(胃\) is the angular displacement.

These equations are instrumental in solving a wide range of problems related to rotational motion in various fields, from mechanical engineering to astrophysics. They reveal how the angular velocity of an object changes under the influence of an angular acceleration over time and how these changes affect the object鈥檚 rotation angle.

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Most popular questions from this chapter

Attention! About face! Compute an order-of-magnitude estimate for the moment of inertia of your body as you stand tall and turn about a vertical axis through the top of your head and the point halfway between your ankles. In your solution state the quantities you measure or estimate and their values.

Following Thanksgiving dinner your uncle falls into a deep sleep, sitting straight up facing the television set. A naughty grandchild balances a small spherical grape at the top of his bald head, which itself has the shape of a sphere. After all the children have had time to giggle, the grape starts from rest and rolls down without slipping. It will leave contact with your uncle's scalp when the radial line joining it to the center of curvature makes what angle with the vertical?

A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height \(h\). If they are released from rest and roll without slipping, which object reaches the bottom first? Verify your answer by calculating their speeds when they reach the bottom in terms of \(h\)

An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s. The engine's rotation slows with an angular acceleration of magnitude \(80.0 \mathrm{rad} / \mathrm{s}^{2} .\) (a) Determine the angular speed after \(10.0 \mathrm{s}\) (b) How long does it take the rotor to come to rest?

A wheel \(2.00 \mathrm{m}\) in diameter lies in a vertical plane and rotates with a constant angular acceleration of \(4.00 \mathrm{rad} / \mathrm{s}^{2}\) The wheel starts at rest at \(t=0,\) and the radius vector of a certain point \(P\) on the rim makes an angle of \(57.3^{\circ}\) with the horizontal at this time. At \(t=2.00 \mathrm{s}\), find (a) the angular speed of the wheel, (b) the tangential speed and the total acceleration of the point \(P,\) and \((c)\) the angular position of the point \(P\)
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