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Chapter 9: Problem 16

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

Short Answer

Expert verified
(a) The total angle through which the wheel turned before stopping can be found after carrying out the calculations in Step 2. (b) The time at which the wheel stopped can be calculated from Step 3. (c) The acceleration as the wheel slowed down can be determined using the equation discussed in Step 4.

Step by step solution

01

Calculate the Angular Distance Before Circuit Breaker Trips

The wheel is under constant angular acceleration of \(30.0 rad/s^2\) up to \(t=2.00s\). We can use the formula for the distance under constant acceleration: \(S = ut + 0.5 a t^2\), where \(u\) is the initial velocity (24 rad/s), \(a\) is the acceleration (30 rad/s^2) and \(t\) is the time (2s). By substituting the known values, we get the angular distance before the circuit breaker trips.
02

Calculate Total Angular Distance

After the circuit breaker trips, the wheel turned an additional angular distance of 432 rad while slowing down. By adding this to the angular distance calculated in Step 1, the total angle through which the wheel turned before stopping can be determined.
03

Finding When the Wheel Stops

The time it takes for the wheel to stop after the circuit breaker tripped can be found using the equation \(S = vt - 0.5 a t^2\). Here, \(v\) is the velocity at \(t=2s\), \(a\) is the acceleration we are trying to find, \(S = 432 rad\) is the distance, and \(t\) is the time. One can solve this equation for \(t\) once the acceleration is known (Step 4).
04

Calculate the Slowing Down Acceleration

To find the acceleration as the wheel slowed down, put values from step 1, step 2 and step 3 in the equation \(v = u + at\), where \(u\) is the initial velocity (initial velocity before the circuit tripped \(= u + a.t\)), \(v = 0 rad/s\) (final velocity, as wheel stops), \(a\) is the acceleration and \(t\) is the time to stop after the circuit breaker tripped (from Step 3). Solve this equation to find \(a\).\)

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Key Concepts

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

Angular Velocity
Angular velocity is all about how fast something is rotating. Think of it as how quickly an angle is changing over time. It's measured in radians per second ( rad/s ). In our exercise about the grinding wheel, the wheel begins with an angular velocity of 24 rad/s. This means every second, the wheel rotates through 24 radians, equivalent to roughly 3.8 complete rotations per second, because there are 2蟺 radians in each full rotation.
  • Initial angular velocity, when we start timing, is crucial. It sets the stage for all the calculations that follow.
  • As angular acceleration takes effect, this angular velocity changes, propelling the wheel faster until a circuit breaker stops it.
  • When analyzing problems of angular motion, initial angular velocity is often the starting point for computing distances and times.
Understanding the initial angular velocity helps us determine later states of motion and is fundamental to solving any problem in angular motion.
Angular Acceleration
Angular acceleration refers to how quickly the angular velocity changes. It's measured in radians per second squared ( rad/s^2 ). If you picture a car accelerating down a highway, angular acceleration is like that initial push that makes the car speed up.
In the context of our problem, the wheel starts with a constant angular acceleration of 30 rad/s虏. This means its speed is increasing by 30 rad/s every second for the first 2 seconds, which is while the circuit is up.
  • Angular acceleration in our scenario causes the wheel's angular velocity to increase during the allowed 2 seconds.
  • After the circuit breaker trips, the wheel still has an angular acceleration, but in the opposite direction, making it decelerate until it comes to a halt.
  • The equations of motion for rotational movement involve this acceleration to calculate further aspects like stopping time and distance.
Knowing how angular acceleration influences the motion is key to untangling any complex angular motion problem.
Angular Distance
Angular distance is basically how far something has rotated. It's measured in radians and tells us how much angle has been swept over a period. Imagine tracking the hands of a clock over a few hours; that's akin to measuring angular distance.
In the exercise with the grinding wheel, it turned a certain number of radians in two scenarios: initially while accelerating and later while decelerating as it coasted to a stop.
  • At first, the wheel covers a calculated distance under constant acceleration before the breaker trips.
  • After the breaker trips, the wheel coasts through an additional 432 rad as it slows to a stop.
  • To find the total angular distance, you add both these distances together.
This total angular distance gives an overall picture of the wheel's journey from start to stop, showcasing both periods of motion: acceleration and deceleration.

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Most popular questions from this chapter

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

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.

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00 \mathrm{rad} / \mathrm{s}\) at \(t=0,\) increases linearly with time, and is \(+4.00 \mathrm{rad} / \mathrm{s}\) at \(t=7.00 \mathrm{~s}\). We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00 \mathrm{~s}\) ?

A roller in a printing press turns through an angle \(\begin{array}{llll}\theta(t) & \text { given } & \text { by } & \theta(t)=\gamma t^{2}-\beta t^{3}, & \text { where } & \gamma=3.20 \mathrm{rad} / \mathrm{s}^{2} & \text { and }\end{array}\) \(\beta=0.500 \mathrm{rad} / \mathrm{s}^{3} .\) (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)
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