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A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

Short Answer

Expert verified
The angular speed becomes zero at 12 seconds.

Step by step solution

01

Understand the Relationship

Angular momentum changes due to applied torque. The relationship is given by Newton's second law in rotational form: \[\tau = \frac{dL}{dt},\]where \(\tau\) is torque, \(L\) is angular momentum, and \(t\) is time.
02

Set Up Equation with Given Values

We know the initial angular momentum is \(L_0 = 600 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\) and the torque applied is \(\tau = 50 \, \mathrm{N} \cdot \mathrm{m}\). We're looking for time \(t\) when the angular momentum is zero, meaning \(L = 0\). Thus,\[50 = \frac{d(600)}{dt}.\]
03

Solve for Time

Rearrange the equation for time \(t\):\[dL = -\tau \, dt \600 = 50 \, t \t = \frac{600}{50} = 12 \, \text{seconds}.\]Here, the negative sign accounts for the reduction in angular momentum due to the torque applied counter to its direction.
04

Interpret the Result

The torque applied counteracted the initial angular momentum, so the rotation ceases when the angular momentum reaches zero. It takes \(12\) seconds for the torque to completely stop the wheel's initial rotation.

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Key Concepts

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

Understanding Torque
Torque is a fascinating and central concept in the study of rotational motion. It's essentially a force that causes an object to rotate around an axis. In simpler terms, think of torque as the rotational equivalent of linear force.

When a torque is applied to an object, like a wheel or a seesaw, it affects how and when the object spins. The strength of this effect is determined by two main factors:
  • The magnitude of the force applied.
  • The distance from the axis of rotation at which the force is applied, known as the lever arm.
The equation for torque \( \tau \) is:\[\tau = r \times F \times \sin(\theta)\]Where:
  • \( r \) is the lever arm distance.
  • \( F \) is the force applied.
  • \( \theta \) is the angle between the force and the lever arm.
Torque is measured in Newton-meters (N·m), a combination of the units for force (Newtons) and distance (meters). In this context, applying a torque of \(50 \mathrm{~N} \cdot \mathrm{m}\) implies enough force is exerted at a certain distance to influence the wheel's rotational motion.
Newton's Second Law for Rotation
Newton's second law for rotation is similar to his famous second law of motion for linear dynamics. It states that the change in angular momentum of an object is equal to the torque applied to it. Mathematically, it's expressed as:
\[\tau = \frac{dL}{dt}\] Where:
  • \( \tau \) is the torque applied.
  • \( \frac{dL}{dt} \) represents the rate of change of angular momentum \( L \).
This formula helps us understand how rotational motion changes over time when a torque is applied. It shows that the greater the torque applied to an object, the faster its angular momentum changes.

In the exercise, this equation is central to solving the problem of when the angular speed of a rotating wheel becomes zero. By applying the known torque of \(50 \mathrm{~N} \cdot \mathrm{m}\), we calculate how long it should take for the wheel's rotation to slow to a stop, starting with its initial angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^2/\mathrm{s}\).
Understanding Angular Speed
Angular speed refers to how quickly an object rotates or revolves around a particular point or axis. It's the angle turned by the object in a unit of time and is typically measured in radians per second \(\text{rad/s}\).

Angular speed (\(\omega\)) conveys vital information about rotational motion, similar to how linear speed tells us how fast something moves in a straight line. Formulaically, it's given by:\[\omega = \frac{\Delta \theta}{\Delta t}\]Where:
  • \( \Delta \theta \) is the change in angular displacement in radians.
  • \( \Delta t \) is the time period over which the displacement happens.
When torque is applied to an object with initial angular momentum, it influences the object's angular speed by either accelerating or decelerating the rotation.

In the problem provided, the goal was to determine the time needed for the wheel's angular speed to reach zero. As torque acts against the initial direction of rotation, it gradually reduces angular momentum and, thus, the angular speed of the wheel, until it comes to a stop.

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Most popular questions from this chapter

At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has the position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity is given by \(\vec{v}=\left(-6.0 t^{2} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}\) for \(t \geq 0\) in seconds. About the origin, what are (a) the particle's angular momentum \(\vec{L}\) and \((\mathrm{b})\) the torque \(\vec{\tau}\) acting on the particle, both in unit-vector notation and for \(t>0 ?\) About the point \((-2.0 \mathrm{~m},-3.0 \mathrm{~m}, 0)\), what are (c) \(\vec{L}\) and (d) \(\vec{\tau}\) for \(t>0 ?\)

At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m}\). It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Three particles of mass \(m=23 \mathrm{~g}\) are fastened to three rods of length \(d=12 \mathrm{~cm}\) and negligible mass. The rigid assembly rotates around point \(O\) at the angular speed \(\omega=0.85 \mathrm{rad} / \mathrm{s}\). About \(O\), what are (a) the rotational inertia of the assembly, (b) the magnitude of the angular momentum of the middle particle, and (c) the magnitude of the angular momentum of the asssembly?

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11-18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

A solid ball rolls smoothly from rest (starting at height \(H=6.0 \mathrm{~m}\) ) until it leaves the horizontal section at the end of the track, at height \(h=2.0 \mathrm{~m}\). How far horizontally from point \(A\) does the ball hit the floor?

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