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Chapter 10: Problem 10

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg}\) . A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axe exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

Short Answer

Expert verified
(a) Angular acceleration: 34.8 rad/s虏, Cord acceleration: 8.7 m/s虏. (b) Axle force: 40.0 N leftward. (c) Direction of axle force changes.

Step by step solution

01

Find the Moment of Inertia

The moment of inertia \( I \) for a solid disk (or wheel) about its center is given by \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius. Substitute in the given values: \( I = \frac{1}{2} \times 9.20 \, \mathrm{kg} \times (0.250 \, \mathrm{m})^2 = 0.2875 \, \mathrm{kg\cdot m^2} \).
02

Compute Angular Acceleration

The torque \( \tau \) acting on the wheel due to the force exerted on the cord is given by \( \tau = r F \), where \( F \) is the force applied. Here, \( \tau = 0.250 \, \mathrm{m} \times 40.0 \, \mathrm{N} = 10.0 \, \mathrm{N\cdot m} \). Angular acceleration \( \alpha \) can be found using \( \tau = I \alpha \). Solving for \( \alpha \), we have \( \alpha = \frac{\tau}{I} = \frac{10.0 \, \mathrm{N\cdot m}}{0.2875 \, \mathrm{kg\cdot m^2}} \approx 34.8 \, \mathrm{rad/s^2} \).
03

Find Linear Acceleration of the Cord

The linear acceleration \( a \) of a point on the edge of the wheel is related to the angular acceleration \( \alpha \) by \( a = r \alpha \). Hence, \( a = 0.250 \, \mathrm{m} \times 34.8 \, \mathrm{rad/s^2} = 8.7 \, \mathrm{m/s^2} \).
04

Determine Axle Force Effect

We know that the center of mass of the wheel does not accelerate horizontally (Newton's first law implies the sum of horizontal forces is zero). The pull exerts a horizontal force of 40.0 N. Therefore, the axle must exert an equal and opposite horizontal force to balance it out.
05

Discuss Effects of Vertical Pull

For part (c), if the pull were upward, the components of forces acting on the wheel would change. Specifically: (a) The angular acceleration and linear acceleration remain the same since they depend only on the magnitude of the force and not its direction. (b) The axle force's direction would change (it would provide an upward force).

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Key Concepts

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

Moment of Inertia
In physics, the moment of inertia (\( I \)) is a measure of an object's resistance to changes in its rotation. It's a bit like mass is for linear motion.
For a solid disk or wheel, the moment of inertia is given by:\[I = \frac{1}{2} m r^2\] where \( m \) is the mass of the object and \( r \) is its radius.
  • In the given problem, the mass of the wheel is \( 9.20 \, \mathrm{kg} \) and its radius is \( 0.250 \, \mathrm{m} \).
  • Substituting these values, we calculate \( I = 0.2875 \, \mathrm{kg\cdot m^2} \).
The moment of inertia tells us how much "rotational inertia" the wheel has, which is crucial for understanding how it will respond to applied torques.
Torque
Torque (\( \tau \)) is the rotational equivalent of force. It measures how much a force acting on an object causes that object to rotate. In simple terms, it's what makes things spin. The formula for torque is:\[\tau = r F\] where \( r \) is the distance from the axis of rotation to where the force is applied, and \( F \) is the force itself.

In the exercise:
  • The radius \( r \) is \( 0.250 \, \mathrm{m} \).
  • The force applied is \( 40.0 \, \mathrm{N} \).
  • Thus, the torque is \( 10.0 \, \mathrm{N\cdot m} \).
Torque is crucial because it drives the wheel's rotation, and allows us to find its angular acceleration鈥攁 measure of how fast its spin speed is increasing.
Linear Acceleration
Linear acceleration (\( a \)) is the rate at which an object's velocity changes in a straight line. Although we are dealing with rotational motion, linear acceleration relates directly to what's happening at the edge of a spinning object. For a rotating object, the linear acceleration of a point on its edge is tied to its angular acceleration (\( \alpha \)) by the equation:\[a = r \alpha\]
  • Here, \( r \) (radius) is \( 0.250 \, \mathrm{m} \).
  • The angular acceleration \( \alpha \) was calculated as \( 34.8 \, \mathrm{rad/s^2} \).
  • Therefore, the linear acceleration is \( 8.7 \, \mathrm{m/s^2} \).
This means that the edge of the wheel, where the cord is unspooling, moves with this linear acceleration.
Newton's Laws
Newton's Laws of Motion provide the foundation for understanding how forces affect the motion of objects. In relation to our exercise:
  • Newton's First Law: An object remains at rest or in uniform motion unless acted on by a force. In this scenario, it implies that the horizontal forces must balance out, as the wheel won't move horizontally.
  • Newton's Second Law: States that \( F = ma \). It's about how forces result in acceleration, which is evident in how a torque leads to the angular acceleration of the wheel.
  • Newton's Third Law: For every action, there is an equal and opposite reaction. Here, the cord exerts a force on the wheel, and the axle provides an equal and opposite force to prevent the wheel from moving horizontally.
These laws are crucial for understanding not only why the wheel spins but how forces are distributed and balanced in the whole system.

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Most popular questions from this chapter

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

In a spring gun, a spring of force constant 400 \(\mathrm{N} / \mathrm{m}\) is com- pressed 0.15 \(\mathrm{m}\) . When fired, 80.0\(\%\) of the elastic potential energy stored in the spring is eventually converted into the kinetic energy of a \(0.0590-\mathrm{kg}\) uniform ball that is rolling without slipping at the base of a ramp. The ball continues to roll without slipping up the ramp with 90.0\(\%\) of the kinetic energy at the bottom converted into an increase in gravitational potential energy at the instant it stops. (a) What is the speed of the ball's center of mass at the base of the ramp? (b) At this position, what is the speed of a point at the top of the ball? (c) At this position, what is the speed of a point at the bottom of the ball? (d) What maximum vertical height up the ramp does the ball move?

A uniform drawbridge 8.00 \(\mathrm{m}\) long is attached to the road- way by a frictionless hinge at one end, and it can be raised by a cable attached to the other end. The bridge is at rest, suspended at \(60.0^{\circ}\) above the horizontal, when the cable suddenly breaks. (a) Find the angular acceleration of the drawbridge just after the cable breaks. (Gravity behaves as though it all acts at the center of mass. 6 ) Could you use the equation \(\omega=\omega_{0}+\alpha t\) to calculate the angular speed of the drawbridge at a later time? Explain why. (c) What is the angular speed of the drawbridge as it becomes horizontal?

A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is supported from the ceiling (Fig. 10.51 ). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 1.20 \(\mathrm{m} .\)
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