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Chapter 9: Problem 68

A platform is rotating at an angular speed of 2.2 rad/s. A block is resting on this platform at a distance of 0.30 m from the axis. The coefficient of static friction between the block and the platform is 0.75. Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

Short Answer

Expert verified
1.52 meters

Step by step solution

01

Understand the Forces

The block is subject to two main forces: the static friction force and the centripetal force. The frictional force must be equal or greater than the centripetal force to keep the block from sliding off.
02

Static Friction Formula

The static friction force can be calculated using the formula: \[ F_{friction} = \mu_s \cdot m \cdot g \]where \( \mu_s \) is the coefficient of static friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity \((9.8\, m/s^2)\).
03

Centripetal Force Formula

The centripetal force needed to keep the block in circular motion is given by: \[ F_{centripetal} = m \cdot \omega^2 \cdot r \]where \( \omega \) is the angular speed, and \( r \) is the distance from the axis.
04

Set Forces Equal

For the block to remain stationary, set the static friction force equal to the centripetal force:\[ \mu_s \cdot m \cdot g = m \cdot \omega^2 \cdot r \]The mass \( m \) cancels out from both sides, simplifying to:\[ \mu_s \cdot g = \omega^2 \cdot r \]
05

Solve for r

Rearrange the equation to solve for \( r \):\[ r = \frac{\mu_s \cdot g}{\omega^2} \]Substitute the given values: \( \mu_s = 0.75 \), \( g = 9.8\, m/s^2 \), and \( \omega = 2.2\, rad/s \), obtaining:\[ r = \frac{0.75 \times 9.8}{2.2^2} = \frac{7.35}{4.84} \approx 1.52 \, m \]
06

Conclusion

The smallest distance from the axis that the block can be relocated to without sliding off is approximately 1.52 meters.

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Key Concepts

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

Angular Speed
Angular speed is a measure of how fast the platform is rotating. It is represented by the symbol \( \omega \) and the unit used is radian per second (rad/s). For this exercise, the platform's angular speed is given as 2.2 rad/s.
This value indicates how quickly the block on the platform would move around the axis if it were free to slide.

Understanding angular speed is important because it helps to determine how much force is exerted on objects in rotational motion.
  • Relation to Rotational Motion: Angular speed describes the rate at which an object rotates around an axis. It is similar to linear speed, but for rotation.
  • Impact on Force: Higher angular speeds result in larger forces acting on objects, affecting their stability on rotating surfaces.
By knowing the angular speed, we can calculate the centripetal force required to keep the block from sliding off the rotating surface.
Centripetal Force
Centripetal force is the necessary inward force that keeps an object moving in a circular path. In this case, it is the force required to keep the block from flying off the rotating platform.
This force is always directed towards the center of the circle around which the object moves.

The formula to calculate the centripetal force is:
\[ F_{centripetal} = m \cdot \omega^2 \cdot r \]
where:
  • \( m \) is the mass of the block.
  • \( \omega \) is the angular speed of the platform, given as 2.2 rad/s.
  • \( r \) is the distance from the axis of rotation to the block.
This force is crucial for understanding rotational motion, as it determines if the block will maintain its position as the platform spins. By balancing it with the static friction, it ensures the block doesn鈥檛 slip or move unexpectedly.
Moment of Inertia
The concept of moment of inertia, often described as rotational inertia, helps to determine how much an object resists changes to its rotational motion. However, in this exercise, the moment of inertia of the platform is ignored, simplifying calculations.

This focus allows us to better understand other forces, like friction and centripetal force, without the added complexity of rotational resistance.
  • Importance in Rotations: In general, moment of inertia influences how quickly an object can start or stop rotating. Similar to how mass affects linear motion, moment of inertia affects rotational motion.
  • Simplification in the Problem: Ignoring the platform's moment of inertia lets us focus solely on the forces acting on the block, like the friction and centripetal forces.
Consequently, this omission simplifies solving the problem, making it more straightforward to calculate needed forces and factors affecting the block's stability.
Coefficient of Static Friction
The coefficient of static friction, denoted as \( \mu_s \), quantifies the frictional force that prevents the block from sliding off the rotating platform. It is a dimensionless number that represents the ratio between the maximum static frictional force and the normal force exerted by the surface.

In this exercise, the coefficient is given as 0.75.
  • Formula Overview: The static friction force can be calculated using:
    \[ F_{friction} = \mu_s \cdot m \cdot g \]
    where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
  • Balancing Forces: For the block to remain stationary, static friction must counteract the centripetal force. Therefore, the setup of \( \mu_s \cdot m \cdot g = m \cdot \omega^2 \cdot r \) is crucial to find the maximum distance \( r \).
Understanding \( \mu_s \) explains how much friction is available to counteract other forces, clarifying why the block remains steady as the platform spins.

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Most popular questions from this chapter

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a \(30.0^{\circ}\) angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is 0.650. Find the smallest value for the angle , such that the lower end of the board does not slide along the ground.

A small 0.500-kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m. The angular speed is 6.28 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m},\) an angular speed of 76.0 \(\mathrm{rad} / \mathrm{s}\) , and a moment of inertia of 0.615 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 6.40 s. \((\text { a) Find the magnitude of the angular deceleration of the }\) cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.
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