/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 337 The member \(O A\) rotates about... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 337

The member \(O A\) rotates about a horizontal axis through \(O\) with a constant counterclockwise angular velocity \(\omega=3\) rad/sec. As it passes the position \(\theta=0,\) a small block of mass \(m\) is placed on it at a radial distance \(r=18\) in. If the block is observed to slip at \(\theta=50^{\circ},\) determine the coefficient of static friction \(\mu_{s}\) between the block and the member.

Short Answer

Expert verified
The coefficient of static friction is approximately 0.442.

Step by step solution

01

Identify Rotational Motion Variables

The block slips at \( \theta = 50^{\circ} \), and the angular velocity is given as \( \omega = 3 \) rad/sec. The initial radial distance of the block is \( r = 18 \) in. Convert the angular position \( \theta \) from degrees to radians: \( \theta = 50^{\circ} = \frac{5\pi}{18} \) radians.
02

Calculate Centripetal Force

The centripetal force required to keep the block moving in a circle is given by \( F_c = m \cdot r \cdot \omega^2 \). Substituting the given values, we get the equation \( F_c = m \cdot 18 \cdot 3^2 = 162m \) (ensure units are consistent with calculations).
03

Identify Frictional Force Needed

The maximum frictional force before slipping occurs is given by \( f_{max} = \mu_s \cdot N \), where \( N \) is the normal force. Here, \( N = m \cdot g \cdot \cos(\theta) \), because the component of weight perpendicular to the arm at \( \theta = 50^{\circ} \) is \( N = m \cdot g \cdot \cos\left(\frac{5\pi}{18}\right) \).
04

Equate Forces at the Slip Point

At the point of slippage, the frictional force equals the centripetal force provided by friction: \( \mu_s \cdot m \cdot g \cdot \cos\left(\frac{5\pi}{18}\right) = 162m \). The mass \( m \) cancels out, and we are left with \( \mu_s \cdot g \cdot \cos\left(\frac{5\pi}{18}\right) = 162 \).
05

Solve for Coefficient of Static Friction

Substitute \( g = 32 \) ft/s虏 (converted to inches as \( g = 386 \) in/s虏) into the equation to get \( \mu_s \cdot 386 \cdot \cos\left(\frac{5\pi}{18}\right) = 162 \). Solve for \( \mu_s \), yielding \[ \mu_s = \frac{162}{386 \cdot \cos\left(\frac{5\pi}{18}\right)} \approx 0.442. \]

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Key Concepts

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

Rotational Motion
Rotational motion refers to the movement of an object about a specific axis. This motion occurs in a circular path, and the object remains at a constant distance from the center of rotation. In the given exercise, member \( OA \) rotates with a constant angular velocity. Understanding rotational motion is key to solving problems where objects rotate around a point. Think of a merry-go-round: all points on the ride move in circular paths around its center.
  • Each point follows a circular trajectory.
  • Angular position and velocity describe the motion.
For this exercise, identifying that the block slips at an angular position \( \theta = 50^{\circ} \), lets us recognize the point where rotational dynamics and other forces come into play.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path. It acts towards the center of the circle, ensuring the object doesn't fly off in a straight line. In our problem, the block on the rotating member requires an inward force to stay in its circular path. This inward force here is provided by the friction between the block and the rotating member.
  • Its magnitude is determined by the formula \( F_c = m \cdot r \cdot \omega^2 \).
  • In the problem, \( r = 18 \) in, and \( \omega = 3 \) rad/sec.
Calculating, we find that the centripetal force equals \( 162m \) units of force (with appropriate unit conversion for calculations). Recognizing the need for this force helps address why static friction is critical in preventing slippage.
Static Friction
Static friction is the force that resists the initiation of sliding motion between two surfaces. In rotational scenarios, it prevents the block from slipping off due to the rotational force. The challenge is to determine the coefficient \( \mu_s \) that represents this static friction between the block and member.
Since the block only starts to slip when the static friction can no longer counter the centripetal force, the calculations involve:
  • Maximum frictional force \( f_{max} = \mu_s \cdot N \).
  • Normal force \( N \) derived from the object鈥檚 weight \( N = m \, g \, \cos(\theta) \).
Realizing that \( \mu_s \) defines the stickiness of the two surfaces helps us solve the puzzle when the centripetal and frictional forces balance, just before slippage.
Angular Velocity
Angular velocity describes how quickly an object rotates around an axis and is typically measured in radians per second. For our problem, the member \( OA \) rotates with a constant angular velocity of \( \omega = 3 \) rad/sec.
Angular velocity links linear and rotational motion:
  • Related to linear speed by the radius of rotation.
  • Constant angular velocity implies a consistent rotational speed.
By understanding angular velocity, we gauge how swiftly an angle changes during motion, crucial for determining the dynamic forces that act on rotating bodies like the block in the exercise.

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Most popular questions from this chapter

Just after launch from the earth, the space-shuttle orbiter is in the \(37 \times 137-\mathrm{mi}\) orbit shown. At the apogee point \(A,\) its speed is \(17,290 \mathrm{mi} / \mathrm{hr}\). If nothing were done to modify the orbit, what would be its speed at the perigee \(P ?\) Neglect aerodynamic drag. (Note that the normal practice is to add speed at \(A,\) which raises the perigee altitude to a value that is well above the bulk of the atmosphere.

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