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Chapter 3: Problem 66

A 0.2 -kg particle \(P\) is constrained to move along the vertical-plane circular slot of radius \(r=0.5 \mathrm{m}\) and is confined to the slot of arm \(O A,\) which rotates about a horizontal axis through \(O\) with a constant angular rate \(\Omega=3\) rad/s. For the instant when \(\beta=\) \(20^{\circ},\) determine the force \(N\) exerted on the particle by the circular constraint and the force \(R\) exerted on it by the slotted arm.

Short Answer

Expert verified
The forces are: N 鈮� 0.229 N, R 鈮� 1.844 N.

Step by step solution

01

Identify Forces Acting on the Particle

The particle is constrained to move in a circular motion due to the slot. Thus, it experiences centripetal force towards the center of the circle and gravitational force acting downwards. We need to resolve these forces to find their resultant in terms of the given variables.
02

Determine the Components of Forces

Centripetal force, necessary for circular motion, is given by \(F_c = m r \Omega^2\) where \(m\) is the mass (0.2 kg), \(r\) is 0.5 m, and \(\Omega\) is 3 rad/s. The gravitational force, \(F_g = m g\) where \(g\) is 9.81 m/s虏.
03

Calculate Centripetal and Gravitational Forces

Substitute the values:\(F_c = 0.2 \times 0.5 \times 3^2 = 0.9 \text{ N}\).\(F_g = 0.2 \times 9.81 = 1.962 \text{ N}\).
04

Analyze Forces with Respect to the Circular Path

At angle \(\beta = 20^\circ\), decompose the forces. The centrifugal force acts radially outward and gravitational force acts downward. Calculate their components along and perpendicular to the circular path to find \(N\) and \(R\).
05

Calculate Force Components in Circular Constraints

Calculate components along the radial direction (toward the center) and perpendicular to it:Along the radial direction: - Component of gravitational force along radial direction: \(F_{g \parallel} = F_g \sin \beta\) - Total radial force \(F_r = F_c - F_{g \parallel}\)Perpendicular to the radial (tangential to the path but we consider interactions with the arm): - Net tangential force is due to the tangential component of gravity: \(F_{g \perp} = F_g \cos \beta\).
06

Solve for Normal Force and Arm Force

The normal force \(N\) is due to the radial force exerted by the circular constraint:\[ N = F_r = F_c - F_{g \parallel} = 0.9 - 1.962 \sin(20^{\circ}) \approx 0.9 - 0.671 = 0.229 \text{ N} \]The force \(R\) by the slotted arm is due to the force in the tangential direction:\[ R = F_{g \perp} = 1.962 \cos(20^{\circ}) \approx 1.962 \times 0.940 = 1.844 \text{ N} \]
07

Compilation of Results

The force \(N\) exerted on the particle by the circular constraint is approximately 0.229 N. The force \(R\) exerted on it by the slotted arm is approximately 1.844 N.

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Key Concepts

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

Centripetal Force
Centripetal force is critical when understanding circular motion. Imagine you are swinging a ball on a string in a circle. The force you feel pulling the ball towards the center is the centripetal force. This force is necessary to keep the object moving in a circle, rather than flying off in a straight line.
It can be calculated using the formula:
  • \( F_c = m r \Omega^2 \)
  • where \( m \) is mass, \( r \) is radius, and \( \Omega \) is angular velocity.
In our original exercise, the particle experiences a centripetal force directed towards the center of its circular path due to its motion in a slot. Knowing how to calculate this force helps us determine other forces acting on the particle. Including effects like gravity, it's essential to isolate these forces to solve overall dynamics problems effectively.
Gravitational Force
Gravitational force is a fundamental concept that affects almost any object on Earth. It is the force that pulls objects towards the center of the Earth. In simplest terms, it is the "weight" of the object. The force can be calculated with the formula:
  • \( F_g = m g \)
  • where \( m \) is the mass of the object and \( g \) is the gravitational acceleration (approximately \( 9.81 \text{ m/s}^2 \) on Earth).
In the case of our exercise, the gravitational force acts downward on the particle in addition to the constraints of its circular path. This force is a significant part of the problem solution as it contributes to the total forces acting on the particle. Gravitational forces must be decomposed into components that align and counter the directions of other forces to determine the net effect on an object.
Circular Motion
Understanding circular motion is vital when analyzing how objects move along a curved path. Circular motion can be uniform if the object moves at a constant speed around a circle or non-uniform if the speed changes.
For an object moving in a circle, the velocity is tangent to the path, and acceleration always points towards the circle's center, known as centripetal acceleration. This concept involves maintaining a constant rate of rotation or angular velocity, as in this exercise.
  • When an object is in circular motion, its centripetal acceleration is \( a_c = \frac{v^2}{r} \), where \( v \) is the tangential speed.
  • We also use \( a_c = r \Omega^2 \) when considering angular velocity \( \Omega \).
The particle in the exercise revolves around a circular path dictated by the arm, showcasing principles of circular motion and centripetal force in action. Understanding how these factors interact is fundamental to solving problems involving dynamics in rotational systems.

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Most popular questions from this chapter

Two barges, each with a displacement (mass) of \(500 \mathrm{Mg},\) are loosely moored in calm water. A stunt driver starts his 1500 -kg car from rest at \(A\), drives along the deck, and leaves the end of the \(15^{\circ}\) ramp at a speed of \(50 \mathrm{km} / \mathrm{h}\) relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at \(B\). Calculate the velocity \(v_{2}\) imparted to barge 2 just after the car has come to rest on the barge. Neglect the resistance of the water to motion at the low velocities involved.

The 1.2 -lb sphere is moving in the horizontal \(x-y\) plane with a velocity of \(10 \mathrm{ft} / \mathrm{sec}\) in the direction shown and encounters a steady flow of air in the \(x\) -direction. If the air stream exerts an essentially constant force of 0.2 lb on the sphere in the \(x\) -direction, determine the time \(t\) required for the sphere to cross the \(y\) -axis again.

Determine the coefficient of restitution \(e\) for a steel ball dropped from rest at a height \(h\) above a heavy horizontal steel plate if the height of the second rebound is \(h_{2}\).

A particle of mass \(m\) is attached to the end of the light rigid rod of length \(L,\) and the assembly rotates freely about a horizontal axis through the pivot \(O .\) The particle is given an initial speed \(v_{0}\) when the assembly is in the horizontal position \(\theta=0 .\) Determine the speed \(v\) of the particle as a function of \(\theta\).

A 90 -lb boy starts from rest at the bottom \(A\) of a 10-percent incline and increases his speed at a constant rate to \(5 \mathrm{mi} / \mathrm{hr}\) as he passes \(B, 50 \mathrm{ft}\) along the incline from \(A .\) Determine his power output as he approaches \(B\).
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