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Chapter 5: Problem 60

A bowling ball weighing 71.2 \(\mathrm{N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80-\mathrm{m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the bowling ball, in magnitude and direction, at this instant? (b) What is the tension in the rope at this instant?

Short Answer

Expert verified
The ball's acceleration is 4.64 m/s虏 upwards; the tension in the rope is 105.9 N.

Step by step solution

01

Analyze Forces Acting on the Ball

As the ball swings through the lowest point of its arc, we must consider two forces: the gravitational force acting downward and the tension in the rope acting upward. These forces dictate the ball's motion.
02

Determine the Gravitational Force

The gravitational force can be calculated using \[ F_g = mg \]where \( m = \frac{71.2\text{ N}}{9.8 \text{ m/s}^2} \approx 7.27 \text{ kg} \). Thus, \( F_g = 71.2 \text{ N} \).
03

Calculate Centripetal Acceleration

The centripetal acceleration is given by \[ a_c = \frac{v^2}{r} \]where \( v = 4.20 \text{ m/s} \) and \( r = 3.80 \text{ m} \). Thus, \[ a_c = \frac{(4.20)^2}{3.80} \approx 4.64 \text{ m/s}^2 \].
04

Determine Net Force and Tension in the Rope

At the lowest point, the net force is the centripetal force, which also considers the rope's tension. Given by\[ F_{net} = F_T - F_g = ma_c \].Rearranging gives \[ F_T = ma_c + F_g \].
05

Calculate Tension

Substitute the known values:\[ F_T = 7.27 \times 4.64 + 71.2 \approx 105.9 \text{ N} \].This is the tension required to sustain the centripetal acceleration while counteracting gravitational pull.
06

Conclude the Ball's Acceleration and Tension

The acceleration of the ball is \( 4.64 \text{ m/s}^2 \) upwards (toward the pivot point of the rope), and the tension in the rope is \( 105.9 \text{ N} \).

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Key Concepts

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

Centripetal Force
When a bowling ball swings as a pendulum, it experiences a force that keeps it moving along a curved path. This force is known as the centripetal force. Unlike gravity, which pulls objects straight downward, centripetal force acts towards the center of the ball's circular path.
  • Centripetal force is not an actual force applied to the object but results from the net effect of other forces like tension and gravity.
  • It is given by the formula \( F_c = m a_c \), where \( m \) is the mass of the object and \( a_c \) is the centripetal acceleration.
The centripetal acceleration is crucial for maintaining this force. Using the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity and \( r \) the radius of the circle, we can calculate how quickly the bowling ball accelerates towards the path's center. This concept helps explain why, during the swing, the ball remains tied to a circular trajectory as opposed to moving in a straight line.
Gravitational Force
Gravitational force is a fundamental concept in physics that affects all objects with mass. When considering a pendulum like a swinging bowling ball, gravity acts as the force pulling the ball downward toward Earth's center.
  • Gravity is defined by \( F_g = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
  • This force is constant and acts vertically downward regardless of the pendulum's position in its swing.
In our exercise, we calculated the gravitational force exerted on the bowling ball as \( 71.2 \text{ N} \). This force plays a crucial role in the pendulum's motion, influencing both the speed and the path of its swing.Understanding gravitational pull is essential in analyzing the net forces acting on the ball and consequently the tension in the rope.
Tension in Rope
Tension in the rope is the force exerted by the rope on the bowling ball to keep it from falling and allowing it to swing. This tension is not just opposing gravity but also providing the necessary centripetal force for the circular motion.
  • The tension force \( F_T \) can be determined by adding the gravitational force \( F_g \) and the required centripetal force \( F_c \).
  • Mathematically, it is expressed as \( F_T = ma_c + F_g \).
In our case, using the mass of the ball \( 7.27 \text{ kg} \), centripetal acceleration \( 4.64 \text{ m/s}^2 \), and gravitational force \( 71.2 \text{ N} \), the tension is calculated to be approximately \( 105.9 \text{ N} \).This represents the force that ensures the pendulum swings properly without the rope breaking. Appreciating how tension operates with other forces enables us to predict pendulum behavior accurately in physics scenarios.

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Most popular questions from this chapter

You find it takes 200 \(\mathrm{N}\) of horizontal force to move an empty pickup truck along a level road at a speed of 2.4 \(\mathrm{m} / \mathrm{s}\) . You then load the pickup and pump up its tires so that its total weight increases by 42\(\%\) while the coefficient of rolling friction decreases by 19\(\%\) . Now what horizontal force will you need to move the pickup along the same road at the same speed? The speed is low enough that you can ignore air resistance.

Losing Cargo. A \(12.0-\mathrm{kg}\) box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{s}=0.19\) and \(\mu_{k}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

A small block with mass \(m\) rests on a frictionless horizontal tabletop a distance \(r\) from a hole in the center of the table (Fig. 5.79\() .\) A string tied to the small block passes down through the hole, and a larger block with mass \(M\) is suspended from the free end of the string. The small block is set into uniform circular motion with radius \(r\) and speed \(v\) . What must \(v\) be if the large block is to remain motionless when released? figure can't copy

A machine part consists of a thin 40.0 -cm-long bar with small \(1.15-\mathrm{kg}\) masses fastened by screws to its ends. The screws can support a maximum force of 75.0 \(\mathrm{N}\) without pulling out. This bar rotates about an axis perpendicular to it at its center. (a) As the bar is turning at a constant rate on a horizontal frictionless surface, what is the maximum speed the masses can have without pulling out the screws? (b) Suppose the machine is redesigned so that the bar turns at a constant rate in a vertical circle. Will one of the screws be more likely to pull out when the mass is at the top of the circle or at the bottom? Use a free-body diagram to see why. (c) Using the result of part (b), what is the greatest speed the masses can have without pulling a screw?

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m}\) . A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) , (a) What is the minimum coefficient of friction the coefficient of friction (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?
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