/*! 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 58 A bowling ball weighing \(71.2 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 58

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 \(4.20 \mathrm{~m} / \mathrm{s}\). At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

Short Answer

Expert verified
(a) The acceleration of the bowling ball is \(4.62 m/s^2\), directed upward along the rope. (b) The tension in the rope is \(104.29 N\), also directed upward along the rope.

Step by step solution

01

Identify the forces acting

The forces acting on the bowling ball are the tension \(T\), directed upwards along the rope and the weight \(W\) of the ball directed vertically downward. The weight can be calculated as: \(W = m \cdot g\), where \(g\) is the acceleration due to gravity and \(m\) is the mass of the bowling ball. Given the weight, rearrange the equation to find mass: \(m = W / g = 71.2 N / 9.8 m/s^2 = 7.27 kg\). Since the question asks for acceleration and tension when the rope swings through the vertical, the forces on the bowling ball are balanced.
02

Calculate the Centripetal acceleration

When the rope is vertical, the ball follows a circular path. The radial acceleration it experiences is the centripetal acceleration, required for circular motion, and it is given by \(a_c = v^2 / r\), where \(v\) is the speed of the ball and \(r\) is the length of the rope. Substitute the given values: \(a_c = (4.2 m/s)^2 / 3.8 m = 4.62 m/s^2\). This is directed towards the center of the circle (or up along the rope).
03

Calculate the Tension in the rope

By applying the second law of Newton in the vertical direction, we have \(T - W = m \cdot a_c\). Thus Tension \(T = m \cdot a_c + W\). Substitute the values: \(T = 7.27 kg \cdot 4.62 m/s^2 + 71.2 N = 104.29 N\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Centripetal Acceleration
To understand the notion of centripetal acceleration, let's picture circular motion, like a bowling ball swinging as a pendulum. At any point during its swing, there is an acceleration that points towards the center of the circular path. This is what we call centripetal acceleration.

Centripetal acceleration is essential for an object to maintain its circular motion. Mathematically, it's calculated by the formula: \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity of the object and \( r \) is the radius of the circular path. In the problem, when the bowling ball swings through the bottom of its path, it's at its maximum speed and experiences the centripetal acceleration pulling it inward towards the pivot point, where the rope meets the ceiling.
Tension in Pendulum Rope
The tension in the pendulum rope is a force exerted by the rope on the pendulum. This force ensures that the ball follows a circular trajectory. When we look at a pendulum at the moment it's vertical during its swing, two key forces are at play: the gravitational force pulling it downwards and the tension in the rope pulling it upwards.

In our scenario, the tension isn't just counteracting gravity. It also provides the centripetal force needed for the ball鈥檚 circular motion. To find the tension at this point, we can use Newton's second law. It's important to remember that the tension must be greater than the weight of the object due to the additional requirement to accelerate the ball towards the center of the circle. The formula we use reflects this: \( T = m \times a_c + W \).
Newton's Second Law of Motion
Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. The law is often written as \( F = m \times a \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

In the case of the pendulum, the net force at the bottom of the pendulum鈥檚 swing is the tension minus the weight of the ball, as they are in opposite directions. So, according to Newton鈥檚 second law, we have \( T - W = m \times a_c \). For the ball to continue in circular motion, the tension provides not just enough force to balance the weight but also the extra force needed for centripetal acceleration. This shows the beauty of Newton鈥檚 second law: it can be applied to objects at rest, in linear motion, and, as seen here, in circular motion too.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Block \(B\) has mass \(5.00 \mathrm{~kg}\) and sits at rest on a horizontal, frictionless surface. Block \(A\) has mass \(2.00 \mathrm{~kg}\) and sits at rest on top of block \(B\). The coefficient of static friction between the two blocks is \(0.400 .\) A horizontal force \(\vec{P}\) is then applied to block \(A .\) What is the largest value \(P\) can have and the blocks move together with equal accelerations?

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.

A box with mass \(m\) is dragged across a level floor with coefficient of kinetic friction \(\mu_{\mathrm{k}}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_{\mathrm{k}}, \theta,\) and \(g,\) obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90 \mathrm{~kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

A small remote-controlled car with mass \(1.60 \mathrm{~kg}\) moves at a constant speed of \(v=12.0 \mathrm{~m} / \mathrm{s}\) in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of \(5.00 \mathrm{~m}\) (Fig. E5.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point \(A\) (bottom of the track) and (b) point \(B\) (top of the track)?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation

Oscillations

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.