/*! 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 64 (III) An object of mass \(m\) is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(III) An object of mass \(m\) is constrained to move in a circle of radius \(r\). Its tangential acceleration as a function of time is given by \(a_{\tan }=b+c t^{2},\) where \(b\) and \(c\) are constants. If \(v=v_{0}\) at \(t=0,\) determine the tangential and radial components of the force, \(F_{\text {tan }}\) and \(F_{\mathrm{R}}\), acting on the object at any time \(t>0\).

Short Answer

Expert verified
Tangential Force: \( F_{\tan} = m(b + ct^2) \); Radial Force: \( F_{R} = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \).

Step by step solution

01

Identify Known Quantities and Equations

We are given the tangential acceleration, \(a_{\tan} = b + ct^2\). We also know \(v = v_0\) at \(t = 0\), and we seek to find the tangential and radial components of the force. The tangential force can be found using \(F_{\tan} = ma_{\tan}\).
02

Find Tangential Velocity as a Function of Time

The tangential acceleration is \(a_{\tan} = \frac{dv}{dt}\), so integrate to find the velocity:\[\int dv = \int (b + ct^2) \, dt.\]This results in \(v = v_0 + bt + \frac{ct^3}{3}\) by applying the initial condition \(v = v_0\) at \(t = 0\).
03

Calculate Tangential Force

Use the expression for tangential acceleration in the force equation:\[F_{\tan} = ma_{\tan} = m(b + ct^2).\]
04

Find Radial Acceleration and Force

The radial acceleration is given by \(a_R = \frac{v^2}{r}\). Substitute the expression for tangential velocity we found:\[a_R = \frac{(v_0 + bt + \frac{ct^3}{3})^2}{r}\].Then the radial force is:\[F_{R} = ma_R = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r}\].

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 91Ó°ÊÓ!

Key Concepts

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

Tangential Acceleration
Tangential acceleration is the rate of change of tangential velocity with respect to time. In circular motion, it's crucial because it dictates how quickly the speed along the path changes, not the direction. For the given problem, the tangential acceleration is stated as a function of time: \[ a_{\text{tan}} = b + ct^2 \] where \(b\) and \(c\) are constants. This expression means acceleration can change over time due to the \(ct^2\) term. This scenario indicates a non-linear acceleration path, which is common in more complex circular motion problems.
  • At \(t=0\), the acceleration equals \(b\).
  • As time increases, \(ct^2\) causes the acceleration to grow, showing an increasing tangential effect.
Understanding how tangential acceleration works helps predict how an object accelerates along the circle's path. This insight is necessary to calculate the tangential force exerted on the object as time passes, which is given by:\[ F_{\text{tan}} = ma_{\text{tan}} = m(b+ct^2) \] This equation highlights that the tangential force varies directly with mass and the rate of tangential acceleration.
Radial Force
Radial force, also known as centripetal force, is vital for understanding the forces that act perpendicular to the motion in circular paths. It's responsible for keeping the object moving in a circle by constantly changing the object's direction to point towards the center of the circle. In our problem, the radial acceleration is defined as:\[ a_R = \frac{v^2}{r} \] where \(v\) is tangential velocity and \(r\) is the circle's radius. Radial acceleration is not constant because it's dependent on the square of velocity, which changes as a function of time due to tangential acceleration. After finding the tangential velocity function:\[ v = v_0 + bt + \frac{ct^3}{3} \] we substitute it into the radial acceleration equation:\[ a_R = \frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \] We then compute the radial force:\[ F_{R} = ma_R = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \] The radial force changes over time since both \(v\) and \(a_R\) depend on time, showing a dynamic relation in circular motion. Understanding this concept helps to appreciate how continuously changing forces result in uniform circular motion.
Integration for Velocity
In the context of acceleration and velocity, integration plays an essential role in determining the velocity function when given the acceleration function. For this exercise, the tangential acceleration function:\[ a_{\text{tan}} = b + ct^2 \] Being given the condition that at time \(t=0\), the velocity \(v = v_0\), we integrate the acceleration with respect to time to find velocity:\[ \int dv = \int (b + ct^2) \ dt \] Upon integration, we get:\[ v = v_0 + bt + \frac{ct^3}{3} \]This integration reveals a velocity function that encapsulates the initial velocity and the effects of increasing tangential acceleration over time with terms involving time to the first and third power. Integration transforms our understanding from a static view of instantaneous change to a more dynamic view of movement over time. This step is crucial in physics problems as it allows us to transition from isolated data points to a continuous function, offering a complete understanding of an object's motion in a circular path.

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

(II) A \(975-\mathrm{kg}\) sports car (including driver) crosses the rounded top of a hill (radius = 88.0 \(\mathrm{m} )\) at 12.0 \(\mathrm{m} / \mathrm{s}\) . Determine \((a)\) the normal force exerted by the road on the car, \((b)\) the normal force exerted by the car on the \(72.0-\mathrm{kg}\) driver, and \((c)\) the car speed at which the normal force on the driver equals zero.

The sides of a cone make an angle \(\phi\) with the vertical. A small mass \(m\) is placed on the inside of the cone and the cone, with its point down, is revolved at a frequency \(f\) (revolutions per second) about its symmetry axis. If the coefficient of static friction is \(\mu_{\mathrm{s}}\), at what positions on the cone can the mass be placed without sliding on the cone? (Give the maximum and minimum distances, \(r\), from the axis).

(III) A small block of mass \(m\) rests on the rough, sloping side of a triangular block of mass \(M\) which itself rests on a horizontal frictionless table as shown in Fig. 5-41. If the coefficient of static friction is \(\mu,\) determine the minimum horizontal force \(F\) applied to \(M\) that will cause the small block \(m\) to start moving up the incline.

(I) A jet plane traveling \(1890 \mathrm{~km} / \mathrm{h}(525 \mathrm{~m} / \mathrm{s})\) pulls out of a dive by moving in an arc of radius \(4.80 \mathrm{~km} .\) What is the plane's acceleration in \(g\) 's?

(III) A particle rotates in a circle of radius \(3.80 \mathrm{~m}\). At a particular instant its acceleration is \(1.15 \mathrm{~m} / \mathrm{s}^{2}\) in a direction that makes an angle of \(38.0^{\circ}\) to its direction of motion. Determine its speed \((a)\) at this moment and \((b) 2.00 \mathrm{~s}\) later, assuming constant tangential acceleration.

See all solutions

Recommended explanations on Physics Textbooks

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.