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Chapter 10: Problem 72

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

Short Answer

Expert verified
The acceleration \(a\) is \(\frac{F}{2M}\) and the friction force \(f\) is \(\frac{F}{2}\).

Step by step solution

01

Identify Forces and Analyze the System

We start by analyzing the forces acting on the lawn roller. Since a horizontal force \(F\) is applied and it rolls without slipping, the torque due to the frictional force provides the rotational motion. The forces we consider include the applied force \(F\), frictional force \(f\), and the gravitational force \(Mg\).
02

Apply Newton's Second Law for Translation

Newton's second law for linear motion states that the net force equals mass times acceleration: \(F - f = Ma\), where \(a\) is the linear acceleration of the center of mass.
03

Apply Newton's Second Law for Rotation

The torque \(\tau\) due to friction causes the roller to rotate: \(fR = I\alpha\), where \(R\) is the radius of the roller, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. For a thin-walled, hollow cylinder, \(I = MR^2\).
04

Relate Linear and Angular Acceleration

Since the roller rolls without slipping, the linear acceleration \(a\) and angular acceleration \(\alpha\) are related by \(a = R\alpha\). Substituting \(\alpha = \frac{a}{R}\) into the rotational equation gives: \(fR = MR^2 \frac{a}{R}\).
05

Solve for Friction Force

Simplify the torque equation to find \(f = Ma\).
06

Substitute Friction into Translation Equation

Substitute \(f = Ma\) into the translation equation: \(F - Ma = Ma\).
07

Solve for Linear Acceleration

Rearrange \(F - Ma = Ma\) to solve for \(a\):\[a = \frac{F}{2M}\].
08

Calculate Friction Force

Using the equation \(f = Ma\), plug in \(a = \frac{F}{2M}\) to find \(f = M \times \frac{F}{2M} = \frac{F}{2}\).

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Key Concepts

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

Friction Force
Friction is a resistive force. It acts opposite to the direction of motion. In the case of the rolling lawn roller, friction is vital for the rotational motion. Without it, the roller would simply slide instead of rolling.
  • When an object rolls without slipping, static friction is at play, as it ensures the point of contact does not slide across the surface.
  • In the solution, the frictional force provides the necessary torque for rotation, which is described by the equation for torque: \(\tau = fR\).
  • This force is equal to the mass times the linear acceleration, \(f = Ma\), due to its role in the rotational dynamics.
Understanding that friction is not always a negative force but a necessary one for certain motions, like rolling, often helps students appreciate its role. Analyzing forces and torques together gives a more complete picture of the problem at hand.
Rotational Motion
Rotational motion is motion around an axis. For objects like our lawn roller, rotational motion involves understanding how the entire body rotates around a central point. An important aspect to grasp is how the same forces affect not only linear motion but also rotational motion.
  • The torque, a product of the friction force and radius (\(\tau = fR\)), causes the roller to rotate.
  • In rotational motion, the moment of inertia (\(I\)), plays a role akin to mass in linear motion. For a thin-walled, hollow cylinder like the lawn roller, \(I = MR^2\).
  • Newton's second law applies to rotational motion as \(\tau = I\alpha\), allowing you to solve for quantities like angular acceleration.
Linking linear motion with rotational principles helps reinforce understanding of how bodies move in different but interconnected ways. Considering rotational effects is critical when a problem involves turning or rolling.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It is similar to linear acceleration but applies to rotational motion.
  • In the context of the lawn roller, angular and linear accelerations are related through the equation \(a = R\alpha\), where \(\alpha\) is angular acceleration and \(a\) is linear acceleration.
  • This relationship means linear acceleration changes when the roller rotates, establishing a connection between translational and rotational motion.
  • Knowing \(a = R\alpha\) helps in solving for acceleration values, and in our exercise also led us to find \(f = Ma\).
Understanding this relationship is critical for analyzing systems where rotation and translation are related. By examining both components, solving physics problems involving rotational dynamics becomes more intuitive.

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Most popular questions from this chapter

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 \(\mathrm{N}\) is applied to the rim of the wheel. The wheel has radius 0.120 \(\mathrm{m} .\) Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.42). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N} .\) What is the radius of the circle when the string breaks?

A uniform ball of radius \(R\) rolls without slipping between two rails such that the horizontal distance is \(d\) between the two contact points of the rails to the ball. (a) In a sketch, show that at any instant \(v_{\mathrm{cm}}=\omega \sqrt{R^{2}-d^{2} / 4} .\) Discuss this expression in the limits \(d=0\) and \(d=2 R\) . (b) For a uniform ball starting from rest and descending a vertical distance \(h\) while rolling without slipping down a ramp, \(v_{\mathrm{cm}}=\sqrt{10 g h / 7}\) . Replacing the ramp with the two rails, show that $$ v_{\mathrm{cm}}=\sqrt{\frac{10 g h}{5+2 /\left(1-d^{2} / 4 R^{2}\right)}} $$ In each case, the work done by friction has been ignored. (c) Which speed in part (b) is smaller? Why? Answer in terms of how the loss of potential energy is shared between the gain in translational and rotational kinetic energies. (d) For which value of the ratio \(d / R\) do the two expressions for the speed in part (b) differ by 5.0\(\% ?\) By 0.50\(\% ?\)
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