/*! 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 7 A rolling body of mass \(m=4 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A rolling body of mass \(m=4 \mathrm{~kg}\), radius \(R\) and radius of gyration \(k=R / \sqrt{3}\) is placed as a plank which moves with an acceleration \(a_{0}=1 \mathrm{~ms}^{-2}\). Find the frictional force acting on the body if it rolls without sliding. (in \(\mathrm{N}\) ).

Short Answer

Expert verified
The frictional force acting on the body is \( \frac{4}{3} \text{ N} \).

Step by step solution

01

Understanding the Problem

We have a body with mass \( m = 4 \) kg, moving with acceleration \( a_0 = 1 \text{ m/s}^2 \). It rolls without sliding on a plank. We need to find the frictional force acting on it.
02

List Known Quantities

The known quantities are: mass \( m = 4 \text{ kg} \), acceleration \( a_0 = 1 \text{ m/s}^2 \), and radius of gyration \( k = \frac{R}{\sqrt{3}} \).
03

Use the Concept of Rolling Without Slipping

For a body rolling without slipping, the acceleration of the center of mass \( a = a_0 \). Hence, \( a = 1 \text{ m/s}^2 \). Also, for rolling without slipping, \( a = R \cdot \alpha \), where \( \alpha \) is angular acceleration.
04

Calculate the Moment of Inertia

The moment of inertia \( I \) of the body is expressed in terms of the radius of gyration \( k \) as \( I = mk^2 \). Substituting \( k = \frac{R}{\sqrt{3}} \), we have: \[ I = m \left( \frac{R}{\sqrt{3}} \right)^2 = \frac{mR^2}{3} \]
05

Apply Newton's Second Law for Rotation

The net torque \( \tau \) about the center of mass is provided by friction: \( \tau = fR \). By Newton's second law for rotation, \( \tau = I \alpha \). Thus: \[ fR = \frac{mR^2}{3} \alpha \] and \( \alpha = \frac{a}{R} \), simplifying to: \[ f = \frac{mR \alpha}{3} = \frac{mR \cdot a/R}{3} = \frac{ma}{3} \]
06

Calculate the Frictional Force

Substitute the known values into the equation for friction: \[ f = \frac{4 \times 1}{3} = \frac{4}{3} \text{ N} \] Thus, the frictional force \( f \) is \( \frac{4}{3} \text{ N} \).

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Key Concepts

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

Frictional Force
When a body rolls without sliding, like the object in this exercise, it means there is a perfect balance between the linear and rotational movements. Frictional force plays a crucial role in this balance. It's the force that prevents the object from sliding and helps it roll.

In the case of the rolling body, the frictional force acts at the point of contact with the surface. This force provides the necessary torque that enables the rotation of the body. It's important to note that the frictional force does not do work because the point of contact is momentarily at rest as the body rolls.

The frictional force can be calculated using the known values of mass and acceleration. In this example, using the formula derived from Newton's second law for rotation, the frictional force is found to be:\[f = \frac{ma}{3}\]

Substituting in the given values of mass and acceleration, the frictional force is calculated to be \(\frac{4}{3} \text{ N}\). This value shows how much force is needed to keep the body rolling without slipping, perfectly balancing forces and torques.
Moment Of Inertia
Moment of inertia is a concept that quantifies how much torque is needed for a body to achieve a certain angular acceleration. It's analogous to mass in linear motion but for rotational motion.

The moment of inertia depends on how the mass of the object is distributed relative to the axis of rotation. For our object in the exercise, the radius of gyration is provided as \(k = \frac{R}{\sqrt{3}}\). This enables the calculation of the moment of inertia \(I\):

\[I = mk^2 = m\left(\frac{R}{\sqrt{3}}\right)^2 = \frac{mR^2}{3}\]

The moment of inertia is a crucial factor in determining the angular acceleration of the body when subjected to a torque. Here it is used to establish the relationship between frictional force and angular acceleration, which ensures the body rolls without slipping.
Newton's Second Law for Rotation
Newton's second law for rotation is similar to its linear counterpart, detailing how a rotational motion occurs. It states that the net torque acting on a body is equal to the product of its moment of inertia and angular acceleration.

Mathematically, this is expressed as \( \tau = I \alpha \), where \(\tau\) is the net torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. This principle helps in understanding how forces cause changes in the rotational state of a body.

In the context of this problem, the frictional force works as the torque to produce angular acceleration. The equation \(fR = \frac{mR^2}{3} \left(\frac{a}{R}\right)\) is a direct application of this law, connecting the linear acceleration to the angular one. It elegantly ties together mass, acceleration, and radius, showing how they interplay to achieve rolling motion without slipping. The frictional force is vital for maintaining this rotational dynamic, as it provides the needed torque for the body to rotate.

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Most popular questions from this chapter

A horizontal turn table in the form of a disc of radius \(r\) carries a gun at \(G\) and rotates with angular velocity \(\omega_{0}\) about a vertical axis passing through the centre \(O\). The increase in angular velocity of the system if the gun fires a bullet of mass \(m\) with a tangential velocity \(v\) with respect to the gun is (moment of inertia of gun \(+\) table about \(O\) is \(I_{0}\) ) (1) \(\frac{m v r}{I_{0}+m r^{2}}\) (2) \(\frac{2 m v r}{I_{0}}\) (3) \(\frac{v}{2 r}\) (4) \(\frac{m v r}{2 I_{0}}\)

A body is rolling without slipping on a horizontal plane. The rotational energy of the body is \(40 \%\) of the total kinetic energy. Identify the body. (1) Ring (2) Hollow eylinder (3) Solid cylinder (4) Hollow sphere

A solid cylinder of mass \(M\) and radius \(R\) is resting on a horizontal platform (which is parallel to the \(x-y\) plane) with its axis fixed along \(Y\)-axis and free to rotate about its axis. The platform is given a motion in the \(X\)-direction given by \(x=A \cos (\omega t)\). There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is (1) \(\frac{M \omega^{2} A R}{3}\) (2) \(\frac{M \omega^{2} A R}{2}\) (3) \(\frac{2}{3} \times M \omega^{2} A R\) (4) the situation is not possible

A ring of radius \(R\) is first rotated with an angular velocity \(\omega_{0}\) and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is \(\mu\). Time after which its angular speed is reduced to half is (I) \(\frac{\omega_{0} \mu R}{2 g}\) (2) \(\frac{\omega_{0} g}{2 \mu R}\) (3) \(\frac{2 \omega_{0} R}{\mu g}\) (4) \(\frac{\omega_{0} R}{2 \mu g}\)

A uniform rod of mass \(m\) and length \(l\) is placed over a smooth horizontal surface along the \(y\)-axis and is at rest as shown in figure. An impulsive force \(F\) is applied for a small time \(\Delta t\) along \(x\)-direction at point \(A\). The \(x\)-coordinate of end \(A\) of the rod when the rod becomes parallel to \(x\)-axis for the first time is [initially, the coordinate of centre of mass of the rod is \((0,0)]\) (1) \(\frac{\pi l}{12}\) (2) \(\frac{l}{2}\left(1+\frac{\pi}{12}\right)\) (3) \(\frac{l}{2}\left(1-\frac{\pi}{6}\right)\) (4) \(\frac{l}{2}\left(1+\frac{\pi}{6}\right)\)

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