/*! 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 35 A force \(F\) acts tangentially ... [FREE SOLUTION] | 91Ó°ÊÓ

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A force \(F\) acts tangentially at the highest point of a disc of mass \(m\) kept on a rough horizontal plane. If the disc rolls without slipping, the acceleration of centre of the disc is (1) \(\frac{2 F}{3 m}\) (2) \(\frac{10 F}{7 m}\) (3) zero (4) \(\frac{4 F}{3 m}\)

Short Answer

Expert verified
The acceleration of the center of the disc is \(\frac{2F}{3m}\) (Option 1).

Step by step solution

01

Understand the Problem

We have a disc of mass \(m\) on a rough horizontal surface. A tangential force \(F\) is applied at the top. Since the disc rolls without slipping, we need to find the acceleration of its center of mass.
02

Analyze Rolling Without Slipping

For rolling without slipping, the point of contact with the surface has zero velocity and there is a relationship between translational and rotational motion: \( a = r\alpha \), where \(a\) is the linear acceleration of the center of mass, \(r\) is the radius of the disc, and \(\alpha\) is the angular acceleration.
03

Apply Newton's Second Law - Translation

Apply Newton's second law for translation: \( F - f = ma \), where \(f\) is the frictional force.
04

Apply Newton's Second Law - Rotation

Apply Newton's second law for rotation: \( Fa - fR = I\alpha \), where \(I\) is the moment of inertia of the disc about its center, which is \( \frac{1}{2}mR^2 \). Using \( \alpha = \frac{a}{R} \), we have \( FR - fR = \frac{1}{2}mR^2 \cdot \frac{a}{R} \).
05

Solve the Equations

Combine the translational \( F - f = ma \) and rotational \( FR - fR = \frac{1}{2}mRa \) equations. Replace \(fR\) with its equivalent from the rotational equation: \(F \cdot R - \frac{1}{2}mRa = fR \). This gives us \(F - \frac{f}{R} = \frac{3}{2}ma\) and substituting back into the equation for translation we solve for \(a\): \[F = ma + \frac{1}{2}ma\Rightarrow a = \frac{2F}{3m}.\]
06

Confirm Consistency with Non-slipping

The solution must satisfy the rolling without slipping condition, which integrates naturally across steps and confirms that the answer aligns with fundamental dynamics.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that relates force, mass, and acceleration. In simple terms, it states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is represented as \[ F = ma \] Where:
  • \( F \) is the net force applied to the object.
  • \( m \) is the mass of the object.
  • \( a \) is the acceleration.
When dealing with rolling motion, it’s important to consider both the linear and rotational effects of forces. In the context of the disc in the problem, a force \( F \) applied tangentially causes both translational movement of the disc's center of mass and rotational movement around its axis.
Combining both linear and rotational elements ensures that the dynamics of rolling without slipping are accurately represented.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion, much like mass is a measure of resistance to changes in linear motion. For a disc rolling without slipping on a surface, the moment of inertia is crucial in determining how easily it can be set into rotational motion under the influence of a tangential force.
The moment of inertia \( I \) for a disc about its center is given by:\[ I = \frac{1}{2}mR^2 \]Where:
  • \( m \) is the mass of the disc.
  • \( R \) is the radius of the disc.
This formula shows that the moment of inertia increases with the mass of the disc and the square of its radius. Thus, heavier and larger discs will have more inertia and resist changes to their motion more strongly. Understanding moment of inertia is key to analyzing how a disc spins and rolls.
Frictional Force
Frictional force is the force that resists the sliding of the disc and is essential for rolling without slipping. In this context, friction acts at the point of contact between the disc and the surface, preventing any slipping and ensuring that motion is transferred smoothly from linear to rotational.
When a tangential force \( F \) is applied, friction acts opposite. It helps in transferring some of the linear force into rotational movement, satisfying the condition for rolling without slipping. If there were no friction, the disc would slide instead of rolling.
  • Static friction is what enables rolling without slipping.
  • It contributes to the net force and affects the equations used to calculate the acceleration of the disc's center of mass.
Analyzing friction in this problem is vital because it ensures the correct relationship between translational and rotational dynamics.
Angular Acceleration
Angular acceleration \( \alpha \) refers to the rate at which the angular velocity of the disc changes over time. It plays a significant role in describing how the disc spins as it rolls without slipping.
Angular acceleration is linked to linear acceleration \( a \) through the relationship: \[ a = R \alpha \] Where:
  • \( a \) is the linear acceleration of the disc's center of mass.
  • \( R \) is the radius of the disc.
  • \( \alpha \) is the angular acceleration.
This relationship is important because it shows how changes in the rotational speed of the disc relate directly to its linear motion. In our problem, we must ensure that the value of \( \alpha \) derived from our forces and moments aligns with the linear acceleration to ensure consistent rolling without slipping. Understanding angular acceleration helps in accurately predicting the disc's behavior under the influence of the applied force \( F \).

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Most popular questions from this chapter

A horizontal plane supports a fixed vertical cylinder of radius \(R\) and a particle is attached to the cylinder by a horizontal thread \(A B\) as shown in figure. The particle initially rest on a horizontal plane. A horizontal velocity \(v_{0}\) is imparted to the particle, normal to the threading during subsequent motion. Point out the false statements. (1) Angular momentum of particle about \(O\) remains constant. (2) Angular momentum about \(B\) remains constant. (3) Momentum and kinetic energy both remain constant. (4) Kinetic energy remains constant.

A uniform thin stick of mass \(M=24 \mathrm{~kg}\) and length \(L\) rotates on a frictionless horizontal plane, with its centre of mass stationary. A particle of mass \(m\) is placed on the plane at a distance \(x=L / 3\) from the centre of the stick. This stick hits the particle elastically. Find the value of \(m / M\) so that after the collision, there is no rotational motion of the stick.

A smooth uniform rod of length \(L\) and mass \(M\) has two identical beads ( 1 and 2) of negligible size, each of mass \(m\) which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity \(\omega_{0}\) about an axis perpendicular to the rod and is passing through its midpoint. There are no external forces when the beads reach the ends of the rod, the angular velocity of the system is (1) \(\frac{M \omega_{b}}{M+6 m}\) (2) \(\frac{M \omega_{0}}{m}\) (3) \(\frac{M \omega_{0}}{M+12 m}\) (4) \(\omega_{0}\)

A thin uniform rod of mass \(m\) and length \(l\) is kept on a smooth horizontal surface such that it can move freely. At what distance from centre of rod should a particle of mass \(m\) strike on the rod such that the point \(P\) at a distance \(I / 3\) from the end of the rod is instantaneously at rest just after the elastic collision? (1) \(/ / 2\) (2) \(1 / 3\) (3) \(l / 6\) (4) \(l / 4\)

Two bodies with moments of inertia \(I_{1}\) and \(I_{2}\left(I_{1}>I_{2}\right)\) have equal angular momenta. If their kinetic energies of rotation \(\operatorname{are} E_{1}\) and \(E_{2}\), respectively, then (i) \(E_{1}=E_{2}\) (2) \(E_{1}E_{2}\) (4) \(E_{1} \geq E_{2}\)

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