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A solid sphere rests on a horizontal surface. A horizontal impulse is applied at height \(h\) from centre. The sphere starts rolling just after the application of impulse. The ratio \(h / r\) will be (1) \(\frac{1}{2}\) (2) \(\frac{2}{5}\) (3) \(\frac{1}{5}\) (4) \(\frac{2}{3}\)

Short Answer

Expert verified
The ratio \( h / r \) is \( \frac{2}{5} \).

Step by step solution

01

Identify the Given Problem

A solid sphere is resting on a horizontal surface, and an impulse is applied at a height \( h \) from the center. The goal is to find the ratio \( h / r \) where \( r \) is the radius of the sphere.
02

Apply the Impulse-Momentum Theorem

The impulse causes the sphere to have a translational velocity \( v \) and an angular velocity \( \omega \) such that just after the impulse, the sphere starts rolling without slipping. For rolling without slipping, the condition is \( v = r\omega \).
03

Use the Moment of Inertia for a Solid Sphere

For a solid sphere, the moment of inertia \( I \) about its center is \( \frac{2}{5}mr^2 \). The angular impulse is equal to \( I\omega \) and acts at a distance \( h \) from the center, which can be related as \( I\omega = mvr \).
04

Solve the Equations

Since \( v = r\omega \), substitute \( v \) in terms of \( \omega \) into the equation for angular impulse: \( I\omega = m\omega r^2 \). Using \( I = \frac{2}{5}mr^2 \), equate and simplify to find \( h = \frac{2}{5}r \).
05

Calculate the Ratio

The ratio \( \frac{h}{r} \) can be calculated as: \( \frac{h}{r} = \frac{\frac{2}{5}r}{r} = \frac{2}{5} \). Thus, the answer is \( \frac{2}{5} \).

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Key Concepts

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

Understanding Rolling Motion
Rolling motion involves a combination of translational movement along a surface and rotation around an axis. When a sphere rolls without slipping on a plane, every point on the sphere follows a circular path, and the point of contact with the surface is momentarily at rest at each instant in time. This is an important concept because it defines the relationship between the translational velocity of the sphere's center of mass and its angular velocity:
  • Translational velocity (\( v \)) is the speed at which the center of the sphere moves along the surface.
  • Angular velocity (\( \omega \)) is the rate at which the sphere rotates around its axis.
For a sphere rolling without slipping, these two velocities are related by the equation \( v = r\omega \), where \( r \) is the radius of the sphere. This condition ensures that the sphere rolls smoothly, and is crucial for solving problems involving rolling motion as it links linear and angular movements.
Decoding Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational movement. It is similar to mass in linear motion, but instead of considering just mass, moment of inertia also takes into account the distribution of mass relative to the axis of rotation. For a solid sphere, the moment of inertia \( I \) about its center is given by the formula:\[I = \frac{2}{5}mr^2\]where \( m \) is the mass and \( r \) is the radius of the sphere.

The moment of inertia is pivotal in calculating how impulses affect rotational motion. When an impulse applies a force at a height \( h \), it causes both translational and rotational accelerations. The formula \( I\omega = mvr \) links the moment of inertia to angular momentum, thereby helping in deducing how quickly an object will begin rotating when subjected to a force.
Understanding Angular Velocity
Angular velocity (\( \omega \)) indicates how fast something is spinning. It tells you how many angles an object passes through in a given time, measured typically in radians per second. In the context of a rolling sphere, angular velocity is directly tied to the sphere's translational motion, creating a harmonious dance between how fast it spins and how fast it rolls forward.

The relation \( v = r\omega \) is crucial because:
  • It assures us that the rolling sphere covers a certain linear distance while making each full rotation.
  • Helps us determine the speed at which the sphere's center of mass is moving straight along the surface.
Grasping angular velocity is crucial for synchronizing rotational and translational movements—for instance, ensuring a sphere rolls without slipping after a horizontal impulse. This concept ties back to conserving angular momentum and ensuring energy is transmitted efficiently between movement types.

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Most popular questions from this chapter

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) \(\frac{l}{2}\left(1+\frac{\pi}{6}\right)\)

A uniform smooth rod (mass \(m\) and length \(l\) ) placed on a smooth horizontal floor is hit by a particle (mass \(m\) ) moving on the floor, at a distance \(l / 4\) from one end elastically \((e=1)\). The distance travelled by the centre of the rod after the collision when it has completed three revolutions will be (1) \(2 \pi l\) (2) cannot be determined (3) \(\pi l\) (4) none of these

A constant external torque \(\tau\) acts for a very brief period \(\Delta t\) on a rotating system having moment of inertia \(I\), then (1) the angular momentum of the system will change by \(\tau \Delta t\) (2) the angular velocity of the system will change by \(\tau \Delta t / I\) (3) if the system was initially at rest, it will acquire rotational kinetic energy \((\tau \Delta t)^{2} / 2 I\) (4) the kinetic energy of the system will change by \((\tau \Delta t)^{2} / 2 I\)

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 \(\mathrm{N}\) ).

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) \(\omega_{0}\)

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