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A solid sphere, a hollow sphere and a disc, all having the same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. The least time will be taken in reaching the bottom by (1) the solid sphere (2) the hollow sphere (3) the disc (4) all will take the same time

Short Answer

Expert verified
All will take the same time.

Step by step solution

01

Understanding Rotation and Translation Dynamics

To determine which object reaches the bottom of the incline fastest, we must consider both translational and rotational motion. In this case, due to the insufficient friction, the objects will slide down the incline with slipping involved. Hence, the rolling motion isn't purely rolling.
02

Calculating Acceleration for Sliding Objects

For sliding objects, the net force down the incline is given by the component of gravitational force: \[ F = mg \sin(\theta) \]where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the incline angle. This gives accelerations for all objects as: \[ a = g \sin(\theta) \]Since friction is not sufficient, rotational dynamics are not considered in calculating sliding acceleration, meaning all objects have the same acceleration.
03

Comparing Mass Moment of Inertia

Usually, the distribution of mass (moment of inertia) affects how an object rolls down an incline. However, since slipping occurs and pure rolling isn't achieved, moment of inertia does not come into play for determining the time it takes each object to slide to the bottom.
04

Evaluating Time to Reach the Bottom

Since all objects have the same acceleration as derived from sliding down the incline: \[ a = g \sin(\theta) \]and they all start from rest, the time taken to reach the bottom will be the same for each of them. The formula to calculate time is the same for each object starting at initial velocity of zero: \[ t = \sqrt{\frac{2s}{a}} \]where \( s \) is the distance moved along the incline.

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Key Concepts

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

Rotational Dynamics
When we talk about rotational dynamics, we're mainly concerned with how objects rotate about an axis. Key concepts include angular velocity, angular acceleration, and the influence of torque. Unlike regular linear motion, rotational dynamics consider how a force can cause an object to spin, calculating the effects using parameters specific to rotational movement.

For objects on an incline, like the solid sphere, hollow sphere, and disc mentioned, rotational dynamics generally play a crucial role. The spinning movement is usually influenced by forces acting off-center, causing the objects to rotate as they accelerate down the plane. However, in situations where friction isn't sufficient to cause pure rolling, the effects of rotational dynamics are reduced.

In the given exercise, since friction does not allow for pure rolling and slipping occurs, rotational dynamics take a back seat. Thus, the objects primarily slide rather than roll, making their rotational characteristics (like spinning speed or angular momentum) less significant.
Translational Motion
Translational motion describes the movement of an object in space without any rotation. It's all about straight-line movement and how an object's center of mass changes position. In this context, it simplifies to Newton's second law, where force equals mass times acceleration, without any rotational components being considered.

In the scenario of objects on an incline, translational motion is driven by gravity. The force pulling them down is a component of their weight, calculated as \[F = mg \sin(\theta)\], where \( m \) is mass, \( g \) is gravity, and \( \theta \) is the angle of the incline. When slipping occurs, as it does here, this linear force directly dictates their movement down the slope.

The objects have the same mass and experience the same gravitational force component, so they end up having an identical acceleration. Hence, the focus on translational motion explains why all objects reach the bottom at the same time, as each has the same starting conditions and accelerations.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. Think of it as an equivalent to mass in linear motion, but for spinning around an axis. How mass is distributed around this axis affects the moment of inertia.

In this problem, the solid sphere, hollow sphere, and disc each have different moments of inertia due to their shapes. Normally, this would mean they would roll differently down an incline. A solid sphere, with its mass more concentrated towards its center, has a lower moment of inertia than a hollow sphere or disc. This typically allows it to roll faster if they were purely rolling.

However, because the incline lacks sufficient friction for pure rolling, the moment of inertia doesn't impact the time taken to get to the bottom. The objects slide rather than roll, so their rotational resistance is bypassed in this scenario. Therefore, differences in the moment of inertia become irrelevant in determining which objects take the least time to reach the bottom. Instead, their sliding acceleration, which is the same for each, dictates their speed.

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Most popular questions from this chapter

A disc is given an initial angular velocity \(\omega_{0}\) and placed on a rough horizontal surface as shown in figure. The quantities which will not depend on the coefficient of friction is/are (1) the time until rolling begins (2) the displacement of the disc until rolling begins (3) the velocity when rolling begins (4) the work done by the force of friction

A uniform disc is rotating at a constant speed in a vertical plane about a fixed horizontal axis, passing through the centre of the disc. A piece of disc from its rim detaches itself from the disc at the instant when it is on the same horizontal with the centre of the disc and moving upwards. Then about the fixed axis, the angular speed of the (1) remaining disc remains unchanged (2) remaining disc decreases (3) remaining disc increases (4) broken away piece decreases initially and later increases

A yo-yo is placed on a rough horizontal surface and a constant force \(F\), which is less than its weight, pulls it vertically. Due to this (1) friction force acts towards left, so it will move towards left (2) friction force acts towards right, so it will move towards right (3) it will move towards left, so friction force acts towards left (4) it will move towards right so friction force acts towards right

A disc of circumference \(s\) is at rest at a point \(A\) on a horizontal surface when a constant horizontal force begins to act on its centre. Between \(A\) and \(B\) there is sufficient friction to prevent slipping and the surface is smooth to the right of \(B\), \(A B=s\). The disc moves from \(A\) to \(B\) in time \(T\). To the right of \(B\) (1) the angular acceleration of the disc will disappear, linear acceleration will remain unchanged (2) linear acceleration of the disc will increase (3) the disc will make one rotation in time \(T / 2\) (4) the disc will cover a distance greater than \(s\) in further time \(T\)

A loop and a disc roll without slipping with same linear velocity \(v\). The mass of the loop and the disc is same. If the total kinetic energy of the loop is \(8 \mathrm{~J}\), find the kinetic energy of the disc (in \(J\) ).

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