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A disc of mass \(M\) and radius \(R\) rolls without slipping on a A horizontal surface. If the velocity of its centre is \(v_{0}\), then the | toral angular momentum of the disc about a fixed point \(P\) at a height \(3 / 2 R\) above the centre \(C\) (1) increases continuously as the disc moves away (2) decreases continuously as the disc moves away (3) is equal to \(2 M R v_{0}\) (4) is equal to \(M R v_{0}\)

Short Answer

Expert verified
The total angular momentum is \(2 M R v_{0}\).

Step by step solution

01

Understanding the Problem

We need to calculate the total angular momentum of a rolling disc about a point that is above the center of the disc by a distance of \( \frac{3}{2} R \). The disc rolls without slipping, and we know the velocity of its center \( v_0 \).
02

Concepts and Formulas

For a rolling object, angular momentum \( L \) about a given point is the sum of angular momentum due to translational motion and the angular momentum due to rotational motion. For a rolling disc, these can be calculated as \( L = I_C \omega + M v_0 r_ \perp \). The mass \( M \) and radius \( R \) of the disc are important, and the moment of inertia \( I_C \) for a disc about its center is \( \frac{1}{2} M R^2 \). The condition for rolling without slipping is \( \omega = \frac{v_0}{R} \).
03

Angular Momentum from Rotational Motion

We first calculate the angular momentum from the disc's rotational motion around its center, given by \( L_{rotation} = I_C \omega = \frac{1}{2} M R^2 \cdot \frac{v_0}{R} = \frac{1}{2} M R v_0 \).
04

Angular Momentum from Translational Motion

For the translational component, the perpendicular distance from point \( P \) to the line of motion of \( C \) is \( \frac{3}{2} R \). Thus, the angular momentum due to the motion of the center of mass is \( L_{translation} = M v_0 \cdot \frac{3}{2} R = \frac{3}{2} M R v_0 \).
05

Calculate Total Angular Momentum

The total angular momentum \( L \) about point \( P \) is the sum of the rotational and translational angular momentums: \( L = L_{rotation} + L_{translation} = \frac{1}{2} M R v_0 + \frac{3}{2} M R v_0 = 2 M R v_0 \). So, the correct answer is (3).

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Key Concepts

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

Moment of Inertia
Moment of inertia, often symbolized by the letter \( I \), is a crucial concept in rotational dynamics. It can be thought of as the rotational equivalent of mass in linear dynamics. Just as mass measures an object's resistance to linear acceleration, the moment of inertia measures an object's resistance to rotational acceleration. It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation.

For a solid disc, which is a common shape in physics problems, the moment of inertia about its central axis is given by the formula \( I_C = \frac{1}{2} M R^2 \). Here, \( M \) is the mass and \( R \) is the radius of the disc. This formula reveals an important aspect: the mass farther from the axis contributes more to the moment of inertia, stressing the significance of mass distribution.

Understanding moment of inertia is key to solving problems in rotational motion, as it forms part of the angular momentum calculations. By manipulating this variable, we can explore how different mass distributions affect the rotational behavior of objects.
Rolling Motion
Rolling motion is a common type of movement that combines both translational and rotational motion. When an object like a disc or wheel rolls without slipping on a surface, every point on its circumference that comes into contact with the surface has zero velocity relative to the surface at any instant. This condition is known as rolling without slipping.

For objects rolling without slipping, there is a relationship between linear velocity \( v \) and angular velocity \( \omega \). This is given by \( \omega = \frac{v}{R} \), where \( R \) is the radius of the object. This equation ensures that the point of contact does not slip as the object rolls.

When calculating angular momentum in rolling motion, the contribution comes from both rotational and translational components. Thus, understanding the mechanics of rolling motion is important for accurately determining the rotational dynamics of objects on surfaces.
Rotational Motion
Rotational motion refers to the movement of an object about an axis. In physics, this includes the spin of discs, wheels, and even celestial bodies. The angular momentum of an object in rotational motion around an axis is a vector quantity that reflects the distribution of mass and speed with which parts of an object rotate.

For a rolling disc, the rotational motion around its center is just one part of its overall motion. The angular momentum from rotational motion is expressed as \( L_{rotation} = I_C \omega \) which translates to \( \frac{1}{2} M R v_0 \) when applying the no-slip condition. This captures how the disc spins about its axis.

Rotational motion is closely related to translational motion in the context of rolling objects. The combined effect of these motions explains how such objects navigate different paths and react to external forces. Comprehending rotational motion is essential for predicting and understanding the behavior of rotating objects in various physics scenarios.

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Most popular questions from this chapter

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) \(l / 4\)

A solid ball of mass \(m\) and radius \(r\) spinning with angular velocity \(\omega\) falls on a horizontal slab of mass \(M\) with rough upper surface (coefficient of friction \(\mu\) ) and smooth lower surface. Immediately after collision the normal component of velocity of the ball remains half of its value just before collision and it stops spinning. Find the velocity of the sphere in horizontal direction immediately after the impact \((\) given: \(R \omega=5\) ).

Two horizontal discs of different radii are free to rotate about their central vertical axes. One is given some angular velocity, the other is stationary. Their rims are now brought in contact. There is friction between the rims. Then (1) the force of friction between the rims will disappear when the discs rotate with equal angular speeds (2) the force of friction between the rims will disappear when they have equal linear velocities (3) the angular momentum of the system will be conserved (4) the rotational kinetic energy of the system will not be conserved

An impulse \(J\) is applied on a ring of mass \(m\) along a line passing through its centre \(O\). The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is (1) \(\mathrm{J} / \mathrm{m}\) (2) \(J / 2 m\) (3) \(J / 4 m\) (4) \(J / 3 m\)

A thin uniform rod of mass \(m\) and length \(l\) is free to rotate about its upper end. When it is at rest, it receives an impulse \(J\) at its lowest point, normal to its length. Immediately after impact, (1) the angular momentum of the rod is \(J l\) (2) the angular velocity of the rod is \(3 \mathrm{~J} / \mathrm{ml}\) (3) the kinetic energy of the rod is \(3 J / 2 m\) (4) the linear velocity of the midpoint of the rod is \(3 J / 2 m\)

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