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A plank \(P\) is placed on a solid cylinder \(S\), which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of \(P\) to the kinetic energy of \(S\) is (1) \(1: 1\) (2) \(2: 1\) (3) \(8: 3\) (4) \(11: 8\)

Short Answer

Expert verified
The ratio of kinetic energy of \(P\) to \(S\) is \(8:3\).

Step by step solution

01

Understanding the Motion of Cylinder

The solid cylinder rolls without slipping, which means its total kinetic energy includes both translational and rotational components. The translational kinetic energy is due to its center of mass moving linearly, and the rotational kinetic energy is due to the cylinder spinning around its axis.
02

Express Kinetic Energies

For the cylinder's kinetic energy, we have translational kinetic energy as \( \frac{1}{2} m v^2 \), and rotational kinetic energy as \( \frac{1}{2} I \omega^2 \). For a solid cylinder, the moment of inertia \( I = \frac{1}{2} m R^2 \), where \( m \) is the mass of the cylinder, and \( R \) is the radius. Since there is no slipping, \( v = \omega R \), which simplifies the rotational kinetic energy to \( \frac{1}{4} m v^2 \). Thus, the total kinetic energy of the cylinder \( S \) is \( \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2 \).
03

Identify Plank's Kinetic Energy

The plank \( P \) moves with the same linear speed \( v \) as the top surface of the cylinder to avoid slipping. Hence, its kinetic energy is simply its translational kinetic energy: \( \frac{1}{2} m v^2 \).
04

Calculate the Ratio of Kinetic Energies

We compare the kinetic energy of the plank \( P \) to the total kinetic energy of the cylinder \( S \). So the ratio is \( \frac{\frac{1}{2} m v^2}{\frac{3}{4} m v^2} = \frac{2}{3} \).
05

Convert the Ratio to Standard Form

The ratio \( \frac{2}{3} \) can be rewritten as a whole number ratio: \( 8:3 \). Thus, the ratio of kinetic energy of \( P \) to \( S \) is \( 8:3 \).

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Key Concepts

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

Rotational Dynamics
In the realm of physics, understanding how objects rotate is crucial, and this is where rotational dynamics comes in. It is the study of the motion of objects that have rotation around an axis. This concept is not only vital in solving problems involving wheels or gears but also essential in understanding complex systems like the solar system.

Two important elements in rotational dynamics are the angular velocity (\( \omega \)) and angular acceleration. Angular velocity is the rate at which an object spins about an axis, while angular acceleration is the rate of change of angular velocity. The motion of the solid cylinder in the given exercise is a perfect example of rotational dynamics at work.

When the cylinder rolls without slipping, it combines both translational and rotational motion. This no-slip condition maintains a relationship between linear speed (\( v \)) and angular speed (\( \omega \)) by the equation \( v = \omega R \), where \( R \) is the radius of the cylinder. This interrelationship is key, as it links rotational motion to translational motion.
Translational Motion
Translational motion refers to the movement of an object in space from one point to another. Unlike rotational motion, this type of motion is linear and does not involve spinning around an axis. In the context of our exercise, both the plank and the cylinder exhibit translational motion.

The solid cylinder rolls without slipping, which means its center of mass translates along a horizontal plane. This translation contributes to its kinetic energy, specifically the translational kinetic energy component, expressed as:
  • \( \frac{1}{2} m v^2 \) - where \( m \) is mass and \( v \) is velocity.
The plank, placed on top of the cylinder, also moves with the same linear velocity as the cylinder’s top surface to maintain this no-slipping condition, which ensures seamless translational motion for both.

For the plank, its kinetic energy is purely translational since it does not have rotational movement like the cylinder. Understanding these linear properties of motion helps us find the individual kinetic energies and, eventually, the ratio of these energies.
Moment of Inertia
The moment of inertia, sometimes referred to as the rotational inertia, is a measure of an object's resistance to change in its rotational motion around an axis. It's like mass in translational motion, but specifically for rotation.

For the solid cylinder in the exercise, the moment of inertia (\( I \)) is calculated using the formula \( I = \frac{1}{2} m R^2 \).
  • Where \( m \) is the mass of the cylinder.
  • \( R \) is the radius.
The low moment of inertia in this case means that the cylinder doesn’t require much torque to start spinning.

Understanding the moment of inertia is key when calculating rotational kinetic energy, expressed as:
  • \( \frac{1}{2} I \omega^2 \) - shows how the cylinder’s mass distribution and angular velocity contribute to its rotational energy.
By coupling it with the translational kinetic energy, you can find the total kinetic energy of the rolling cylinder. This helps in drawing comparisons and calculating ratios, as seen in the exercise where the cylinder’s kinetic energy is more than purely translational due to its rotational component.

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Most popular questions from this chapter

A small bead of mass \(m\) moving with velocity \(v\) gets threaded on a stationary semicircular ring of mass \(m\) and radius \(R\) kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system? (1) \(\frac{v}{R}\) (2) \(\frac{2 v}{R}\) (3) \(\frac{v}{2 R}\) (4) \(\frac{3 v}{R}\)

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 uniform rod is resting freely over a smooth horizontal plane. A particle moving horizontally strikes at one end of the rod normally and gets stuck. Then (1) the momentum of the particle is shared between the particle and the rod and remains conserved (2) the angular momentum about the mid-point of the rod before and after the collision is equal (3) the angular momentum about the centre of mass of the combination before and after the collision is equal (4) the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle

Two discs, each having moment of inertia \(5 \mathrm{~kg} \mathrm{~m}^{2}\) about its central axis, rotating with speeds \(10 \mathrm{rad} \mathrm{s}^{-1}\) and \(20 \mathrm{rad} \mathrm{s}^{-1}\), are brought in contact face to face with their axes of rotation coincided. The loss of kinetic energy in the process is (1) \(2 \mathrm{~J}\) (2) \(5 \mathrm{~J}\) (3) \(125 \mathrm{~J}\) (4) \(0 \mathrm{~J}\)

A uniform sphere of mass \(m\), radius \(r\) and moment of inertia \(I\) about its centre moves along the \(x\)-axis as shown in figure. Its centre of mass moves with velocity \(=v_{0}\), and it rotates about its centre of mass with angular velocity \(=\omega_{0}\). Let \(\vec{L}=\left(I \omega_{0}+m v_{0} r\right)(-\hat{k})\). The angular momentum of the body about the origin \(O\) is (1) \(\vec{L}\), only if \(v_{0}=\omega_{0} r\) (2) greater than \(\vec{L}\), if \(v_{0}>\omega_{0} r\) (3) less than \(\vec{L}\), if \(v_{0}>\omega_{0} r\) (4) \(\vec{L}\), for all values of \(\omega_{0}\) and \(v_{o}\)

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