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Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.

Short Answer

Expert verified
The solid sphere will acquire a greater angular speed.

Step by step solution

01

Understand Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. For a hollow cylinder about its axis, the moment of inertia is \(I_{cylinder} = mr^2\), and for a solid sphere, it's \(I_{sphere} = \frac{2}{5}mr^2\), where \(m\) is mass and \(r\) is radius. This means the sphere has a smaller moment of inertia.
02

Apply the Equation for Angular Acceleration

The torque \( \tau \) on an object is related to its moment of inertia and angular acceleration \( \alpha \) by the equation \( \tau = I \alpha \). With equal torque on both objects, \( \alpha_{cylinder} = \frac{\tau}{I_{cylinder}} = \frac{\tau}{mr^2} \) and \( \alpha_{sphere} = \frac{\tau}{I_{sphere}} = \frac{5\tau}{2mr^2} \). Calculating these, the sphere has a larger angular acceleration.
03

Determine the Final Angular Speed

The final angular speed \( \omega \) after time \( t \) is given by \( \omega = \alpha t \). Hence, \( \omega_{cylinder} = \frac{\tau}{mr^2}t \) and \( \omega_{sphere} = \frac{5\tau}{2mr^2}t \). Since \( \alpha_{sphere} > \alpha_{cylinder} \), \( \omega_{sphere} > \omega_{cylinder} \).
04

Conclusion

The solid sphere, having a lower moment of inertia, will acquire a greater angular speed compared to the hollow cylinder when torques of equal magnitude are applied. Thus, for equal torque and time, objects with a smaller moment of inertia reach higher angular speeds.

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Key Concepts

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

Torque
Torque can be thought of as the rotational equivalent of linear force. When we apply a torque to an object, it tends to produce rotational motion. Just like force can cause an object to accelerate, torque causes an object to experience angular acceleration. In our scenario, we are applying equal torques to both a hollow cylinder and a solid sphere. But what is this mysterious torque?Torque is defined mathematically as the product of force and the distance from the pivot point at which the force is applied. If you think about using a wrench, the torque is greater if you apply the same force farther from the nut. This is expressed as:\[ \tau = r \, F \]where:
  • \(\tau\) is the torque,
  • \(r\) is the distance from the axis of rotation (the lever arm), and
  • \(F\) is the force applied.
In the exercise, both objects experience the same torque, which means the same rotational force is trying to spin the cylinder and the sphere. But their responses, described by their angular accelerations and angular speeds, depend on their moments of inertia, which we'll explore next.
Angular Acceleration
Angular acceleration is a measure of how quickly an object changes its angular speed. It is similar to linear acceleration, but applies to rotational movement. When you apply a torque to an object, its angular acceleration depends on the torque and the object's moment of inertia (which is its resistance to rotation).Angular acceleration is represented by the Greek letter \( \alpha \) and is determined by the equation:\[ \alpha = \frac{\tau}{I} \]where:
  • \( \alpha \) is the angular acceleration,
  • \( \tau \) is the torque applied, and
  • \( I \) is the moment of inertia.
From the exercise, since the solid sphere has a smaller moment of inertia compared to the hollow cylinder, it means with the same torque, the sphere will have a larger angular acceleration. This happens because the sphere resists changes in its rotational state less than the cylinder. This principle, that objects with lower moments of inertia experience greater angular accelerations under the same torque, is key to understanding rotational dynamics.
Angular Speed
Angular speed is how fast an object spins around its axis and is usually denoted by the Greek letter \( \omega \). It's akin to how fast an object might move in a straight line, except here it's about rotation. In our exercise scenario, after applying equally strong torques to the cylinder and the sphere, we want to see which one ends up spinning faster over a period of time.The relationship between angular speed and angular acceleration can be expressed through time as:\[ \omega = \alpha t \]where:
  • \( \omega \) is the final angular speed,
  • \( \alpha \) is the angular acceleration, and
  • \( t \) is the time for which torque is applied.
With the sphere having greater angular acceleration (as derived from its smaller moment of inertia), it will reach a higher angular speed than the cylinder when the same torque is applied for the same duration. This relationship becomes evident because the smaller the resistance to rotation, the faster an object spins up, showing that less force is needed to reach a higher speed. This concept is crucial in systems where speed and efficiency are desired.

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Most popular questions from this chapter

A hoop of radius \(2 \mathrm{~m}\) weighs \(100 \mathrm{~kg}\). It rolls along a horizontal floor so that its centre of mass has a speed of \(20 \mathrm{~cm} / \mathrm{s}\). How much work has to be done to stop it?

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of \(40 \mathrm{rev} / \mathrm{min} .\) How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to \(2 / 5\) times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

From a uniform disk of radius \(R\), a circular hole of radius \(R / 2\) is cut out. The centre of the hole is at \(R / 2\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

A solid cylinder rolls up an inclined plane of angle of inclination \(30^{\circ}\). At the bottom of the inclined plane the centre of mass of the cylinder has a speed of \(5 \mathrm{~m} / \mathrm{s}\). (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom?

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