/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A spherical solid ball of \(1 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A spherical solid ball of \(1 \mathrm{~kg}\) mass and radius \(3 \mathrm{~cm}\) is rotating about an axis passing through its centre with an angular velocity of \(50 \mathrm{rad} / \mathrm{s}\). The kinetic energy of rotation is (a) \(4500 \mathrm{~J}\) (b) \(90 \mathrm{~J}\) (c) \((9 / 20) \mathrm{J}\) (d) \((9 / 10) \mathrm{J}\)

Short Answer

Expert verified
The kinetic energy is approximately \(0.45 \mathrm{~J}\), closest to option (d) \(\frac{9}{10} \mathrm{~J}\).

Step by step solution

01

Identify the Formula

The kinetic energy of a rotating object is given by the formula \( K = \frac{1}{2} I \omega^2 \). Here, \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
02

Determine Moment of Inertia

For a solid sphere rotating about an axis through its center, the moment of inertia \( I \) is \( \frac{2}{5} m r^2 \). Here, \( m = 1 \mathrm{~kg} \) and \( r = 0.03 \mathrm{~m} \). Substitute to find \( I = \frac{2}{5} \times 1 \times (0.03)^2 = 0.00036 \mathrm{~kg \cdot m^2} \).
03

Calculate Kinetic Energy

Substitute \( I = 0.00036 \mathrm{~kg \cdot m^2} \) and \( \omega = 50 \mathrm{~rad/s} \) into the kinetic energy formula: \( K = \frac{1}{2} \times 0.00036 \times 50^2 \). Simplify this to find \( K = 0.00018 \times 2500 = 0.45 \mathrm{~J} \).
04

Evaluate the Answer

The options provided round the kinetic energy value to one of them. The closest option is (d) \( \frac{9}{10} \mathrm{~J} \), which is 0.45 J.

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Key Concepts

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

Kinetic Energy of Rotation
Kinetic energy in rotational motion refers to the energy possessed by an object due to its rotation. It's crucial for understanding how energy is utilized in systems involving rotation. When an object rotates, its different parts are moving. This movement contributes to the object's mechanical energy. The formula to calculate the kinetic energy of rotation is:
  • \( K = \frac{1}{2} I \omega^2 \)
Where \( K \) is the kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
This formula is similar to the linear kinetic energy formula \( \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. Just like mass is a measure of how an object resists acceleration in linear motion, the moment of inertia quantifies the same for rotational motion. Hence, both equations highlight the role of resisting force whether it's rotational or linear.
Moment of Inertia
Moment of inertia is a fundamental concept in rotational motion. It measures an object's resistance to changes in its rotation. Different objects have different moments of inertia based on their shape, mass distribution, and the axis they rotate around. For a solid sphere rotating about an axis through its center, the formula is:
  • \( I = \frac{2}{5} m r^2 \)
Here, \( m \) is the mass of the sphere and \( r \) is its radius.
In practice, this means two objects with the same shape but different size or mass will have different moments of inertia. A larger or heavier object will resist changes to its rotation more than a smaller or lighter one. This resistance determines how much energy is needed to start, stop, or alter its spinning, similar to how mass works in linear motion.
Angular Velocity
Angular velocity describes how quickly an object rotates or spins around an axis. It is an important factor in calculating the kinetic energy of rotation. Angular velocity is typically measured in radians per second (rad/s). It's a vector quantity, meaning it has both magnitude and direction. The magnitude expresses how fast the rotation occurs, while the direction is determined by the rotational axis.
Understanding angular velocity helps to visualize how fast an object is spinning, but it also plays a significant role in calculating rotational energy. For instance, in our previous exercise, understanding the concept allowed us to substitute the known angular velocity to determine the sphere's total kinetic energy. Essentially, the greater the angular velocity, the more kinetic energy is present, given the other parameters stay constant.
Solid Sphere
The solid sphere is a common object studied in physics, especially when dealing with rotational motion. It's essentially a three-dimensional, perfectly symmetrical object, like a ball. Important properties include its mass, radius, and how its mass is distributed relative to its rotation axis.
The symmetry of a solid sphere allows for simpler calculations of moment of inertia and kinetic energy compared to objects of irregular shapes. The mass is evenly distributed around the axis, providing predictable and uniform rotational characteristics. Solid spheres are often used as examples in physics problems due to their straightforward calculations, making them perfect models to apply rotational concepts like moment of inertia and kinetic energy of rotation effectively.

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Most popular questions from this chapter

A uniform disc of radius \(R\) lies in \(X Y\) -plane with its centre at origin. Its moment of inertia about the axis \(x=2 R\) and \(y=0\) is equal to the moment of inertia about the axis \(y=d\) and \(z=0\), where \(d\) is equal to (a) \(\frac{4}{3} R\) (b) \(\frac{\sqrt{17}}{2} R\) (c) \(\sqrt{13} R\) (d) \(\frac{\sqrt{15}}{2} R\)

A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 \mathrm{sec}-\) onds. The magnitude of this torque is (a) \(3 A_{0} / 4\) (b) \(A_{0}\) (c) \(4 A_{\text {? }}\) (d) \(12 A_{0}\)

Of the two eggs which have identical sizes, shapes and weights, one is raw and other is half boiled. The ratio between the moment of inertia of the raw to the half boiled egg about central axis is (a) \(=1\) (b) \(>1\) (c) \(<1\) (d) Not comparable

A disc of moment of inertia \(I_{1}\) is rotating freely with angular velocity \(\omega_{1}\) when a second, non-rotating disc with moment of inertia \(I_{2}\) is dropped on it gently the two then rotate as a unit. Then the total angular speed is (a) \(\frac{I_{1} \omega_{1}}{I_{2}}\) (b) \(\frac{I_{2} \omega_{1}}{I_{1}}\) (c) \(\frac{I_{1} \omega_{1}}{I_{2}+I_{1}}\) (d) \(\frac{\left(I_{1}+I_{2}\right) \omega_{1}}{I_{2}}\)

A uniform solid sphere rolls on a horizontal surface at \(20 \mathrm{~m} / \mathrm{s}\). It then rolls up an incline having an angle of inclination at \(30^{\circ}\) with the horizontal. If the friction losses are negligible, the value of height \(h\) above the ground where the ball stops is (a) \(14.3 \mathrm{~m}\) (b) \(28.6 \mathrm{~m}\) (c) \(57.2 \mathrm{~m}\) (d) \(9.8 \mathrm{~m}\)

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