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A wheel is at rest. Its angular velocity increases uniformly and becomes \(60 \mathrm{rad} / \mathrm{sec}\) after \(5 \mathrm{sec}\). The total angular displacement is (a) \(600 \mathrm{rad}\) (b) \(75 \mathrm{rad}\) (c) \(300 \mathrm{rad}\) (d) \(150 \mathrm{rad}\)

Short Answer

Expert verified
Hence, the total angular displacement of the wheel is \(150 \, rad\), which corresponds to option (d).

Step by step solution

01

Find Angular Acceleration

As given in the problem, the initial angular velocity \( \omega_0 = 0 \) (since the wheel starts from rest), the final angular velocity \( \omega = 60 \, rad/sec \) and the time \( t = 5 \, sec \). Using the equation \( \omega = \omega_0 + \alpha \cdot t \), we calculate the angular acceleration \( \alpha = (\omega - \omega_0) / t = 60 / 5 = 12 \, rad/sec^2.
02

Calculate Angular Displacement

Substitute \( \omega_0 = 0 \), \( \alpha = 12 \, rad/sec^2 \), and \( t = 5 \, sec \) into the equation of angular displacement \( \theta = \omega_0 \cdot t + 0.5 \cdot \alpha \cdot t^2 \) to find \( \theta = 0 + 0.5 \cdot 12 \cdot (5)^2 = 150 \, rad \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around an axis. In simpler terms, it tells us how fast something spins or revolves. The standard unit for angular velocity is radians per second (rad/s). In rotational motion, angular velocity (\( \omega \)) is analogous to linear velocity in linear motion. Just like linear velocity defines how fast an object moves in a straight line, angular velocity defines how fast an object moves in a circular path.
  • When an object starts spinning from rest, its initial angular velocity (\( \omega_0 \)) is zero.
  • The final angular velocity is the speed reached at a particular moment in time.
  • In the given exercise, the wheel starts off with an angular velocity of 0 rad/s and reaches 60 rad/s after 5 seconds.
This change happens due to angular acceleration, increasing the wheel's spinning speed over time.
Angular Acceleration
Angular acceleration defines the rate at which angular velocity changes with time. It measures how fast an object speeds up or slows down its rotation. Like other forms of acceleration, it can be positive or negative (deceleration), depending on whether the rotation is speeding up or slowing down.The formula for angular acceleration (\( \alpha \)) is:\[ \alpha = \frac{\Delta \omega}{\Delta t} \]Where:
  • \( \Delta \omega \) is the change in angular velocity.
  • \( \Delta t \) is the change in time.
In the exercise, the initial velocity (\( \omega_0 \)) was 0 rad/s and reached 60 rad/s after 5 seconds. Calculating with these values:\[ \alpha = \frac{60 \text{ rad/s} - 0 \text{ rad/s}}{5 \text{ s}} = 12 \text{ rad/s}^2 \]This means the wheel's angular velocity increased by 12 rad/s every second.
Rotational Motion
Rotational motion refers to the movement of a body around a fixed axis. When an object rotates, every point on that object follows a circular path around that axis, creating unique properties in physics like torque, angular speed, and angular momentum.To illustrate, think of a spinning wheel. Each part of the wheel moves in a circle, describing rotational motion. Key aspects to remember:
  • Angular displacement (\( \theta \)) signifies how much an object has rotated during motion, measured in radians.
  • Angular displacement can be calculated using the formula: \[ \theta = \omega_0 \cdot t + 0.5 \cdot \alpha \cdot t^2 \]
In our example, with an initial angular velocity of 0, angular acceleration of 12 rad/s², and time of 5 seconds:\[ \theta = 0 + 0.5 \cdot 12 \cdot (5)^2 = 150 \text{ rad} \]This tells us that the wheel underwent 150 radians of angular displacement in 5 seconds, encapsulating the essence of rotational motion.

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Most popular questions from this chapter

A force of \((2 \hat{i}-4 \hat{j}+2 \hat{k}) N\) acts at a point \((3 \hat{i}+2 \hat{j}-4 \hat{k})\) metre from the origin. The magnitude of torque is (a) Zero (b) \(24.4 \mathrm{~N}-\mathrm{m}\) (c) \(0.244 \mathrm{~N}-\mathrm{m}\) (d) \(2.444 \mathrm{~N}-\mathrm{m}\)

Moment of inertia of uniform rod of mass \(M\) and length \(L\) about an axis through its centre and perpendicular to its length is given by \(\frac{M L^{2}}{12} .\) Now consider one such rod pivoted at its centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass \(M\) moving horizontally at a speed \(v\) strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be (a) \(v / L\) (b) \(2 v / L\) (c) \(3 v / 2 L\) (d) \(6 v / L\)

Five particles of mass \(=2 \mathrm{~kg}\) are attached to the rim of a circular disc of radius \(0.1 \mathrm{~m}\) and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is (a) \(1 \mathrm{~kg} \mathrm{~m}^{2}\) (b) \(0.1 \mathrm{~kg} \mathrm{~m}^{2}\) (c) \(2 \mathrm{~kg} \mathrm{~m}^{2}\) (d) \(0.2 \mathrm{~kg} \mathrm{~m}^{2}\)

Two circular discs \(A\) and \(B\) are of equal masses and thickness but made of metals with densities \(d_{A}\) and \(d_{B}\left(d_{A}>d_{B}\right) .\) If their moments of inertia about an axis passing through centres and normal to the circular faces be \(I_{A}\) and \(I_{B}\), then(a) \(I_{A}=I_{B}\) (b) \(I_{A}>I_{B}\) (c) \(I_{A}=\left\langle I_{B}\right.\)

If the angular momentum of a rotating body is increased by \(200 \%\), then its kinetic energy of rotation will be increased by (a) \(400 \%\) (b) \(800 \%\) (c) \(200 \%\) (d) \(100 \%\)

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