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A horizontal \(800-\mathrm{N}\) merry-go-round is a solid disk of radius \(1.50 \mathrm{m},\) started from rest by a constant horizontal force of \(50.0 \mathrm{N}\) applied tangentially to the edge of the disk. Find the kinetic energy of the disk after \(3.00 \mathrm{s}\)

Short Answer

Expert verified
The kinetic energy of the disk after 3.00 s is 276.12 J.

Step by step solution

01

Calculate the torque

Torque \(\tau\) for the merry-go-round is the force applied multiplied by the radius of the disk. That is \(\tau = F*r = 50.0 N * 1.50 m = 75.0 Nm\).
02

Calculate the moment of inertia

The moment of inertia \(I\) for a solid disk is \(\frac{1}{2}mr^2\). Here, the mass \(m\) is \(800 N / 9.8 m/s^2 = 81.63 kg\) (by using the relation \(F = mg\)). Substitute all the values into the equation to get \(I = 0.5 * 81.63 kg * (1.5 m)^2 = 91.83 kg m^2\).
03

Find out the angular acceleration

Angular acceleration \(\alpha\) can be calculated by using the formula \(\alpha = \tau / I = 75.0 Nm / 91.83 kg m^2 = 0.817 rad/s^2\).
04

Solve for angular velocity

Angular velocity \(\omega\) is given by \(\omega = \alpha t\). Substituting the values from the previous step \(\omega = 0.817 rad/s^2 * 3.00 s = 2.45 rad/s\).
05

Calculate the kinetic energy

We can now find the Kinetic energy using the formula \(0.5 I \omega^2\). Substituting, we get \(0.5 * 91.83 kg m^2 * (2.45 rad/s)^2 = 276.12 J\). Thus, the kinetic energy of the disk after 3.00 s is 276.12 J.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In the context of rotational motion, it concerns the spinning or rotating energy of the object. For a rotating disk, like our merry-go-round, the kinetic energy can be found using the formula:
  • \[ KE = \frac{1}{2} I \omega^2 \]
Here, \(I\) is the moment of inertia, which tells us how the mass is distributed in the disk, and \(\omega\) is the angular velocity, which shows how fast the disk is spinning. It’s important to notice that while linear kinetic energy is focused on straight-line motion, rotational kinetic energy focuses on circular motion. In this problem, after applying force, the disk's kinetic energy increases over time, ultimately reaching 276.12 J after 3 seconds.
Torque
Torque is all about how forces cause objects to rotate. It's like the twisting force you apply to open a jar lid. In mathematical terms, torque \(\tau\) is calculated by:
  • \[ \tau = F \times r \]
Where \(F\) is the force applied, and \(r\) is the distance from the pivot point or axis of rotation. In our exercise, a force of 50.0 N is applied tangentially to the edge of a merry-go-round, creating a torque of 75.0 Nm. This torque is crucial for changing the state of motion from rest to rotating.
Moment of Inertia
The moment of inertia is a concept that can be compared to mass in linear motion. It explains how difficult it is to change the rotational state of an object. For a solid disk, the moment of inertia \(I\) is given by:
  • \[ I = \frac{1}{2} m r^2 \]
Where \(m\) is the mass and \(r\) is the radius. By knowing both the mass and the radius of the merry-go-round, we can compute its moment of inertia, which is 91.83 kg \(m^2\). This property is key in determining how much torque is needed to achieve angular acceleration.
Angular Acceleration
Angular acceleration is about how quickly the angular velocity changes over time. It works similarly to linear acceleration but deals with rotation. The formula for angular acceleration \(\alpha\) is:
  • \[ \alpha = \tau / I \]
Substituting the torque and moment of inertia calculated before, the angular acceleration is 0.817 rad/s\(^2\). This tells us that every second, the rotational speed of the merry-go-round increases by 0.817 rad/s. Understanding angular acceleration helps us understand how fast the object moves from rest to a specific rotational speed.
Angular Velocity
Angular velocity is a measure of how fast something rotates. It’s similar to speed in a straight line, but for rotational movement. The formula to find angular velocity \(\omega\) is:
  • \[ \omega = \alpha t \]
Where \(\alpha\) is the angular acceleration, and \(t\) is the time duration. Using the previously calculated angular acceleration (0.817 rad/s\(^2\)) and the time (3 seconds), we find that \(\omega\) is 2.45 rad/s. This represents the rotational speed of the merry-go-round after 3 seconds. Angular velocity helps in understanding how fast an object is turning at any given point.

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Most popular questions from this chapter

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