To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as
[tex]W=\tau \Delta\theta[/tex]
In the case of rotational kinetic energy we know that
[tex]KE = \frac{1}{2}I\omega^2[/tex]
PART A) [tex]\theta[/tex] is given in revolutions and needs to be in radians therefore
[tex]\theta = 12rev(\frac{2\pi rad}{1rev})[/tex]
[tex]\theta = 24\pi rad[/tex]
Replacing in the work equation we have to
[tex]W=\tau \Delta\theta[/tex]
[tex]W= (25)(24\pi)[/tex]
[tex]W = 1884.95J[/tex]
PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.
[tex]\tau = I \alpha[/tex]
Rearrange for the angular acceleration,
[tex]\alpha = \frac{\tau}{I}[/tex]
[tex]\alpha = \frac{25}{50}[/tex]
[tex]\alpha = 0.5rad/s[/tex]
From the kinematic equations of angular motion we have,
[tex]\omega_f^2=\omega_i^2+2\alpha\theta[/tex]
[tex]\omega_f^2=0+2*0.5*24\pi[/tex]
[tex]\omega_f=\sqrt{0+2*0.5*24\pi}[/tex]
[tex]\omega_f = 8.68rad/s[/tex]
In this way the rotational kinetic energy would be given by
[tex]KE = \frac{1}{2}I\omega_f^2[/tex]
[tex]KE = \frac{1}{2}(50)(8.68)^2[/tex]
[tex]KE = 1883.56J[/tex]
A. The motors does work about 1880 J on the platform
B. The final rotational kinetic energy of the platform is about 1880 J
[tex]\texttt{ }[/tex]
Let's recall Moment of Force as follows:
[tex]\boxed{\tau = F d}[/tex]
where:
Ï„ = moment of force ( Nm )
F = magnitude of force ( N )
d = perpendicular distance between force and pivot ( m )
Let us now tackle the problem !
Given:
constant torque = Ï„ = 25.0 Nm
moment of inertia = I = 50.0 kg.m²
number of rotations = θ = 12.0 rotations = 24π rad
Asked:
A. work = W = ?
B. final rotational kinetic energy = Ek = ?
Solution:
[tex]W = \tau \theta[/tex]
[tex]W = 25.0 \times 24\pi[/tex]
[tex]W = 600 \pi \texttt{ J}[/tex]
[tex]W \approx 1880 \texttt{ J}[/tex]
[tex]\texttt{ }[/tex]
[tex]W = \Delta Ek[/tex]
[tex]W = Ek - Ek_o[/tex]
[tex]600 \pi = Ek - 0[/tex]
[tex]Ek = 600 \pi[/tex]
[tex]Ek \approx 1880 \texttt{ J}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Moment of Force
