Physics, 21.11.2019 19:31 memeE15

Consider a motor that exerts a constant torque of 25.0 n⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . assume that the platform is initially at rest and the torque is applied for 12.0 rotations . neglect friction.
part a
how much work w does the motor do on the platform during this process?
part b
what is the rotational kinetic energy of the platform krot, f at the end of the process described above?

Answers: 3

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Answers

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

$W=\tau \Delta\theta$

In the case of rotational kinetic energy we know that

$KE = \frac{1}{2}I\omega^2$

PART A) $\theta$ is given in revolutions and needs to be in radians therefore

$\theta = 12rev(\frac{2\pi rad}{1rev})$

$\theta = 24\pi rad$

Replacing in the work equation we have to

$W=\tau \Delta\theta$

$W= (25)(24\pi)$

W = 1884.95J

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.

$\tau = I \alpha$

Rearrange for the angular acceleration,

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{25}{50}$

$\alpha = 0.5rad/s$

From the kinematic equations of angular motion we have,

$\omega_f^2=\omega_i^2+2\alpha\theta$

$\omega_f^2=0+2*0.5*24\pi$

$\omega_f=\sqrt{0+2*0.5*24\pi}$

$\omega_f = 8.68rad/s$

In this way the rotational kinetic energy would be given by

$KE = \frac{1}{2}I\omega_f^2$

$KE = \frac{1}{2}(50)(8.68)^2$

KE = 1883.56J

Answer from: elopezhilario6339

A. The motors does work about 1880 J on the platform

B. The final rotational kinetic energy of the platform is about 1880 J

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Further explanation

Let's recall Moment of Force as follows:

$\boxed{\tau = F d}$

where:

τ = 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 = τ = 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:

Part A:

$W = \tau \theta$

$W = 25.0 \times 24\pi$

$W = 600 \pi \texttt{ J}$

$W \approx 1880 \texttt{ J}$

$\texttt{ }$

Part B:

$W = \Delta Ek$

W = Ek - Ek_o

$600 \pi = Ek - 0$

$Ek = 600 \pi$

$Ek \approx 1880 \texttt{ J}$

$\texttt{ }$

Learn moreImpacts of Gravity : Effect of Earth’s Gravity on Objects : The Acceleration Due To Gravity : Newton's Law of Motion: Example of Newton's Law:

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Answer details

Grade: High School

Subject: Physics

Chapter: Moment of Force

Consider a motor that exerts a constant torque of 25.0 n⋅m to a horizontal platform whose moment of

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