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?

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

In the case of rotational kinetic energy we know that

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

Replacing in the work equation we have to

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.

Rearrange for the angular acceleration,

From the kinematic equations of angular motion we have,

In this way the rotational kinetic energy would be given by

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

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

Let's recall Moment of Force as follows:

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:

Grade: High School

Subject: Physics

Chapter: Moment of Force

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this