Torque & Net Work Homework: Find Average Torque & Net Work

[Mowgli]

Homework Statement

An old 33 1/3 rpm turntable may be approximated by a cylindrical disk of mass 0.250kg and a diameter of 31.0cm.

a.) What average torque will bring it up to operating speed in 1.50 seconds starting from rest?

b.) what net work is done on the disk in bringing it up to operating speed?

Homework Equations

The Attempt at a Solution

Can anyone help me get started on this? We never went over torque!

 

[collinsmark]

Can anyone help me get started on this? We never went over torque!

Really? Is this for a homework problem?

Well I guess a first step would be to read up on torque. You'll also want to acquaint yourself with the concept of "moment of inertia." As an interim step to this problem, you will need to find the moment of inertia of a 0.250kg cylindrical disk. Before you're finished, you'll also need to research the concept of rotational kinetic energy.

 

[Mowgli]

Is this correct?

mass m = 0.25 Kg
diameter d = 31 cm
radius r = 15.5 cm = 0.155 m
moment of inertia I = (1/2)mr^2 = 0.003 Kg m^2
(a)
ω1 = 0
n = 33 1/3 = 100/3 = 33.33 rpm = 0.5555 rps
ω2 = 2πn = 2*3.14*0.5555 = 3.48854 rad/s
t = 1.5 s
we know that
torque τ = I α = I (ω2 - ω1)/t = 0.00697708 N m
(b)
work done w = change in rotational kinetic energy
= (1/2)*I*(ω2)^2
= 0.5*0.003*3.48854*3.48854
= 0.0182 J

 

[collinsmark]

Is this correct?

mass m = 0.25 Kg
diameter d = 31 cm
radius r = 15.5 cm = 0.155 m
moment of inertia I = (1/2)mr^2 = 0.003 Kg m^2
(a)
ω1 = 0
n = 33 1/3 = 100/3 = 33.33 rpm = 0.5555 rps
ω2 = 2πn = 2*3.14*0.5555 = 3.48854 rad/s
t = 1.5 s
we know that
torque τ = I α = I (ω2 - ω1)/t = 0.00697708 N m
(b)
work done w = change in rotational kinetic energy
= (1/2)*I*(ω2)^2
= 0.5*0.003*3.48854*3.48854
= 0.0182 J

'Looks okay to me.

 

[rwisz]

Sure, I can help you get started on this problem. Let's start by defining some terms and equations that will be useful in solving this problem.

Torque is a measure of the twisting force that is applied to an object. It is calculated by multiplying the force applied to the object by the distance from the point of rotation. The equation for torque is T = F x r, where T is torque, F is force, and r is the distance from the point of rotation. In this problem, the force will be the average force applied to the turntable and the distance will be the radius of the turntable.

Net work is the total amount of work done on an object. It is calculated by multiplying the force applied to the object by the distance the object moves. The equation for net work is W = F x d, where W is net work, F is force, and d is distance. In this problem, the force will be the average force applied to the turntable and the distance will be the distance the turntable moves, which is related to its angular velocity and time.

To solve this problem, we will need to use the equations for torque and net work, as well as the equation for angular acceleration, which is a = α x r, where a is angular acceleration, α is angular velocity, and r is the radius of the turntable.

a) To find the average torque, we need to find the average force applied to the turntable. Since we know the mass and diameter of the turntable, we can calculate its moment of inertia, which is a measure of how difficult it is to change an object's rotational motion. The equation for moment of inertia for a disk is I = 1/2 x m x r^2, where I is moment of inertia, m is mass, and r is radius. Plugging in the values given in the problem, we get I = 1/2 x 0.250kg x (0.155m)^2 = 0.00382 kgm^2.

Next, we can use the equation for angular acceleration to find the average force applied to the turntable. Since the turntable starts from rest and reaches operating speed in 1.50 seconds, we can use the equation α = Δω/Δt, where α is angular acceleration, Δω is the change in angular velocity, and Δt is the change in time. The angular velocity