Calculating Torque for Spinning a 250g Plastic Disk on an Electric Motor in 4.6s

[logan]

Homework Statement

A 250 g , 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 1500 rpm in 4.6 s ?

Homework Equations

ET=iα

The Attempt at a Solution

I just really don't know where to start

 

[haruspex]

ET=iα

Start by describing what the variables in that equation represent. E.g. "If a torque (variable name) acts on a mass with ..."

 

[logan]

Torque = Inertia x angular acceleration

 

[haruspex]

Torque = Inertia x angular acceleration

Ok, so can you calculate the angular acceleration?

 

[logan]

to get angular acceleration you would have to do angular acceleration=Torque/Inertia and we don't know the torque

 

[haruspex]

to get angular acceleration you would have to do angular acceleration=Torque/Inertia and we don't know the torque

No, you can calculate the angular acceleration from the information given.

0 to 1500 rpm in 4.6 s

 

[logan]

so a=w/t
1500/4.6=326.08

 

[haruspex]

so a=w/t
1500/4.6=326.08

Units?

 

[logan]

rad/s2

 

[haruspex]

rad/s2

does rpm divided by seconds give rad/s²?

 

[logan]

so you have to convert the 1500 to rads?
so that would be 157
157/4.6=34013 rad

 

[haruspex]

so you have to convert the 1500 to rads?

You have to convert from rpm to units involving radians, but not merely radians. Radians per ...?

157/4.6=34013 rad

Eh?

 

[logan]

radians per second

 

[haruspex]

radians per second

Ok. What about this part:

157/4.6=34013 rad

?
Can you correct that now?

 

[logan]

so 157/4.6=34 radians per second?

 

[haruspex]

so 157/4.6=34 radians per second?

You have 157 rad/s, and you are dividing by 4.6 s. What units result?

 

[logan]

Is it rad/s squared

 

[haruspex]

Is it rad/s squared

Rad/s², yes.

Next, can you calculate the moment of inertia of the disc?

 

[logan]

ok so that would be I=ta
4.6x34=156.4

 

[haruspex]

I=ta

No, as you wrote the equation is τ=Iα, where τ is torque, I is moment of inertia and α is angular acceleration.
Do not confuse τ, torque, with t, time.

You have information about the disc. Use that find its moment of inertia.

 

[logan]

I=t/a
so t is 4.6 but how do you get the torque to solve for inertia

thanks for helping me out so much I am just so lost with this again thank you

 

[haruspex]

I=t/a
so t is 4.6 but how do you get the torque to solve for inertia

thanks for helping me out so much I am just so lost with this again thank you

Go back and read the last sentence of post #20. How did I say you were to find the moment of inertia?

You don't seem to understand how equations work. In the equation τ=Iα, not only do you not know τ, you do not know any other way to find τ. So you must keep that equation in reserve for finding τ after you have already found I and α by other means. You cannot use the same equation twice, to find two different unknowns in it.

What information do you have about the disc itself?

 

[logan]

you know that the disk is 250 g and 25-cm-diameter
and it goes from 0 to 1500 rpm in 4.6 s

 

[haruspex]

the disk is 250 g and 25-cm-diameter

That is the information you have about the disc itself, i.e. as an object. From that you can calculate its moment of inertia. Have you not been taught any equations for that?

 

[logan]

no my professor only went over Torque = Inertia x angular acceleration because he thinks by making the homework hard the test will be easy for us

 

[haruspex]

no my professor only went over Torque = Inertia x angular acceleration because he thinks by making the homework hard the test will be easy for us

Then see http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html

 

[CWatters]

Wikipedia also has a handy table providing equations for the Moment of Inertia for objects of different shapes...
https://en.wikipedia.org/wiki/List_of_moments_of_inertia#Moments_of_inertia

 

[CWatters]

Perhaps worth pointing out the similarity between the equations for linear and rotational motion..

Linear...
Force = mass * acceleration

Rotation...
Torque = Moment of Inertia * angular acceleration

Both mass and moment of inertia depend on the stuff the object is made of and the dimensions.
Pretty much all the linear equations of motion can be adapted to rotation.

 

[logan]

so Ix would be .976
1/4(250)(.125)squared

 

[haruspex]

so Ix would be .976
1/4(250)(.125)squared

Try to keep everything in SI units. You converted cm to m ok, but you also need to convert the g to kg.
The ¼ is for rotation of a disc about a diameter, I_x or I_y at the link I posted. For this problem you want rotation about the axis perpendicular to the disc, I_z.

 

[logan]

got it was 6.7x10^-2
thanks haruspex and cwatters

 

[CWatters]

Just for info I made it 6.45x10^-2Nm

 

[haruspex]

Just for info I made it 6.45x10^-2Nm

I get 6.67x10^-2.

 

[CWatters]

Oops yes found my mistake. Agree with your figure.