Calculating Torque of a Solid Cylinder

[Schu]

I am attempting to the following question:
How large is the torque that act on an armature?

Particulars: (can be approximated as a solid cylinder)
radius of .081 m
length of .124 m
mass of 13.13 kg
Accelerated from REST to operating speed of 3530 rpm in 5.57 second

I know T = F*r
I'm not sure where to start.

 

[pt176900]

You are given the mass, the radius, and a change in velocity over a given time interval.

Think of the problem another way:
T = F x r
from Newton's 2nd law: F = m*a
a = dv/dt
so you have: T = m dv/dt x r
start by converting the given velocity into meters per second, then calculating the acceleration. from there it is plug and chug

 

[Doc Al]

Newton's 2nd law - for rotational motion

I know T = F*r
I'm not sure where to start.

You'll need to apply Newton's 2nd law for rotational motion:
\tau = I \alpha

You are given all the information needed to calculate I (the rotational inertia) and \alpha (the rotational acceleration).

 

[Corneo]

Just to make sure you know that \tau = r \times F not \tau = F \times r, it makes a difference. And ||\tau||= rF\sin \phi

 

[Schu]

Now ya'll have me confused again, I thought I was all set with the first formula,
T = f * r. Now where does Newtons Second Law fit into it to get the rotational movement.

Take me step by step if you would.
Thanks for the help so far

 

[Doc Al]

Good point, Corneo. (But it won't matter for this particular problem.)

 

[Doc Al]

Now ya'll have me confused again, I thought I was all set with the first formula,
T = f * r. Now where does Newtons Second Law fit into it to get the rotational movement.

T = r X F is true, but not helpful in this problem. Are you given the force? No.

This is just the rotational analog to an "F = ma" problem. Instead of m, you calculate I; instead of a, you calculate α. Then apply T = I α.

Give it a shot.

 

[Schu]

Ok check to see if I am OK on this;
we know T = I α
I = 1/2 mr^2 and α = a / r
so plug it in
I = (1/2) 13.13 * .081^2 = .043072965
α = (29.9425 / 5.57)/.081 = 66.36633641
Multiply the two together for T and you get 2.8585 Nm

How'd I do??

 

[Doc Al]

Looks good. Note that you can calculate \omega directly from the rpm, then use it to calculate \alpha = \omega/t. Just convert rpm to radians/sec.