Solving Rotational Dynamics: 4.0 Nm Torque, 4.0 kg Cylinder, 2.0 s

[kingyof2thejring]

Hello there, I've got a problem here.
A solid cylinder of radius 0.8 m and mass 4.0 kg is at rest. A 4.0 N m torque is applied to the cylinder about an axis through its centre.

Calculate the angular velocity of the cylinder, in rad s-1 , after the torque has been applied for 2.0 s.

how do i go about solving this problem. what should i be considering first.
thanks in advance

 

[FredGarvin]

Think of Newton's second law for rotational systems. The linear version is F=ma. What is the rotational equivilent?

 

[physicsmy]

torque=(m.r^2).angular acceleration
since torque=F.r=ma.r
where a=r.angular acceleration.

 

[johnw188]

torque=(m.r^2).angular acceleration
since torque=F.r=ma.r
where a=r.angular acceleration.

I'm really not sure what you just wrote there, though I have a feeling you're missing something. If I'm correct, you have the form of the equation as

$$\sum \tau = (m r^2) \alpha$$

It also looks like you have your definition of torque mixed up. Torque is defined as such: $$\tau = R cross s$$. (what exactly is the tex for cross product x as opposed to curly mathematical x?) Since the cross product of R and s is ||R|| ||s|| sin(angle between r and s), if R and s are at a 90 degree angle it reduces down to the $$\tau = F r$$ that you stated, but only if they meet at a 90 degree angle. (You can, of course, take components to find the perpundicular and parallel forces ;) )

However, your statement that the sum of the torques is equal to mr^2*alpha is false in all but one case. The proper general form is

$$\sum \tau = I \alpha$$

where I is the moment of inertia of the body.

 

[lightgrav]

Somehow, you got this far into your physics course and you STILL ignore the Sum:
$$\Sigma \vec{F} = m \vec{a}$$ , not just F = ma .

In rotation, rotational Inertia depends on shape, which can change! Learn:
$$\Sigma \vec{\tau} = \frac{\Delta \vec{L}}{\Delta t} = sometimes = I \vec{\alpha}$$ (if shape is constant).
each $$\vec{\tau} = \vec{r} \times \vec{F}$$ helps to cause the effect, on the right-hand-side.
(NEVER ignore the Delta, which operates on the L and the t).
In your case, $$L = I \omega$$ , where solid disk (cylinder) Inertia is
(look up a table in your book. BOOKMARK it. Shapes differ!) I = ½ M R^2 .