How Does Torque Affect Angular Acceleration in Rotational Motion?

[Swagger]

A uniform 2.9-kg cylinder can rotate about an axis through its center at O. The forces applied are: F1 = 2.4 N, F2 = 3.9 N, F3 = 6.1 N, and F4 = 4.6 N. Also, R1 = 11.2 cm and R3 = 4.0 cm. Find the magnitude and direction (+: counterclockwise; -: clockwise) of the angular acceleration of the cylinder. (see attachment)

I'm having a hard time with this one. I know that:
Torque (net) = Inertia * Angular Acceleration
and
Torque = Force * Radius

I've got:
T(F1)=0.2688 Nm
T(F2)=-0.4368 Nm
T(F3)=-0.0244 Nm
T(F4)=0 Nm

Tnet=-0.1924 Intertia=0.0181888
and the angular accerleration= -10.58 rad/s^2 (INCORRECT)

What am I doing wrong?

 

[nrqed]

A uniform 2.9-kg cylinder can rotate about an axis through its center at O. The forces applied are: F1 = 2.4 N, F2 = 3.9 N, F3 = 6.1 N, and F4 = 4.6 N. Also, R1 = 11.2 cm and R3 = 4.0 cm. Find the magnitude and direction (+: counterclockwise; -: clockwise) of the angular acceleration of the cylinder. (see attachment)

I'm having a hard time with this one. I know that:
Torque (net) = Inertia * Angular Acceleration
and
Torque = Force * Radius

I've got:
T(F1)=0.2688 Nm
T(F2)=-0.4368 Nm
T(F3)=-0.0244 Nm
T(F4)=0 Nm

Tnet=-0.1924 Intertia=0.0181888
and the angular accerleration= -10.58 rad/s^2 (INCORRECT)

What am I doing wrong?

I can't see the figure yet so I can't tell. But the torque is F times r times sin theta. That may (or may not, depending on what direction the forces are acting) explain a discrepancy.

Pat

 

[Doc Al]

A uniform 2.9-kg cylinder can rotate about an axis through its center at O. The forces applied are: F1 = 2.4 N, F2 = 3.9 N, F3 = 6.1 N, and F4 = 4.6 N. Also, R1 = 11.2 cm and R3 = 4.0 cm. Find the magnitude and direction (+: counterclockwise; -: clockwise) of the angular acceleration of the cylinder. (see attachment)

The diagram seems to show a single radius. What are R1 and R2? What's the radius of the cylinder?

I'm having a hard time with this one. I know that:
Torque (net) = Inertia * Angular Acceleration

OK.

and
Torque = Force * Radius

Don't forget the angle that the force is applied.

Torque = Force * Radius * sin(angle)

 

[Swagger]

Don't forget the angle that the force is applied.

Torque = Force * Radius * sin(angle)

All angles a 90deg, that would just be multiplying by 1, right? Also, is my F3 torque correct? I did .004 * 6.1

 

[nrqed]

I can't see the figure yet so I can't tell. But the torque is F times r times sin theta. That may (or may not, depending on what direction the forces are acting) explain a discrepancy.

Pat

Your T(F3) is off by a factor of 10. That will probably do it.

 

[Doc Al]

All angles a 90deg, that would just be multiplying by 1, right?

Perhaps you are defining things differently, but only F1 and F2 are applied at 90 degrees to the radial line through the point of application.

 

[Swagger]

Your T(F3) is off by a factor of 10. That will probably do it.

UGH! That was it...I got it. Thanks!

 

[consemiu21]

The diagram seems to show a single radius. What are R1 and R2? What's the radius of the cylinder?



OK.

Don't forget the angle that the force is applied.

Torque = Force * Radius * sin(angle)

O k Doc I am just having trouble get my (T1), T(2)..., my numbers are off, so what is the formula for that?

 

[Doc Al]

O k Doc I am just having trouble get my (T1), T(2)..., my numbers are off, so what is the formula for that?

Formula for what? You just quoted the formula for finding the torque.

Show what you did.