Calculating Angular Speed of a Disk with Applied Torque and Constant Force

[ljucf]

Homework Statement

A disk of radius 0.41 m and moment of inertia 2.8 kg·m² is mounted on a nearly frictionless axle. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 52 N.

What is the magnitude of the torque?
torque = 21.32 N·m

After a short time the disk has reached an angular speed of 4 radians/s, rotating clockwise. What is the angular speed 0.63 seconds later?
angular speed = ? radians/s

Homework Equations

torque = RF_T

Rotational angular momentum = (MR²/2)ω

The Attempt at a Solution

The new angular momentum is the old angular momentum plus the angular impulse, torque times time interval.

(2.8/2)4 + (21.32)(.63) = 19.03 radians/s

However, this answers is wrong, and I can't figure out what I am doing wrong.

 

[ehild]

The moment of inertia is given, it is 2.8 kgm². Why did you divide it by 2?

 

[ljucf]

The moment of inertia is given, it is 2.8 kgm². Why did you divide it by 2?

According to my textbook, Rotational angular momentum = (MR²/2)ω

Therefore, I divided it by two. Is that incorrect?

 

[ehild]

MR²/2 is the moment of inertia of a homogeneous disk. The mass is not given. The moment of inertia is given as 2.8 kgm². The angular momentum is moment of inertia times the angular speed.

 

[ljucf]

MR²/2 is the moment of inertia of a homogeneous disk. The mass is not given. The moment of inertia is given as 2.8 kgm². The angular momentum is moment of inertia times the angular speed.

So is the Rotational angular momentum = Iω

 

[ehild]

So is the Rotational angular momentum = Iω

Yes, the angular momentum of a rotating body is Iω.

 

[ljucf]

Yes, the angular momentum of a rotating body is Iω.

So now I take (2.8)(4) + 21.32(.63) and divide that by moment of inertia?

 

[ehild]

So now I take (2.8)(4) + 21.32(.63) and divide that by moment of inertia?

Yes.

 

[ljucf]

Yes.

Thank you for your help!

 

[ehild]

You are welcome:)