Angular Velocity of 10 N m Applied Disk - 2 Revolutions

[tica86]

A torque of 10 N m is applied to a 10 kg disk of radius 0.2 m initially at rest. What is the angular acceleration of the disk?

I used I=1/2mr^2 so
1/2*10*0.2^2=.2

10Nm/0.2= 50 rad/s^2?

(With regard to the previous problem) The applied torque of 10 N m gets the disk rotating starting from rest. What is the angular velocity of the disk about an axis through its center, after it has completed two revolutions?

 

[tiny-tim]

hi tica86!

(have a theta: θ an omega: ω and an alpha: α and try using the X² icon just above the Reply box )

A https://www.physicsforums.com/library.php?do=view_item&itemid=175" of 10 N m is applied to a 10 kg disk of radius 0.2 m initially at rest. What is the angular acceleration of the disk?

I used I=1/2mr^2 so
1/2*10*0.2^2=.2

10Nm/0.2= 50 rad/s^2?

looks good!

(With regard to the previous problem) The applied torque of 10 N m gets the disk rotating starting from rest. What is the angular velocity of the disk about an axis through its center, after it has completed two revolutions?

you have ω_i α and t, and you want θ …

(that's like having v_i a and t, and wanting s)

so use the angular version of one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations

 

[tica86]

hi tica86!

(have a theta: θ an omega: ω and an alpha: α and try using the X² icon just above the Reply box )


looks good!


you have ω_i α and t, and you want θ …

(that's like having v_i a and t, and wanting s)

so use the angular version of one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations

Ok, so I tried 4*3.14=12.56
12.56^2/2*50rad/s^2= 1.57
square root=12.60??
Is that correct?

 

[tiny-tim]

hi tica86!

(have a square-root: √ and a pi: π and please use the X² tag just above the Reply box )

12.56^2/2*50rad/s^2= 1.57
square root=12.60??

uhh? what formula are you using?

 

[rwisz]

The angular velocity of the disk can be calculated using the formula ω = √(2τ/I), where τ is the applied torque and I is the moment of inertia. Plugging in the given values, we get ω = √(2*10 N m/(1/2*10 kg*0.2^2 m^2)) = 10 rad/s. After completing two revolutions, the disk has traveled a distance of 4π radians, which is equivalent to 2 complete revolutions. Therefore, the final angular velocity of the disk would be 20 rad/s.