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Angular Momentum of a sanding disk

  • Thread starter Destrio
  • Start date
212
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A sanding disk with rotational inertia 1.22x10^-3 kgm^2 is attached to an electric drill whose motor delivers a torque of 15.8 Nm
a) find angular momentum
b) find angular speed of the disk 33.0ms after the motor is turned on.

L = Iω
τ = Iα

we can find angular acceleration with what we are given α = τ/I
but I'm completely stumped on how to find L without having time.
any hints?

thanks
 

Doc Al

Mentor
44,815
1,078
Here's a hint: You are given the time! :smile:
 
14
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Here's a hint: You are given the time! :smile:
lol man I stared at that over and over again and thought, "I must be misreading this...wheres the trick?"

But seriously, yes you are given the time. its "33.0ms" part
 
212
0
Haha, thanks
I realized from that that I can use the time to solve the first part of my problem.

L = Iω

L =τω/α
ω = αt
L =τt

that works

τt = Iω
ω = τt/I
which gives me 427m/s
but i need the answer in rev/min
im okay with seconds to minutes, but how do i convert meters to revolutions?
2pi?
 
14
0
If you can find its angular acceleration, then ( just like linear kinematics... ) its angular velocity is equal to its angular acceleration multiplied by the duration of the acceleration.
 

Doc Al

Mentor
44,815
1,078
τt = Iω
ω = τt/I
which gives me 427m/s
The units for ω are radians/sec, not m/s. One revolution equals [itex]2 \pi[/itex] radians.
 
212
0
woo i got it!

thanks
 

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