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Angular speed/torque

  • #1
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Homework Statement


The Hubble space telescope has a maximum diameter (cylinder) of 4.20 m and a moment of inertia of 77,217 kg-m2 about its axis. A reaction motor applies a force of 18.0 N tangentially to the edge of the cylinder for 5.00 seconds. (a) What is the angular speed of the telescope after the 5 second interval? (b) How much work did the reaction motor do on the telescope?

Homework Equations


81a69207104f00baaabd6f84cafd15a0.png
= Iα
W = Fd

The Attempt at a Solution


I don't know where to really start with this problem. Can anyone give me a few clues?
 

Answers and Replies

  • #2
SteamKing
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Take the first equation, ∑T = I α. You're given I and you know the diameter of the telescope and the force applied to it. What's the resultant torque? Do you know how to calculate torque?

Once you calculate torque, what's the angular acceleration of the telescope? If the scope accelerates for 5 sec. from rest, what's the angular speed?
 
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  • #3
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Take the first equation, ∑T = I α. You're given I and you know the diameter of the telescope and the force applied to it. What's the resultant torque? Do you know how to calculate torque?

Once you calculate torque, what's the angular acceleration of the telescope? If the scope accelerates for 5 sec. from rest, what's the angular speed?
Okay I found the torque by multiplying the applied force (18.0 N) by the radius (2.1 m). I then divided that by the moment of inertia to give me the angular acceleration. Then multiplied the angular acceleration by 5 seconds. Is that correct?
 
  • #4
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Take the first equation, ∑T = I α. You're given I and you know the diameter of the telescope and the force applied to it. What's the resultant torque? Do you know how to calculate torque?

Once you calculate torque, what's the angular acceleration of the telescope? If the scope accelerates for 5 sec. from rest, what's the angular speed?
Should I go about giving direction or is that unnecessary here?
 
  • #5
SteamKing
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Should I go about giving direction or is that unnecessary here?
The problem doesn't care in which direction the telescope rotates, only how fast.
Okay I found the torque by multiplying the applied force (18.0 N) by the radius (2.1 m). I then divided that by the moment of inertia to give me the angular acceleration. Then multiplied the angular acceleration by 5 seconds. Is that correct?
Yes.
 
  • #6
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The problem doesn't care in which direction the telescope rotates, only how fast.

Yes.
Okay, so how would I find the work? I know that W=Fd - the force that the reaction motor applies on the cylinder is 18.0 N, but what about the distance?
 
  • #7
SteamKing
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Okay, so how would I find the work? I know that W=Fd - the force that the reaction motor applies on the cylinder is 18.0 N, but what about the distance?
Since the telescope is rotating, the distance will be the angular displacement which occurs over that 5-sec. interval.

E = τ ⋅ θ

θ - angular displacement, radians
τ - torque, N-m
E - energy or work, N-m
 
  • #8
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Since the telescope is rotating, the distance will be the angular displacement which occurs over that 5-sec. interval.

E = τ ⋅ θ

θ - angular displacement, radians
τ - torque, N-m
E - energy or work, N-m
How would I find the angular displacement? Would it just be multiplying the angular speed I found by 5 seconds?
 
  • #9
SteamKing
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How would I find the angular displacement? Would it just be multiplying the angular speed I found by 5 seconds?
No, because the telescope was undergoing a constant angular acceleration during that time.

Just like you can calculate the linear distance traveled while undergoing a linear acceleration, there are equivalent formulas to calculate the angular displacement made while undergoing angular displacement. These are the angular equivalents of the well-known SUVAT equations.


upload_2016-4-3_16-50-26.png

 
  • #10
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No, because the telescope was undergoing a constant angular acceleration during that time.

Just like you can calculate the linear distance traveled while undergoing a linear acceleration, there are equivalent formulas to calculate the angular displacement made while undergoing angular displacement. These are the angular equivalents of the well-known SUVAT equations.

Okay so I got for a) 0.00245 rad/s. For b), I got angular displacement as 0.006125 radians. I multiplied that by (18.0 N)(2.1 m) and I got 0.232 J. Is that correct?
 
  • #11
SteamKing
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Okay so I got for a) 0.00245 rad/s. For b), I got angular displacement as 0.006125 radians. I multiplied that by (18.0 N)(2.1 m) and I got 0.232 J. Is that correct?
These results look correct to me.
 

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