# Angular speed/torque

1. Apr 2, 2016

### reminiscent

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
= Iα
W = Fd

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

2. Apr 2, 2016

### SteamKing

Staff Emeritus
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?

Last edited by a moderator: Apr 19, 2017
3. Apr 2, 2016

### reminiscent

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. Apr 2, 2016

### reminiscent

Should I go about giving direction or is that unnecessary here?

5. Apr 2, 2016

### SteamKing

Staff Emeritus
The problem doesn't care in which direction the telescope rotates, only how fast.
Yes.

6. Apr 3, 2016

### reminiscent

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. Apr 3, 2016

### SteamKing

Staff Emeritus
Since the telescope is rotating, the distance will be the angular displacement which occurs over that 5-sec. interval.

E = τ ⋅ θ

τ - torque, N-m
E - energy or work, N-m

8. Apr 3, 2016

### reminiscent

How would I find the angular displacement? Would it just be multiplying the angular speed I found by 5 seconds?

9. Apr 3, 2016

### SteamKing

Staff Emeritus
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.

10. Apr 3, 2016

### reminiscent

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. Apr 3, 2016

### SteamKing

Staff Emeritus
These results look correct to me.