# TORQUE A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm .

1. Jan 9, 2008

### lettertwelve

TORQUE A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm.....

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

A cylindrical fishing reel has a mass of 0.65 kg and a radius of 4.3 cm. A friction clutch in the reel exerts a restraining torque of 1.3 N·m if a fish pulls on the line. The fisherman gets a bite, and the reel begins to spin with an angular acceleration of 66 rad/s2.

(a) Find the force of the fish on the line.
______N
(b) Find the amount of line that unwinds in 0.50 s.
______m

3. The attempt at a solution

for the fishing reel, I=m*r^2

Net torque equals I*α
I will assume that the torque from the line is at the edge of the radius:
F*4.3/100-1.3=I*66
where F is the force of the fish
F=32.1 N

θ=.5*α*t^2

θ=.5*66*.5^2
d=2*π*r*θ
=2*3.14*4.3*8.25/100
2.23 m

but it says those answers are wrong. what do i do?

2. Jan 9, 2008

### Staff: Mentor

The rotational inertia of a cylinder is $(1/2)mr^2$. (Assuming that it's a solid cylinder.)

Last edited: Jan 9, 2008
3. Jan 11, 2008

### lettertwelve

but i dont understand why the second answer is wrong, either

4. Jan 11, 2008

### Staff: Mentor

So far, so good.
No, distance just equals r*θ. (Since theta is in radians.)

5. Jan 11, 2008

### lettertwelve

i'm sorry, i still don't understand...

6. Jan 11, 2008

### Staff: Mentor

Last edited by a moderator: Apr 23, 2017
7. Jan 11, 2008

### lettertwelve

Last edited by a moderator: Apr 23, 2017
8. Jan 11, 2008

### Staff: Mentor

I don't see anything wrong with the method you used in part a (except for that moment of inertia).

What do you think you left out?

9. Jan 11, 2008

### lettertwelve

I feel i need to rearrange the whole thing, according to the equation you gave me.
So if I=.5mr^2, how would i set it up to that....
I am really confused.
Also i'm partly a perfectionist so i need to start from the beginning hah, just a habit i picked up..

10. Jan 11, 2008

### Staff: Mentor

I would set it up exactly as you did:
$$\tau_{net} = I \alpha$$

$$Fr -\tau_{rest} = I \alpha$$

$$Fr = I \alpha + \tau_{rest} = (0.5mr^2) \alpha + \tau_{rest}$$

And so on...

11. Jan 11, 2008

### lettertwelve

using .5mr^2*a*Trest i get 1.34.......

12. Jan 11, 2008

### Staff: Mentor

For F*r, I presume? Now solve for F.

13. Jan 11, 2008

### lettertwelve

it worked out perfectly. thank you!!!!