Respuesta :
Explanation:
The given data is as follows.
Pulling force on the reel is F = T = 2.3 N
Mass of cylinder (m) = 0.82 kg
Radius of cylinder (r) = 0.045 m
Formula for torque pulling force on the cylinder is as follows.
[tex]\tau = Fr[/tex]
= [tex]I \times \alpha[/tex]
Moment of inertia of the cylinder (I) is as follows.
I = [tex]\frac{mr^{2}}{2}[/tex]
[tex]\alpha[/tex] = angular acceleration of the cylinder
Hence,
Fr = [tex](\frac{mr^{2}}{2}) \alpha[/tex]
or, [tex]\alpha = \frac{2F}{mr}[/tex]
= [tex]\frac{2 \times 2.3 N}{0.82 kg \times 0.045 m}[/tex]
= 124.66 [tex]rad/s^{2}[/tex]
Amount of line pulled is h and it is in a times of 0.2 sec.
Now, linear acceleration is calculated as follows.
a = [tex]r \alpha[/tex]
= [tex]0.045 m \times 124.66 rad/s^{2}[/tex]
= 5.61 [tex]m/s^{2}[/tex]
Now, relation between h and acceleration as follows.
h = [tex]v_{o}t + \frac{1}{2}at^{2}[/tex]
here, [tex]v_{o}[/tex] = 0
Hence, calculate the value of h as follows.
h = [tex]v_{o}t + \frac{1}{2}at^{2}[/tex]
= [tex]0 + \frac{1}{2} \times 5.61 m/s^{2} \times (0.2s)^{2}[/tex]
= 0.1121 m
Thus, we can conclude that acceleration of the fishing reel is 5.61 [tex]m/s^{2}[/tex] and it will pull 5.61 [tex]m/s^{2}[/tex] line from the reel in 0.20 s.
Answer:
[tex]\alpha=12.47\ rad.s^{-2}[/tex]
[tex]l=0.011223\ m=11.223\ mm[/tex]
Explanation:
Given:
- radius of cylinder, [tex]r=0.045\ m[/tex]
- mass of cylinder, [tex]m=0.82\ kg[/tex]
- time observation, [tex]t=0.2\ s[/tex]
- force of pull by the fish, [tex]F=2.3\ N[/tex]
Angular acceleration is given as:
[tex]\tau=I.\alpha[/tex] ............(1)
Now as, [tex]\tau=F.r[/tex] ...................(2)
where:
[tex]\tau=[/tex] torque
[tex]I=[/tex]moment of inertia
[tex]\alpha=[/tex] angular acceleration.
From eq (2)
[tex]\tau=2.3\times 0.045[/tex]
[tex]\tau=0.1035\ N.m[/tex]
For cylinder:
[tex]I=\frac{1}{2} m.r^{2}[/tex]
[tex]I=0.5\times 0.82\times 0.045^2[/tex]
[tex]I=8.3025\times 10^{-4}\ kg.m^2[/tex]
Now using eq. (1):
[tex]0.1035=8.3025\times 10^{-4}\times \alpha[/tex]
[tex]\alpha=12.47\ rad.s^{-2}[/tex]
Revolution made in 0.2 seconds:
[tex]\theta=\omega_i.t+\frac{1}{2} \alpha.t^2[/tex]
where:
[tex]\omega_i=[/tex] initial angular velocity = 0
[tex]\theta=0+0.5\times 12.47\times 0.2^2[/tex]
[tex]\theta=0.2494\ rad[/tex]
Hence the length of reel pulled out:
[tex]l=\theta \times r[/tex]
[tex]l=0.2494\times 0.045[/tex]
[tex]l=0.011223\ m=11.223\ mm[/tex]
