Question

A large fish strikes bait cast by a fisherman with a rod and reel. The fish creates tension in the line of 63.7 N. How many revolutions of the reel occur in 0.8 s if the reel has a moment of inertia of 4.9 × 10-3 kg·m2 and the radial distance of the line from the reel’s axis is 3.4 cm? Assume the reel has negligible friction. (The units for your answer should be rev) I got 0.0001146 rev but it is incorrect, what did I do wrong?

Answer 1

Given moment of inertia=4.9*10^-3 kg-m2.

Now, torque=r*F*sin, where r is position of the point where force is applied, F is force and is the angle between r and F.

Here,r=3.4 cm = 0.034 m, F=63.7 N, =90 deg.

So, torque=0.034*63.7 sin 90 =2.1658 N-m.

Angular acceleration = (torque)/(moment of inertia) = 2.1658/(4.9*10^-3) = 442 rad/s2.

Now, angular displacement = i*t+1/2**t^2, wherei is initial angular velocity, t is time and is angular acceleration.

Here,i= 0 rad/s, t=0.8 s and =442 rad/s2

So, angular displacement = 1/2*442*0.8*0.8 = 141.44 radians=141.44/(2) revolutions = 22.5 revolutions.