Respuesta :
ANSWER:
Moment of inertia = 0.8375 kg*m²
Net torque = 1.45 Nm
Frictional force applied tangentially = 2.84 N
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 3.22 kg
Radius (r) = 51 cm = 0.51 m
Angular velocity (ω) = 5.2 rad/s
Time (t) = 3 s
The first thing is to determine the moment of inertia:
[tex]\begin{gathered} I=mr^2 \\ \\ \text{ We replacing:} \\ \\ I=(3.22)(0.51^2) \\ \\ I=0.8375\text{ kg}\cdot\text{m}^2 \end{gathered}[/tex]To calculate the torque, we need to first calculate the tangential acceleration, just like this:
[tex]\begin{gathered} \alpha=\frac{\Delta\omega}{\Delta t} \\ \\ \text{ We replacing:} \\ \\ \alpha=\frac{5.2-0}{3}=1.73\text{ rad/s}^2 \\ \\ \text{ Therefore:} \\ \\ \tau=I\alpha=0.8375\cdot1.73 \\ \\ \tau=1.45\text{ Nm} \end{gathered}[/tex]Finally, frictional force applied tangentially we calculate it just like this:
[tex]\begin{gathered} \tau=F\cdot r \\ \\ \text{ We replacing:} \\ \\ 1.45=F\cdot0.51 \\ \\ F=\frac{1.45}{0.51} \\ \\ F=2.84\text{ N} \end{gathered}[/tex]