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Sagot :
ANSWER:
0.079 J
STEP-BY-STEP EXPLANATION:
The average torque exerted due to friction force is:
[tex]\tau=I\alpha[/tex]Here, I is the moment of inertia and α is the angular acceleration.
The mass of wheel is concentrated at the rim. Therefore, the moment of inertia of wheel is,
[tex]I=m\cdot r^2[/tex]Here, m is the mass of wheel and r is the radius of wheel.
replacing:
[tex]\begin{gathered} I=2.07\cdot0.3^2 \\ I=0.1863\text{ kg}\cdot s^2 \end{gathered}[/tex]The angular speed is decreased from 4 rev/s to 0 rev/s in 59 sec. Therefore, the angular acceleration is calculated as:
[tex]\begin{gathered} \alpha=\frac{0-4}{59} \\ \alpha=0.068\cdot\frac{rev}{\sec}\cdot\frac{2\pi\text{ rad/rev}}{1rev/sec^2} \\ \alpha=-0.426rad/sec^2 \end{gathered}[/tex]The negative sign represents that the wheel is de-accelerating.
Replacing:
[tex]\begin{gathered} \tau=I\alpha \\ \tau=0.1863\cdot-0.426 \\ \tau=-0.079\text{J} \end{gathered}[/tex]The negative sign represents the direction of torque i.e. clockwise direction. Thus, the magnitude of average torque is 0.079 J.
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