Respuesta :
Answer:
a.[tex]\tau=200J[/tex] b.[tex]\alpha=0.44 \frac{rad}{s^2}[/tex] c. [tex]\alpha=0.33\frac{rad}{s^2}[/tex] d. The angular acceleration when sitting in the middle is larger.
Explanation:
a. The magnitude of the torque is given by [tex]\tau=rF\sin \theta[/tex], being r the radius, F the force aplied and [tex]\theta[/tex] the angle between the vector force and the vector radius. Since [tex]\theta=90^{\circ}, \, \sin\theta=1[/tex] and so [tex]\tau=rF=2m100N=200Nm=200J[/tex].
b. Since the relation [tex]\tau=I\alpha[/tex] hols, being I the moment of inertia, the angular acceleration can be calculated by [tex]\alpha=\frac{\tau}{I}[/tex]. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is [tex]I=\frac{1}{2}Mr^2[/tex], being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by [tex]I_p=m_pr_p^{2}[/tex], being [tex]m_p[/tex] the mass of the person and [tex]r_p^{2}[/tex] the distance from the person to the center. Given all of this, we have
[tex]\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}[/tex].
c. Similar equation to b, but changing [tex]r_p=2m[/tex], so
[tex]alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}[/tex].
d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.
