Answer:
The frictional torque is [tex]\tau = 0.2505 \ N \cdot m[/tex]
Explanation:
From the question we are told that
The mass attached to one end the string is [tex]m_1 = 0.341 \ kg[/tex]
The mass attached to the other end of the string is [tex]m_2 = 0.625 \ kg[/tex]
The radius of the disk is [tex]r = 9.00 \ cm = 0.09 \ m[/tex]
At equilibrium the tension on the string due to the first mass is mathematically represented as
[tex]T_1 = m_1 * g[/tex]
substituting values
[tex]T_1 = 0.341 * 9.8[/tex]
[tex]T_1 = 3.342 \ N[/tex]
At equilibrium the tension on the string due to the mass is mathematically represented as
[tex]T_2 = m_2 * g[/tex]
[tex]T_2 = 0.625 * 9.8[/tex]
[tex]T_2 = 6.125 \ N[/tex]
The frictional torque that must be exerted is mathematically represented as
[tex]\tau = (T_2 * r ) - (T_1 * r )[/tex]
substituting values
[tex]\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )[/tex]
[tex]\tau = 0.2505 \ N \cdot m[/tex]
