Gravitino
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Two blocks of masses [tex]M_1[/tex] and [tex]M_2[/tex] passes through a pulley (See the Fig. attached.). Assuming the friction between block and the incline is zero, and neglecting the mass of string find the net torque about the center of mass of the pulley.
I solved this problem with two different ways. The answers are different. Please, let me know, where I am wrong.
1) [tex]\tau=\left(M_1-M_2\sin\theta\right)gR[/tex] because the net torque is due to external forces.
2) On the other hand, [tex]\tau=\left(T_1-T_2\right)R[/tex]. And the tensions [tex]T_1[/tex] and [tex]T_2[/tex] are found from
[tex]M_1g-T_1=M_1a[/tex]
[tex]-M_2\sin\theta g+T_2=-M_2a[/tex]
The solution gives different answer. In the case, [tex]a=0[/tex], the two solution matches.
Which one is correct?
I solved this problem with two different ways. The answers are different. Please, let me know, where I am wrong.
1) [tex]\tau=\left(M_1-M_2\sin\theta\right)gR[/tex] because the net torque is due to external forces.
2) On the other hand, [tex]\tau=\left(T_1-T_2\right)R[/tex]. And the tensions [tex]T_1[/tex] and [tex]T_2[/tex] are found from
[tex]M_1g-T_1=M_1a[/tex]
[tex]-M_2\sin\theta g+T_2=-M_2a[/tex]
The solution gives different answer. In the case, [tex]a=0[/tex], the two solution matches.
Which one is correct?
