Two masses are suspended from a pulley. The pulley has a massof 0.2 kg, a radius

Question

Two masses are suspended from a pulley. The pulley has a massof 0.2 kg, a radius of 0.15 m, and a constant torque of 0.35 Nm dueto the friction between the rotating pulley and its axle. What isthe magnitude of the acceleration of the suspended masses if m1 =0.4 kg and m2 = 0.8 kg? (Neglect the mass of the string.) Here is a link with the diagram: http://www.physics.brocku.ca/Courses/1P21/problems/WB-p08056.html I know the answer is 1.2 m/s^2, but I can't seem to get itright. Thanks. Two masses are suspended from a pulley. The pulley has a massof 0.2 kg, a radius of 0.15 m, and a constant torque of 0.35 Nm dueto the friction between the rotating pulley and its axle. What isthe magnitude of the acceleration of the suspended masses if m1 =0.4 kg and m2 = 0.8 kg? (Neglect the mass of the string.) Here is a link with the diagram: http://www.physics.brocku.ca/Courses/1P21/problems/WB-p08056.html I know the answer is 1.2 m/s^2, but I can't seem to get itright. Thanks.

Explanation / Answer

The moment of inertia of pulley isI=MR2=0.2*0.152=0.0045kgm2 Let a be the acceleration of the blocks then, For the block m1 we have as, T1-m1g=m1a For m2 we have as, m2g-T2=m2a Adding we get by it, (m2-m1)g-(T2-T1)=(m1+m2)a But we have by as, (T2-T1)R=I= (T2-T1)=0.35/0.15=2.33N (0.8-0.4)*9.8-2.33=(0.8+0.4)*a a=1.322m/s2 Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework." Let a be the acceleration of the blocks then, For the block m1 we have as, T1-m1g=m1a For m2 we have as, T1-m1g=m1a For m2 we have as, m2g-T2=m2a Adding we get by it, (m2-m1)g-(T2-T1)=(m1+m2)a But we have by as, (T2-T1)R=I= (T2-T1)=0.35/0.15=2.33N (0.8-0.4)*9.8-2.33=(0.8+0.4)*a a=1.322m/s2 Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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