Two masses with mass M and 3M (M=1kg) are setup 3M---------@ as shown in the fig

Question

Two masses with mass M and 3M (M=1kg) are setup            3M---------@ as shown in the figure with the mass M on a surface withcoeeficient             | kinetic friction = 0.2. the pulley has rotational inertia I=0.05kgm2  and          | radius r=5cm. the mass M is intially at rest. the string doesnotslip                M on the pulley, the pulley's axis is frictionless. Find acceleration of the mass M. you can express it in termsof M, mk, I, r, g and numberical constants or find its value! Two masses with mass M and 3M (M=1kg) are setup            3M---------@ as shown in the figure with the mass M on a surface withcoeeficient             | kinetic friction = 0.2. the pulley has rotational inertia I=0.05kgm2  and          | radius r=5cm. the mass M is intially at rest. the string doesnotslip                M on the pulley, the pulley's axis is frictionless. Find acceleration of the mass M. you can express it in termsof M, mk, I, r, g and numberical constants or find its value!

Explanation / Answer

Let tension of the horizontal string be T1. vertical string beT2. ----- acceleration of M. a1=T1/M. acceleration of 3M. a2=(3Mg-T2)/3M=g-T2/3M. angular acceleartion of pulley. =(T2-T1)*R/I ----- we have that a1=a2=*R. so we have. T1/M=g-T2/3M so 3T1+T2=3Mg T1/M=(T2-T1)*R^2/I. so T1*I=(T2-T1)*R^2*M so T1*(I+MR^2)=T2*MR^2. subtitute T2 with T1. 3T1+T1*(I+MR^2)/(MR^2)=3Mg. 4T1+T1*I/(MR^2)=3Mg. so T1*(4+I/(MR^2))=3Mg. so T1/M=3g/(4+I/(MR^2))=a

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.