Question #247910

A pulley is mounted on a frictionless axis and then the pulley is attached to the higher end of the inclined surface . A massless cord is wrapped around the pulley while its other end is tied to a 1.0 kg object. The pulley has a radius of 20 cm and moment of inertia of 0.4 kg m². The angle of the surface is 37° and the coefficient of kinetic friction is 0.2. The object is released from rest. The pulley starts to rotate as the object moves and the cord unwinds. Determine the acceleration of the object and the tension in the cord.


Expert's answer

By Newton's second law, write forces that participate in linear motion of the block:


ma=−μmgcos⁡θ+mgsin⁡θ−T.ma=-\mu mg\cos\theta+mg\sin\theta-T.

By Newton's second law for rotational motion, write torques that participate in rotation of the pulley:


Iα=Tr, where α=ar.I\alpha=Tr,\text{ where }\alpha=\frac ar.

Make substitution:


Iar=[mg(sin⁡θ−μcos⁡θ)−ma]r, Iar=mgr(sin⁡θ−μcos⁡θ)−mar, a(Ir+mr)=mgr(sin⁡θ−μcos⁡θ), a=mgr2(sin⁡θ−μcos⁡θ)I+mr2=0.39 m/s2. T=mg(sin⁡θ−μcos⁡θ)−ma=3.94 N.I\frac ar=[mg(\sin\theta-\mu\cos\theta)-ma]r,\\\space\\ I\frac ar=mgr(\sin\theta-\mu\cos\theta)-mar,\\\space\\ a\bigg(\frac Ir+mr\bigg)=mgr(\sin\theta-\mu\cos\theta),\\\space\\ a=\frac{mgr^2(\sin\theta-\mu\cos\theta)}{I+mr^2}=0.39\text{ m/s}^2.\\\space\\ T=mg(\sin\theta-\mu\cos\theta)-ma=3.94\text{ N}.


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