Question

A pulley is mounted on a frictionless axis and then the pulley is attached to the higher end of the inclined surface as shown in the figure. 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.

Answer 1

Answer:

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

         ma=−μmgcosθ+mgsinθ−T.

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

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

Make substitution:

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}.Ira​=[mg(sinθ−μcosθ)−ma]r, Ira​=mgr(sinθ−μcosθ)−mar, a(rI​+mr)=mgr(sinθ−μcosθ), a=I+mr2mgr2(sinθ−μcosθ)​=0.39 m/s2. T=mg(sinθ−μcosθ)−ma=3.94 N.

Explanation:

Answer 2

The block, according to Newton's second law:

"ma=-\\mu mg\\cos\\theta+mg\\sin\\theta-T."

By Newton's second law for rotational motion:

"I\\alpha=Tr."

Write the equation that relates the angular and tangential acceleration, substitute, and solve for tension:

"a=\\alpha r:\\\\\\space\\\\\nI\\frac ar=[mg(\\sin\\theta-\\mu\\cos\\theta)-ma]r,\\\\\\space\\\\\nI\\frac ar=mgr(\\sin\\theta-\\mu\\cos\\theta)-mar,

\\\\\\space\\\\\na\\bigg(\\frac Ir-mr\\bigg)=mgr(\\sin\\theta-\\mu\\cos\\theta),\\\\\\space\\\\\na=\\frac{mgr^2(\\sin\\theta-\\mu\\cos\\theta)}{I-mr^2}=0.48\\text{ m\/s}^2.\\\\\\space\\\\\nT=mg(\\sin\\theta-\\mu\\cos\\theta)-ma=3.85\\text{ N}."

Explanation: