Question #246539
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
1
Expert's answer
2021-10-05T10:07:37-0400

According to Newton's second law for the block:


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

By Newton's second law for rotational motion:


Iα=Tr.I\alpha=Tr.

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


a=αr: Iar=[mg(sinθμcosθ)ma]r, Iar=mgr(sinθμcosθ)mar, a(Irmr)=mgr(sinθμcosθ), a=mgr2(sinθμcosθ)Imr2=0.48 m/s2. T=mg(sinθμcosθ)ma=3.85 N.a=\alpha r:\\\space\\ 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}=0.48\text{ m/s}^2.\\\space\\ T=mg(\sin\theta-\mu\cos\theta)-ma=3.85\text{ N}.



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