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In a lab experiment a light flexible string is wrapped around a solid cylinder with mass M = 9.50 kg and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. A mass m = 3.00kg is tied at the free end of the string and release and the mass moves downward. Find the magnitude of the acceleration of the object of mass m.

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Respuesta :

Answer:[tex]3.79 m/s^2[/tex]

Explanation:

Given

Mass [tex]M=9.5 kg[/tex]

[tex]m=3 kg[/tex]

Net Force is equivalent to [tex]\sum F=ma [/tex]

with tension T in the string    

For mass [tex]m[/tex]

[tex]mg-T=ma[/tex]

[tex]T=mg-ma--------1[/tex]

For cylinder

[tex]T\cdot R=I\times \alpha [/tex]

I for solid cylinder is [tex]\frac{2}{5}MR^2 , and \alpha =\frac{a}{R}[/tex]

thus [tex]T=\frac{Ma}{2}----2[/tex]

Substitute the value of T we get

[tex]\frac{Ma}{2}=mg-ma[/tex]

[tex]a(\frac{M}{2}+m)=mg[/tex]

[tex]a=\frac{mg}{\frac{M}{2}+m}[/tex]

[tex]a=3.79 m/s^2[/tex]

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