Question

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.

Answer 1

Answer:
`3.79 m/s^2`

Step-by-step explanation:

Given

Mass
`M=9.5 kg`

`m=3 kg`

Net Force is equivalent to
`\sum F=ma`

with tension T in the string

For mass
`m`

`mg-T=ma`

`T=mg-ma--------1`

For cylinder

`T\cdot R=I* \alpha`

I for solid cylinder is
`(2)/(5)MR^2 , and \alpha =(a)/(R)`

thus
`T=(Ma)/(2)----2`

Substitute the value of T we get

`(Ma)/(2)=mg-ma`

`a((M)/(2)+m)=mg`

`a=(mg)/((M)/(2)+m)`

`a=3.79 m/s^2`