Question

A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string. When the system is released from rest, you determine that the stone reaches a speed of 3.40 m/s after having fallen 2.40 m. What is the mass of the cylinder?

Answer 1

Answer:18.5 kg

Explanation:

Given

diameter $d=75 cm$

mass of stone $m=3 kg$

velocity of stone $v=3.4 m/s$

height fallen $h=2.40 m$

using

$v^2-u^2=2as$

$(3.4)^2=2\times a\times 2.4$

$a=2.4 m/s^2$

Tension of string will Provide Torque to cylinder

$T\times r=I\times \alpha$

where $I=moment\ of\ inertia$

$\alpha =angular\ acceleration$

$T=\frac{Mr^2}{2}\times \frac{a}{r}$

$T=\frac{Ma}{2}$

and $mg-T=ma$

$mg-ma=T$

Put value of T

$mg-ma=\frac{Ma}{2}$

$mg=a(m+\frac{M}{2})$

$3\times 9.8=2.4\cdot (m+\frac{M}{2})$

$12.25=3+\frac{M}{2}$

$9.25=\frac{M}{2}$

$M=18.5 kg$