Question

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on a frictionless pin. The free end of the cable is tied to a 4.9 kg box that is initially 2.9 meters above the floor.
Determine the angular acceleration of the cylinder. (+ CCW)

Answer 1

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass,$m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration,$\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$