Question

A cylindrical 4.31 kg pulley with a radius of 0.294 m is used to lower a 6.27 kg bucket into a well. The bucket starts from rest and falls for 3.55 s. The acceleration of gravity is 9.8 m/
`s^(2)`

What is the angular acceleration of the cylindrical pulley?

Answer in units of rad/
`s^(2)`

Answer 1

`24.81\ \text{rad/s}^2`

Step-by-step explanation:

M = Mass of cylinder = 4.31 kg

R = Radius of cylinder = 0.294 m

m = Mass of bucket = 6.27 kg

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

`\alpha` = Angular acceleration

a = Acceleration =
`\alpha R`

I = Moment of inertia of cylinder =
`(MR^2)/(2)`

The force balance of the system is

`mg-T=ma\\\Rightarrow T=m(g-a)`

For the disk

`TR=I\alpha\\\Rightarrow m(g-a)R=(1)/(2)MR^2\alpha\\\Rightarrow \alpha=(g)/((MR)/(2m)+R)\\\Rightarrow \alpha=(9.8)/((4.31* 0.294)/(2* 6.27)+0.294)\\\Rightarrow \alpha=24.81\ \text{rad/s}^2`

The angular acceleration of the pulley is
`24.81\ \text{rad/s}^2`.