Problem 2: A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.11 m. Attached to the disk are four low mass rods of radius b = 0.14 m, each with a small mass m = 0.4 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 21 N for a time of ∆t=0.2 s. (Mom. of inertia of a disk I_(disk, CM)=1/2 MR^2 and point mass is I_(point mass)=mb^2.) Part 1. What is the angular speed of the apparatus? Part 2: What is the angle through which the apparatus turns in radians or degrees?

the angular speed of the apparatus

the angle through which the apparatus turns in radians or degrees is :  

Explanation:

Given that :

mass of uniform disk M = 1.2 kg

Radius R = 0.11 m

lower radius (b) = 0.14 m

small mass (m) = 0.4 kg

Force (F) = 21 N

time (∆t) =0.2 s

Moment of Inertia of =

=

= 0.00726 kgm²

Point mass   = mb²

But since four low rods are attached ; we have :

= 4 × mb²

= 4  × 0.4 (0.14)²

= 0.03136 kgm²

Total moment of Inertia =   +

= (0.00726 + 0.03136) kgm²

= 0.03862 kgm²

Assuming ∝ = angular acceleration = constant;

Then; we can use the following kinematic equations

T = FR

T = 2.1 × 0.11 N

T = 2.31 N

T = I × ∝

2.31 = 0.03862 × ∝

∝ =

∝ = 59.81 rad/s²

Using the formula to determine the angular speed of the apparatus; we have:

        since  

∴ the angular speed of the apparatus

b) Using the formula :

Thus, the angle through which the apparatus turns in radians or degrees is :  

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