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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?

Respuesta :

ajeigbeibraheem

Answer:

the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

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 [tex]I_{disk, CM}[/tex] = [tex]\frac{1}{2}MR^2[/tex]

= [tex]\frac{1}{2}*1.2*0.11^2[/tex]

= 0.00726 kgm²

Point mass  [tex]I_{point \ mass}[/tex] = mb²

But since four low rods are attached ; we have :

[tex]I_{point \ mass}[/tex] = 4 × mb²

= 4  × 0.4 (0.14)²

= 0.03136 kgm²

Total moment of Inertia =  [tex]I_{disk, CM}[/tex] + [tex]I_{point \ mass}[/tex]

= (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 × ∝

∝ = [tex]\frac{2.31}{0.03862}[/tex]

∝ = 59.81 rad/s²

Using the formula [tex]\omega_f = \omega_i + \alpha \delta T[/tex] to determine the angular speed of the apparatus; we have:

[tex]\omega_f =0 + 59.81*0.2[/tex]         since  [tex]( w_i \ is \ at \ rest ; the n\ w_i = 0 )[/tex]

[tex]\omega_f =11.962 \ rad/s[/tex]

∴ the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]

b) Using the formula :

[tex]\theta = \omega_i t + \frac{1}{2}*\alpha*(t)^2\\\\\theta = 0 *0.2 + \frac{1}{2}*59.81*(0.2)^2 \\ \\ \theta = 29.905 *(0.2)^2 \\ \\ \theta = 1.1962 rads \ \ \ ( to \ degree; \ we \ have) \\ \\ \theta = 68.54^0[/tex]

Thus, the angle through which the apparatus turns in radians or degrees is :   [tex]\theta = 68.54^0[/tex]

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