Question

The weight is used to turn the gear system shown below. Gear A has a mass of 15 kg and a radius of gyration of 100 mm about its center of mass. Gear B and Drum C have a combined mass of 25 kg and a radius of gyration about their mass center of 165 mm. If the system starts from rest, determine the speed of the 55-kg weight after it has been lowered 3 m, the angular velocity of each gear, and the linear velocity of the meshing teeth between Gear A and Gear B. 100 mm 150 mm 200 mm D

Answer 1

Given

Gear A Mass Ma = 15kg

Radius of gyration Ra= 100 mm

Gear B and drum Mass Mb = 25 kg

Radius of gyration Rb= 165 mm

Inertia of Gear A = 0.5 * 15* (0.1²) = 75000 kg mm².=0.075 kg m².

Inertia of Gear A = 0.5 * 25* (0.165²) = 340312.5 kg mm².=0.340313 kg m².

Total Inertia of the system

Ia / n + Ib  

n= 150/200 = 0.75

Itot = 0.075/0.75 + 0.0340313 = 0.440313 kg m².

Tension in the string = Ts = m*g - m* a

Mis mass of object

Now

torque at drum

Ts = Itot * $\alpha$

m*g -m*a = Itot * $\alpha$

$\alpha$ =a/Rb

m*g -m*a = Itot * a/Rb

55* 9.81 - 55 * a = 0.440313 * a / 0.165

539.55	-	55*a	=	2.668561	a
		539.55	=	2.668561	a	+	55*a
		539.55	=	57.66856	a
		a	=	9.356051	m/s

a=9.36 m/s

We know distance travelled = 0.5 * a * t² = 0.5 * 9.36 * t²​​​​​​​ = 3

Given 3 m distance

0.5 * 9.36 * t²​​​​​​​ = 3

t= sqrt (3 / (0.5 * 9.36)) = 0.8 seconds

We know the radius of Drum , = 0.1

perimeter = Pi * D = pi * (2* 0.1 ) = 0.63 m

No of times drum rorates to reach 3 m = 3/ 0.63 = 4.775 rotations and in 0.8 seconds

RPS = 1/0.8 * 4.778 = 5.96 ( angular speed of gear B)

The gear B radius = 0.165

Perimeter = 1.04 m

The gear rotates 5.96 times in a seconds so linear speed = 5.96 * 1.04 = 6.2 m/s ( Linear speed of gear B)

Linear speed of gear A and B must be same

Angular speed of Gear A = RPS Gear B / 0.75 ( n) = 5.96/0.75 = 7.95 RPS

Answers

Linear speed of teeth of gear A and B = 6.2 m/s

Angular Speed of Gear A = 7.95 RPS = 477 RPM

Angular speed of Gear B = 5.96 RPS = 358 RPM

Thanks