# The large gear (shown below) has a mass of 10 kg and a radius of gyration...?

The large gear (shown below) has a mass of 10 kg and a radius of gyration if 200 mm, while the small gear has a mass of 3 kg and radius of gyration of 80 mm. The system is at rest when a couple M of a magnitude of 6 N-m is applied rotating the small gear counter-clockwise. Neglecting friction, determine (a) the number of revolutions executed by the small gear before its angular velocity reaches 600 rpm and (b) the tangential force which the small gear exerts on the large gear. Note: The large gear has a radius of 250 mm and the small gear has a radius of 100 mm.

### 2 Answers

- NCSLv 72 years agoFavorite Answer
Ooh, this is a little nasty. And there may very well be a more efficient route to a solution, but here goes...

ω = 600rev/min * 2π rads/rev * 1min/60s = 62.8 rad/s

Let the angular acceleration of the small gear be αA and the acceleration of the large gear be αB.

The velocity of the gears where they mesh must be the same, so

αA*rA = αB*rB

αA = αB*rB/rA = αB*(250mm/100mm) = αB*2.5

For the small gear,

net torque = IA*αA = applied torque - F*rA

where F is the inter-gear force.

Substitute for IA, αA, applied torque, and rA:

3kg*(0.080m)² * αB*2.5 = 6N·m - F*0.100m

0.048kg·m² * αB = 6N·m - F*0.100m

For the large gear,

net torque = IB*αB = F*rB

10kg*(0.200m)² * αB = F*0.250m

αB * 1.6kg·m = F

Plug F into the first equation:

0.048kg·m² * αB = 6N·m - αB*1.6kg·m*0.100m

0.208kg·m² * αB = 6 N·m

αB = 28.8 rad/s²

and so

F = 28.8rad/s² * 1.6kg·m² = 46.2 N ◄ (b)

Then

αA = 28.8rad/s² * 2.5 = 72.1 rad/s²

ω² = ω₀² + 2*αA*ΘA

and so

(62.8 rad/s)² = 0² + 2 * 72.1rad/s² * Θ

which solves to

Θ = 27.3 rads = 4.35 revs ◄ (a)

Does this all fit?

small gear:

IA*αA = applied torque - F*rA

3kg*(0.080m)² * 72.1rad/s² = applied torque - 46.2N*0.100m

1.38 N·m = applied torque - 4.62N·m

applied torque = 6.0 N·m

is TRUE

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- Anonymous2 years ago
56/7/120 .