Question

A steady 3.0-lb force is applied normal to the handle of the hand-operated grinder. The gear inside the housing with its shaft and attached handle together weigh 3.56 lb and have a radius of gyration about their axis of 2.75 in. The grinding wheel with its attached shaft and pinion (inside housing) together weigh 1.09 lb and have a radius of gyration of 2.22 in. If the gear ratio between gear and pinion is 7:1, calculate the speed N of the grinding wheel after 7 complete revolutions of the handle starting from rest.

Answer 1

the speed N of the grinding wheel after 7 complete revolutions of the handle starting from rest = 2355.5 rpm

Step-by-step explanation:

Applying principles of conservation of energy;

Mθ = (1/2 Ighω²₁)_gear housing + (1/2 Iph ω²)_pinion housing

where ;

Mθ = Gain in potential energy due to restoring force

1/2 Ighω²₁ = kinetic energy due to rotation of gear housing

1/2 Iph ω² = kinetic energy due to rotation of pinion housing

Now; we can rewrite the equation as :

F×I×θ = (1/2 mr²ω²₁)_gear housing + (1/2 mr²ω²)_pinion housing

Since; mass moment of inertia (I) is not given; we assume it to be 5.5

So; from the question; plugging our values into the above equation; we have:

3.0 × ( 5.5/12) × 7 × 2π = [ 1/2 ( 3.56/32.2) × (2.75/12)² ×( ω/7)² + 1/2

(1.09/32.2)×(2.22/12)²× ω²]

60.48 = 4.147× 10⁻⁴ ω² + 5.793 × 10⁻⁴ ω²

60.48 = 9.94 × 10⁻⁴ ω²

ω² = 60.48 / 9.94 × 10⁻⁴

ω² = 60845.07

ω = √60845.07

ω = 246.67 rad/s

speed of grinding wheel

ω = 2πN/60

N = 246.67 ×60/2π

N = 2355.5 rpm

Thus, the speed N of the grinding wheel after 7 complete revolutions of the handle starting from rest = 2355.5 rpm