Question

B G 300 mm P $\omega$ = 6 rad/s $\alpha$ = 5 rad/s$^2$ 75 mm

          B
G
300 mm
P
$\omega$ = 6 rad/s
$\alpha$ = 5 rad/s$^2$
75 mm

B
G
300 mm
P
ω = 6 rad/s
α = 5 rad/s^2
75 mm

Added by Rosario V.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Animesh Raj

Step 1

First, let's calculate the moment of inertia of the fan blade about its center of gravity at G. The moment of inertia is given by the formula I = m * k^2, where m is the mass and k is the radius of gyration. Plugging in the values, we have I = 2.5 kg * (225 mm)^2 Show more…

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The single blade PB of this industrial fan has a mass m = 2.5 kg and a radius of gyration about its center of gravity at G of kG = 225 mm. Find the internal forces on the fan blade at P at the moment and position shown. Use this sign convention for the internal forces acting on PB at a section cut at P: +M = clockwise; +V = to the left; +P = downward. B G 300 mm P Ï‰ = 6 rad/s Î± = 5 rad/s^2 75 mm