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3 years ago

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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

constant tangential force of 290 N applied to its edge causes the wheel to have an angular acceleration of 0.814 rad/s2.What is the mass of the wheel?

1 answer:

Flura [38]3 years ago

8 0

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

m

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

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