Finding the Moment of Inertia of...
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Finding the Moment of Inertia of...

[From: ] [author: ] [Date: 11-05-24] [Hit: ]
= 22 * 0.3 = 6.M of I = 6.6 / 10 = 0.......
A cord is wrapped around a fixed pulley of mass 12 kg and radius 30 cm. When a 22 N force is applied to the cord, the pulley accelerates uniformly from rest to an angular speed of 50 rad/s in 5 s. Determine the moment of inertia of the pulley if it is free to rotate about its center. Ignore friction at the axle.

I thought about simply calculating the moment of inertia of a disk, but then how does the Force that's given play into this? Is it torque = Moment of inertia * alpha? Any help would be greatly appreciated!

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We have, Torque = Force * Radius

Here, Force = 22 N and Radius = 30 cm = 0.3 m

Therefore, Torque = 6.6 N-m

Also, Torque = Moment of Inertia * Angular acceleration

Now, Angular acceleration = (Final velocity - Intial Velocity)/(Time taken for the change)

which means, Angular acceleration = (50 - 0)/5 = 10 radian/second

and Torque = 6.6 N-m (Already Derived)

Hence, Moment of Inertia = Torque/Angular acceleration = 6.6/10 = 0.66 kg-m^2

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α = 50/5 = 10 rad/sec²
Torque Q = F*r = 22*.30 = 6.6 N∙m

From α = Q/I we have I = Q/α = 6.6/10

I = .66 kg∙m²

Note that the mass is not needed for this problem

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u=0; v= 50; t =5
v=u+at
50 = 0 + 5a
a= 10 rad/s2

Torque = force * radius
= 22 * 0.3 = 6.6 Nm
= Moment of inertia * a

M of I = 6.6 / 10 = 0.66 kg m^2 ---answer
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