? Question 1 (15 points) Retake question F1 F2 F3 2.5 m 2...

? Question 1 (15 points) Retake question F1 F2 F3 2.5 m 2 m 3 m M = ? The 144 kg see saw (which can be approximated as a slender rod with a total length of 10 m) is pinned at its center of mass, but is free to rotate about the pin joint. Based on the diagram and the magnitudes of the forces given below, determine the direction and magnitude of the moment, $M$, that must be applied to achieve an angular acceleration of $\alpha = 1.41 \text{ rad/s}^2$ in the counter-clockwise direction. F? = 124 N F2 = 151 N F3 = 226 N

Transcript

00:01 In the question number 6, it has been given that this whole rigid system is revolving about point p where the axis is perpendicular to the paper.

00:13 So, if we calculate the moment of inertia and equate with the rotational moment of kinetic energy with the given kinetic energy, we can find the angular velocity.

00:31 So let's first calculate the...

00:34 So let's first calculate the moment of inertia.

00:41 Now we know that since these rods are massless and this mass m and capital 2m are acting as a particle or point mass.

00:55 So moment of inertia will be mr square for a point mass.

01:04 If this is the axis of rotation and this is the radius of rotation, let's say this is o, this is p.

01:11 O, this is the radius of rotation and a mass is situated at point p of magnitude m.

01:21 The mass, the magnitude of this mass is m.

01:26 So moment of inertia will be equals to i equals to mr square.

01:30 Now in this case, for this body, let's say for this capital mass m, the moment of inertia will be ml square.

01:45 Similarly for this mass also same so here it will be 2m.

01:50 Plus now for this one it will be 2m into l by 2 whole square.

02:01 So solving this we will get ml square by 2 so ml square by 2 will be the moment of inertia for this 2m mass.

02:12 Now for this one, we have to multiply it by 2 so for combined moment of inertia for 2m and 2m will be ml squared so overall for all masses the moment of inertia about point p will be equals to 3 ml square now in this question m is given as 2 kg l is given a 0 .2 meter and we know that half i omega square that is a rotation kinetic energy since there is no any kind of linear velocity is present that means there is no any linear kinetic energy is involved so because this point p is fixed and about this this whole masses are rotating so we can say that half i omega square equals to total kinetic energy and that is 2 .6 jule so now let's solve it so half into i now i is 3 ml square into omega square equals to 2 .6 so half into 3 into mass is 2 into l square that is 0 .2 whole square into omega square equals to 2 .6 solving for omega we will get omega equals to 4 .65 radian per second.

03:56 So the angular velocity will be omega will be 4 .65 radian per second...

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