Question

The drawing shows two identical systems of objects; each consists of the same three small balls connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. The same force of magnitude F is applied to the same ball in each system (see the drawing). The masses of the balls are m₁ = 9.4 kg, m2 = 6.8 kg, and m3 = 7.1 kg. The magnitude of the force is F = 476 N. (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the 4.00-m rod. What is the angular velocity of each system after 5.79 s? (a) IA = i IB = i (b) TA = TB = i (c) WA = WB = i + i i m 3.00 m 3.00 m 2 Axis 4.00 m 4.00 m 5.00 m 5.00 m M3 System A Axis System B ```

          The drawing shows two identical systems of objects; each consists of the same three small balls connected by massless rods. In both
systems the axis is perpendicular to the page, but it is located at a different place, as shown. The same force of magnitude F is applied
to the same ball in each system (see the drawing). The masses of the balls are m₁ = 9.4 kg, m2 = 6.8 kg, and m3 = 7.1 kg. The magnitude
of the force is F = 476 N. (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b)
Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force
moves with the system and always points along the 4.00-m rod. What is the angular velocity of each system after 5.79 s?
(a) IA = i
IB = i
(b) TA =
TB = i
(c) WA =
WB =
i
+
i
i
m
3.00 m
3.00 m
2
Axis
4.00 m
4.00 m
5.00 m
5.00 m
M3
System A
Axis
System B
```

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the drawing shows two identical systems of objects each consists of the same three small balls connected by massless rods in both systems the axis is perpendicular to the page but it is loca 86078

Added by Brianna G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mike Gaerlan

04/09/2020

Step 1

$I_A = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2$ $I_A = (9.4 kg)(5.00 m)^2 + (6.8 kg)(3.00 m)^2 + (7.1 kg)(0 m)^2$ $I_A = (9.4 kg)(25 m^2) + (6.8 kg)(9 m^2) + 0$ $I_A = 235 kg \cdot m^2 + 61.2 kg \cdot m^2$ $I_A = 296.2 kg \cdot m^2$ Show more…

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The drawing shows two identical systems of objects; each consists of the same three small balls connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. The same force of magnitude F is applied to the same ball in each system (see the drawing). The masses of the balls are m₁ = 9.4 kg, m2 = 6.8 kg, and m3 = 7.1 kg. The magnitude of the force is F = 476 N. (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the 4.00-m rod. What is the angular velocity of each system after 5.79 s? (a) IA = i IB = i (b) TA = TB = i (c) WA = WB = i + i i m 3.00 m 3.00 m 2 Axis 4.00 m 4.00 m 5.00 m 5.00 m M3 System A Axis System B ```

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Transcript

00:01 Okay, so we're given these two systems.

00:02 So for part a, we need to find the moment of inertia about the given axis.

00:09 So for the first system here, the first system here, we can just use the moment of inertia of the two particles here.

00:21 So the moment of inertia is just going to be so the mass of this one first.

00:31 Which is equal to 6 kilograms, and then times the distance here, which is 3 .00 meters squared.

00:44 So the moment of, oh, sorry, and then plus the, so the moment of ratio due to the third particle, which is going to be equal to 7 .00 kilograms times 5 .00 meters squared, so the total moment of inertia, and for this one, it's going to be zero because the radius is zero.

01:09 So now the moment of inertia total is going to be equal to 200, 229 kilogram meters squared.

01:25 Now that's for the first system.

01:29 Now for the second system, we're going to get now the axis of rotation is here.

01:35 So we're going to look at these two particles.

01:37 So it's going to be first mass 1, which is 9 kilograms, times its distance to the axis of rotation, which is 5 meters squared, and then plus mass 2, which was 6 kilograms times its distance, which is 4 .00 meters squared, which gives us a total moment of inertia of 321 kilograms squared.

02:11 Part b, we need to find the torque acting on each system.

02:18 So in this first scenario here, the force is acting on a perpendicular distance of three meters.

02:28 So, okay, so part b, the, i'll call that torque one.

02:34 So torque one is going to be the force, which has a magnitude of 424 newtons times the perpendicular distance which is 3 .00 meters giving a total torque of 1 ,270 newton meters and this is going in a clockwise direction...

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