Question

A system of rigid body consists of three small balls (m1 = 4 kg, m2 = 5 kg, m3 = 6 kg) which are connected by massless rigid rod, is placed on a flat, frictioness table. The axis of rotation coincides with m1 and is perpendicular to the table, as shown in the figure. At t = 0, a force F with magnitude 20 N directed vertically downward acts on m2, causing the system which was initially at rest to rotate. Calculate: a) moment of inertia of the system with respect to axis of rotation, b) the magnitude of the torque acting on the system at t = 0, c) the magnitude of angular acceleration of the system at t = 0

          A system of rigid body consists of three small balls (m1 = 4 kg,
m2 = 5 kg, m3 = 6 kg) which are connected by massless rigid rod, is
placed on a flat, frictioness table. The axis of rotation coincides
with m1 and is perpendicular to the table, as shown in the figure.
At t = 0, a force F with magnitude 20 N directed vertically
downward acts on m2, causing the system which was initially at
rest to rotate. Calculate: a) moment of inertia of the system with
respect to axis of rotation, b) the magnitude of the torque
acting on the system at t = 0, c) the magnitude of angular
acceleration of the system at t = 0
        
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Added by Christopher P.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A system of rigid body consists of three small balls (m1 = 4 kg, m2 = 5 kg, m3 = 6 kg) which are connected by massless rigid rod, is placed on a flat, frictioness table. The axis of rotation coincides with m1 and is perpendicular to the table, as shown in the figure. At t = 0, a force F with magnitude 20 N directed vertically downward acts on m2, causing the system which was initially at rest to rotate. Calculate: a) moment of inertia of the system with respect to axis of rotation, b) the magnitude of the torque acting on the system at t = 0, c) the magnitude of angular acceleration of the system at t = 0
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Transcript

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00:01 Hello student, to solve the given question for part a, let us find the moment of inertia that will be calculated as m1 multiply by distance is 0 plus m2 multiply by the distance 3 square, that is r square, plus m3 multiply by distance 5 square.
00:19 Now plug in the value for masses here, we get the first term will be 0 plus 5 into 9 plus 6 into 25.
00:28 So from here we get the final answer for the moment of inertia that will be 195 kilogram meter square.
00:37 Similarly, using the relation for torque, that is force into distance, we get the value for part b.
00:46 Torque will be calculated here as 20 multiply by distance 3.
00:52 So from here we get the final answer for torque equals 60 newton meters.
00:58 For part c we can use the relation...
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