Explanation Given: mass of Disk A, m a = 4 kg radius of disk A, r A = 0.3 m Initial angular velocity of disk B = 300 rpm = 300 60 ×2π = 10π mass of disk B, m B = 1.6 kg radius of disk B, r B = 0.18 m Coefficient of kinetic friction = 0.35 Concept used: When the two disks come in contact, a friction force between them comes into play and it causes the disk B to start rotating while accelerating with a certain angular acceleration in anti-clockwise direction. The reaction of the friction force on disk B will be acting on disk A, such that while it will still be rotating in clockwise direction, but with a certain. angular deacceleration. This will continue till the tangential velocity of both the disks become equal. At that point, V A = V B For both the disks the net tangential force that is either accelerating or deaccelerating them is the friction force between them. Magnitude of friction force pulling between disks = μ k × N , Where N is equal to the weight of disk B . Further, Mass moment of inertia for a disk is given by- I = 1 2 mr 2 , where r is the radius of disk The tangential friction force acting on a disk will provide the angular acceleration/deacceleration to the disk. Therefore, Summation of moments applied on it = Mass Moment of inertia × angular acceleration Σ M = Iα Calculation: Mass moment of inertia of disk A = 1 2 m A r A 2 = 1 2 (4kg)×(0 .3m) 2 = 0 .18 kg-m 2 Mass moment of inertia of disk B = 1 2 m B r B 2 = 1 2 (1 To determine ii. TDhe reaction at support C.

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ISBN: 9781259977237

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Chapter 16.1, Problem 16.43P

To determine

i.

The angular acceleration of disk A and B.

To determine

ii.

TDhe reaction at support C.

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Chapter 16.1, Problem 16.42P

Chapter 16.1, Problem 16.44P

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i. The angular acceleration of disk A and B.

Solution Summary: The author explains the angular acceleration of disks A and B, and the friction force acting on them.