Explanation Given information: The radius of the drum is 60 mm and the mass is 6 kg . The radius of the disk is 120 mm and the radius of gyration is 90 mm . The force is 20 N . The following figure represents the disk rolls without sliding. Figure-(1) Write the expression for the linear acceleration. a = r α …… (I) Here, the angular acceleration is α , the linear acceleration is a and the radius of the disk is r . Write the expression for the moment of the inertia. I = m k 2 …… (II) Here, the radius of gyration is k , the moment of inertia is I and the mass is m . Write the expression for the moment at point C. P × r = m a r + I α …… (III) Here, the force is P . Calculation: Substitute 90 mm for k and 6 kg for m in Equation (II). I = ( 6 kg ) ( 90 mm ) 2 = ( 6 kg ) ( 90 mm ( 1 m 1000 mm ) ) 2 = 0.0486 kg ⋅ m 2 Substitute 0.0486 kg ⋅ m 2 for I , 0.12 α m for a , 6 kg for m , 20 N for P and 120 mm for r in Equation (III). ( 20 N ) × 120 mm = ( 6 kg ) ( 0.12 α m ) ( 120 mm ) + ( 0.0486 kg ⋅ m 2 ) α [ ( 20 N ) × 120 mm ( 1 m 1000 mm ) ] = [ ( 6 kg ) ( 0.12 α m ) ( 120 mm ( 1 m 1000 mm ) ) ] + ( 0 To determine (b) The minimum value of the coefficient of static friction.

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 16.2, Problem 16.98P

To determine

(a)

The angular acceleration of the disk.

The acceleration of G.

To determine

(b)

The minimum value of the coefficient of static friction.

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Chapter 16.2, Problem 16.97P

Chapter 16.2, Problem 16.99P

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Chapter 16 Solutions

Vector Mechanics for Engineers: Dynamics

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(a) The angular acceleration of the disk. The acceleration of G.

Solution Summary: The following figure represents the disk rolls without sliding. Write the expression for the linear acceleration.