(a) Explanation Given information: The mass ( m ) of each cylinder A and B is 6 kg. The radius ( r ) of each cylinder A and B is 125 mm. The angular velocity ( ω A ) of cylinder A is 30 rad/s counterclockwise. Calculation: Sketch the system with components of velocity as shown in Figure (1). Consider the point C is the instantaneous center of the cylinder A . Write the equation of velocity ( v A B ) of the given system. v A B = r ω B Here, r is the radius of the cylinder and ω B is the angular velocity of the cylinder. Write the equation of the angular velocity ( ω A ) of the cylinder A . ω A = v A B 2 r = r ω B 2 r = 1 2 ω B Write the equation of the velocity ( v ¯ A ) of the cylinder A . v ¯ A = r ω A = r ( 1 2 ω B ) = 1 2 r ω B (1) Find the mass moment of inertia ( I ¯ ) of the each cylinder. I ¯ = 1 2 m r 2 Substitute 6 kg for m and 125 mm for r . I ¯ = 1 2 ( 6 ) ( 125 mm × 1 m 1 , 000 mm ) 2 = 0.046875 kg ⋅ m 2 Consider the principle of impulse and momentum. Syst Momenta 1 + Syst Ext Imp 1y2 = Syst Momenta 2 Sketch the impulse and momentum diagram of the cylinder B as shown in Figure (2). Refer Figure (2), Take moments about B . I ¯ ( ω B ) 1 − P t r = I ¯ ( ω B ) 2 P t r = I ¯ ( ω B ) 1 − I ¯ ( ω B ) 2 P t r = I ¯ ( ( ω B ) 1 − ( ω B ) 2 ) (2) Here, P is the tension in the belt, ( ω B ) 1 is the initial angular velocity of the cylinder, and ( ω B ) 2 final angular velocity of the cylinder B . Sketch the impulse and momentum diagram of the cylinder A as shown in Figure (3). Take moment about C in clockwise direction. I ¯ ( ω A ) 1 + m ( v A ) 1 r + 2 P t r + m g t r = I ¯ ( ω A ) 2 + m ( v A ) 2 r I ¯ ( ω A ) 1 + m ( v A ) 1 r − I ¯ ( ω A ) 2 − m ( v A ) 2 r + 2 P t r + m g t r = 0 I ¯ ( ( ω A ) 1 − ( ω A ) 2 ) + m r ( ( v A ) 1 − ( v A ) 2 ) + 2 P t r + m g t r = 0 Here, ( ω A ) 1 is the initial angular velocity of the cylinder A , ( ω A ) 2 is the final angular velocity of the cylinder A , ( v A ) 1 is the initial velocity of the cylinder A , and ( v A ) 2 is the final velocity of the cylinder A . Substitute I ¯ ( ( ω B ) 1 − ( ω B ) 2 ) I ¯ ω B for P t r , 1 2 r ( ω B ) 1 for ( v A ) 1 , 1 2 r ( ω B ) 2 for ( v A ) 2 , 1 2 ( ω B ) 1 for ( ω A ) 1 , and 1 2 ( ω B ) 2 for ( ω A ) 2 (b) To determine Find the tension in the portion of belt connecting the two cylinders.

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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

Publisher: McGraw-Hill Education

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Chapter 17.2, Problem 17.75P

(a)

To determine

Find the time required for the angular velocity of cylinder A to be reduced to 5 rad/s.

(b)

To determine

Find the tension in the portion of belt connecting the two cylinders.

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Chapter 17.2, Problem 17.74P

Chapter 17.2, Problem 17.76P

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Two uniform cylinders, each of mass m = 6 kg and radius r = 125 mm, are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder A is 30 rad/s counterclockwise, determine (a) the time required for the angular velocity of cylinder A to be reduced to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

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Two uniform cylinders, each of weight W=14 lb and radius r = 5 in, are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder A is 30 rad/sec counterclockwise, determine (a) the time required for the angular velocity of cylinder A to be reduced to 5 rad/sec, (b) the tension in the portion of belt between the two cylinders. *** Solve using the method of impulse-momentum *** B

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Find the time required for the angular velocity of cylinder A to be reduced to 5 rad/s.

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Find the time required for the angular velocity of cylinder A to be reduced to 5 rad/s.

Find the tension in the portion of belt connecting the two cylinders.

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Chapter 17 Solutions