Explanation Given: The angular acceleration of the disk is α = 8 rad/s 2 . The angular velocity of the disk is ω = 3 rad/s . The radius of the disk is 0.5 m . The disk does not slip at point A . Draw the free body diagram of disk as shown in Figure 1. Write the relation between the linear velocity and angular velocity. ω = v r (I) v = ω r (II) Here, ω is the angular velocity, v is the linear velocity and r is the radial distance from the instantaneous center or permanent center to the point where the linear velocity acting on the object. Refer Figure (1). Write the equation of relative acceleration of point B with respect to the point O . a B = a C + α × r B / C − ω 2 r B / C (III) Here, a B is the acceleration vector of the point B , a C is the acceleration vector of the point C α is the angular acceleration, r B / C is the position vector directed from C to B and ω is the angular velocity. Conclusion: Refer Figure (1). The instantaneous center ( I C ) located at center of the disk ( O ) and shown in Figure (1). Calculate the acceleration of point C . a C = α r Substitute 8 rad/s 2 for α and 0.5 m for r . a C = ( 8 rad/s 2 ) ( 0.5 m ) = 4 m/s 2 Refer Figure (1). Write the position vector ( r B / C ) directed from C to B . r B / C = ( − 0.5 cos 30 ° ) i − ( 0.5 sin 30 ° ) j = − 0.433 i − 0.25 j Refer Figure (1). Express the vector coordinate for all the acceleration components of the system. The acceleration vector B is a B = a B i + a B j The acceleration vector at C is a C = a C i − a C j = − 4 i + 0 j = − 4 i The angular acceleration vector of the disk is α = α k Substitute 8 rad/s 2 for α . α = 8 k Refer Figure (1). Substitute − 4 i for a C , 8 k for α , − 0.433 i − 0.25 j for r B / C and 3 rad/s for ω in Equation (III)

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 16, Problem 7RP

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The acceleration of point B .

Solution Summary: The author illustrates the equation of relative acceleration of point B with respect to the instantaneous center.

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