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Unformatted text preview: 95. We make the unconventional choice of clockwise sense as positive, so that the angular acceleration is positive (as is the linear acceleration of the center of mass, since we take rightwards as positive). (a) We approach this in the manner of Eq. 113 (pure rotation about point P) but use torques instead of energy. The torque (relative to point P) is τ = I Pα , where IP = 1 3 MR 2 + MR 2 = MR 2 2 2 with the use of the parallelaxis theorem and Table 102(c). The torque is due to the Fapp = 12 N force and can be written as τ = Fapp (2 R) . In this way, we find τ = I Pα =
which leads to 2 RFapp 3MR /2
2 3 MR 2 α = 2 RFapp 2 = 4 (12 N ) = 16 rad/s 2 . 3(10 kg)(0.10 m) α= = 4 Fapp 3MR Hence, acom = Rα = 1.6 m/s2. (b) As shown above, α = 16 rad/s2. (c) Applying Newton’s second law in its linear form yields 12 N − f = Macom . Therefore, f = –4.0 N. Contradicting what we assumed in setting up our force equation, ˆ the friction force is found to point rightward with magnitude 4.0 N, i.e., f = (4.0 N)i . bg ...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration, Center Of Mass, Mass

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