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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. 11-3 (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 parallel-axis theorem and Table 10-2(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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