ch10-p069 - 69. We choose positive coordinate directions...

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() 2 11 1 22 2 2(0.024 m)(1.30 rad) (6.20 kg) 9.80 m/s (0.0910 s) 14.0 N. R Tmga Mg t θ §· =− = ¨¸ ©¹ = (d) From the last of the above equations, we obtain the second tension: 42 2 21 2 2 (7.40 10 kg m )(314 rad/s ) 14.0 N 0.024 m 4.36 N. IR I TT M g Rt R t αθ ×⋅ =− = = = 69. We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a 2 = a 1 = R α (for simplicity, we denote this as a ). Thus, we choose rightward positive for m 2 = M (the block on the table), downward positive for m 1 = M (the block at the end of the string) and (somewhat unconventionally) clockwise for positive sense of disk rotation. This means that we interpret given in the problem as a positive-valued quantity. Applying Newton’s second
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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.

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