Question

would be the ac= 18 m/s) ² = 64 mls = 42.67 m/s2 10.3 A drive pulley 6cm in diameter is set toi to rotate, at a rewls. What is the centripetal acceleration of a point on the edge of the puller helt around the pulley. Im= Boobin C = 277 R = 0/2 = 0 / 2 = 0.03m (+27(0.03 m) = 0.19 m 9 reuls = %=0.11. V on = 0.19% = 1.58 mls sec 0.12 sec ac = v= (1.58 m/s) 2 25 0.03m 0.03
this is my work for the wuwstion but the book says the answer should be 95.9. what an i doing wrong?

Answer 1

Yes! There is a little mistake. In these type of questions where rpm is given you should use the following method as described in image

rounding off gives 95.9 cm/s2

9gev Sec amgular peed 2 n grad. N ei 56.52 9uad nec ns radius O,03 cnm n1 eentripatal aueleration 2 0.03 x (56,52 74ds) 2 rw2 95-935