My Na 1 points YF14 9 P079 In the system shown in the figure below, a m- 13.0-kg

Question

My Na 1 points YF14 9 P079 In the system shown in the figure below, a m- 13.0-kg mass is released from rest and falls, causing the uniform M 13.5-kg cylinder of diameter 25.0 cm to turn about a frictionless axle through its center. How far will the mass have to descend to give the cylinder 205 J of kinetic energy? 10. 1/1 points 1 Previous Answers YP149.XP008. Find the required angular speed (in rev/min) of an ultracentrifuge for the radial acceleration of a point 2.57 cm from the xis to equal 575,000g (that is, 575,000 times the acceleration due to gravity). rew/min You need to desyn an industrial turntable that is 67.0 cm in daneter and has a kinetic energy o, olas when tring at trevimin) 22o 4

Explanation / Answer

Set potential energy equal to zero for the position in the figure. So (K+U)i = 0

(K+U)f = 1/2m*v^2-m*g*y +1/2*I*omega^2, but 1/2*i*omega^2 = 285J (given)

From this we can determine the linear v of the edge of the cylinder. This is also the v of the mass.

I = 1/2*M*R^2 and omega = v/R So 1/2*I*omega^2 = 1/2*1/2*13.5kg*v^2 = 205 This means that v^2 = 205/(1/4*13.5) = 60.74 m^2/s^2

Therefore the K of the mass = 1/2*m*v^2 = 1/2*13.5*60.74 = 410 J

Now 410 J + 205 J - m*g*y = 0 This means that y = (410+205)/(9.8*13.5) = 4.65 m

Calculations maybe faulty, procedure is correct.

Hope this helps :)

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