Problem ?unitltem 2398205&enrollment; ID 238319 Friday, April 21 macmillan learn

Question

Problem ?unitltem 2398205&enrollment; ID 238319 Friday, April 21 macmillan learning Physics 2111 Spring 17 (Moulik) Dupage unit 16 Prelocture Checkpoint Homework Homework Homework: Rotational Dynamics Deadline: 100% untn Monday, Apra 17 at 10 oo Non-uniform Cylinder on Incline In our lab, we roll a non-uniform but cylindrically symmetric cylinder down a rough 20' incline. The cylinder rolls without slipping and has mass of a kg, radius 1,7 m, and moment of inertia about its center of mass 7.a kg m 1) What is the acceleration of the cylinders center of mass? 2) What minimum coefficient of friction is required so that there is no stippingh y) what is the ratio of kinetic energy of the center-ofrmass to kinetic energy about the center-of-mass? (KEtrais KEree) translational speed of the center of mass?

Explanation / Answer

(1) friction f will act up along the incline,

Fnet = m a

m g sin20 - f = m a

8 g sin20 - f = 8 a ......(i)

and net torque = I alpha

for rolling without sliiping, alpha = a / r

r f = (I) (a / r)

f = (7.8 / 1.7^2) a

f = 2.70a .......... (ii)

adding (i) and (ii),

8 x 9.8 x sin20 - 2.70a = 8 a

a = 2.51 m/s^2 .........Ans

(2) f = 2.70a = 6.77 N

N = m gcos20 = 8 x 9.8 x cos20 = 73.67 N

u = f / N = 0.092

(3) KE = m v^2 /2 + I w^2 /2

= 8 v^2 /2 + (7.8 / 1.7^2) v^2 /2

= 4 v^2 + 1.35 v^2

KEtrans = 4 v^2

KErot = 1.35 v^2

ratio = KEtrans / KErot = 4/1.35

= 2.96

(4) applying energy conservation,

m g h = KEf

m g h = 4 v^2 + 1.35 v^2

8 x 9.8 x 3.5 sin20 = 5.35 v^2

v = 4.20 m/s

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