A solid cylinder of radius 13.1cm and mass 20.7kg starts from rest to rolls with

Question

A solid cylinder of radius 13.1cm and mass 20.7kg starts from rest to rolls without slipping a distance of 6.16m down the townhome roof that is inclined at 21.5° relative to the horizontal. g= 10N

a) What is in J the initial gravitational potential energy of the cylinder relative to the bottom of the roof?

b) What is in radius/s the angular speed the cylinder about its center when at the bottom of the roof?

c) What is in J its rotational kinetic energy then?

d) What is in J its linear kinetic energy then?

Explanation / Answer

total Initial KE = trans KE + rotation KE

so

Total Grav PE = mgh = 0.5mv^2 + 0.5 IW^2

2gh = v^2 + 0.5 mR^2 *(V^2/r^2)

V = sqrt(4/3 gh)

V^2 = 1.15 * 10 * 6.16 sin 21.5

V = 5.1 m/s

total Grav PE = 1.5 * 5.1* 5.1 = 39.01 J

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2. use the relation as V = r W

so W = V/r

W = 5.1/0.131

W = 39 rad/s

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3. RotationaL KE = 0.5 IW^2

KE = 0.5 * 20.7 * 0.131* 0.131 * 39*39

KE = 270.15 J

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4. KE Linear = 0.5 mv^2

KE linear = 0.5* 20.7 * 5.1* 5.1

KE l = 270.J

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