A solid sphere of mass 25kg and diameter 2ft is sitting at the top of a 35degree incline of length 20ft. When released it rolls without slipping down the incline to the horizontal surface below. After a short time, the sphere rolls up a nearby second incline of 30 degrees where it comes to a stop.
a) what is its speed at the bottom of the first incline
b) what is its kinetic energy at this point?
c) what is its angular speed (in radians/sec) at this point?
d) to what height on the second incline will it roll to when it stops?
e) what is its potential energy at this height on the second incline?
I can't seem to figure this question out... any ideas?
a) what is its speed at the bottom of the first incline
b) what is its kinetic energy at this point?
c) what is its angular speed (in radians/sec) at this point?
d) to what height on the second incline will it roll to when it stops?
e) what is its potential energy at this height on the second incline?
I can't seem to figure this question out... any ideas?
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For a solid sphere, I = k*m*R² where k = 0.4
25 kg = 1.713 slugs
H1 = 20*sin35° = 11.471 ft
a) Vb = √[2gH1/(1+k)] = √[2*32.2*11.471/1.4)] = 22.971 ft/s
b) KE = mgH1 = 632.72 ft∙lb
c) w = Vb/R = 22.971/1 = 22.971 rad/sec
d) H2 = H1 = 11.471 ft
e) PE2 = PE1 = KE = 632.72 ft∙lb
25 kg = 1.713 slugs
H1 = 20*sin35° = 11.471 ft
a) Vb = √[2gH1/(1+k)] = √[2*32.2*11.471/1.4)] = 22.971 ft/s
b) KE = mgH1 = 632.72 ft∙lb
c) w = Vb/R = 22.971/1 = 22.971 rad/sec
d) H2 = H1 = 11.471 ft
e) PE2 = PE1 = KE = 632.72 ft∙lb