Physics Conservation of Energy A hollow sphere of mass m = 1000kg and radius R =
ID: 2242291 • Letter: P
Question
Physics Conservation of Energy
A hollow sphere of mass m = 1000kg and radius R = 25cm is rolling down of an incline plans making an angle beta with horizontal. The thin hoop starts with an initial velocity V0 = 25m/s at a vertical height H and traveling a distance d = 200 m. Calculate the moment of inertia I sphere of the hollow sphere. Calculate the initial Kinetic energy KE0 Of the hollow sphere at the starting point. Calculate the initial Potential Energy PEO at the starting point. Calculate the initial Total Energy EO of the hollow sphere. Calculate the final velocity Vf at the bottom of the incline plan. Calculate the angular acceleration alpha of hollow sphere after 10 s. (tf = 10 s) Calculate the net torque tau produced by the hollow sphere after 10 s (tf = 10 s)Explanation / Answer
a)
I = 2/3 M R^2 = 2/3 * 1000 * (0.25*0.25) = 41.667 = 41.7 Kg.m2
b)
KE0 = 0.5 I w^2 + 0.5 M v^2
= 0.5 I (v/R)^2 + 0.5 M v^2
= 0.5 (2/3 M R^2) (v/R)^2 + 0.5 M v^2
= 0.5 (2/3 M) v^2 + 0.5 M v^2
= (2.5/3) M v^2
= (2.5/3) * 1000 * (25*25)
= 5.2083 x 10^5
= 5.21 x 10^5 J
c)
PE0 = m g h = m g d sin(30) = 1000*9.8*200*sin30 = 9.8e5 = 9.8 x 10^5 J
d)
E0 = 5.2083e5 + 9.8e5 = 1.50 x 10^6 J
e)
conservation of energy:
EKf = Etotal
(2.5/3) M vf^2 = Etotal
(2.5/3) M vf^2 = Etotal
==> (2.5/3) * 1000 * vf^2 = 1.50e6
==> vf = 42.426
==> vf = 42.4 m/s
f)
vf^2 - v0^2 = 2 a d
==> a = (vf^2 - v0^2)/(2 d)
==> a = (42.426*42.426-25*25)/(2*200)
==> a = 2.9374 m/s2
==> alpha = a/R = 2.9374/0.25 = 11.7 rad/s2
g)
torque = I alpha = 41.667*11.7 = 1225.0098 = 490 N.m