A 0.321 kg, 0.27 m radius thin shelled ball rolls (starting at rest) 3.82 m from
ID: 1609942 • Letter: A
Question
A 0.321 kg, 0.27 m radius thin shelled ball rolls (starting at rest) 3.82 m from the top down a 35o 10.0 m long incline without slipping. After the 3.82 m, the incline becomes frictionless for Disk 1 Disk 2 Disk 3 F F F the rest of the board. After 0.8 m of the frictionless unencumbered movement, the ball reaches a frictionless, massless spring with a spring constant of 20 N/m.
a) How fast is the center of mass of the ball going after 3.82 m? (linear speed)
b) How fast is the ball rolling after the 3.82 m? (angular speed)
c) How much does the spring compress?
d) What is the ball's maximum linear speed?
e) What is the balls angular speed with the spring is fully compressed?
Explanation / Answer
(A) Applying energy conservation,
m g h + 0 = 0 + (m v^2 /2 + I w^2 /2 )
m g L sin(theta) = m v^2 /2 + ( 2 m r^2 / 3) ( v/ r)^2 /2
m g L sin(theta) = m v^2 / 2 + m v^2 / 3 = 5 m v^2 / 6
v^2 = 6 x 9.8 x 3.82 x sin35 / 5
v = 5.08 m/s
(B) w = v / r = 5.08 / 0.27
w = 18.8 rad/s
(C) now only ball's linear kinetic energy is trasformed into Spring potential energy.
Applying work - energy theorem,
work done by gravity + work done by spring = change in KE
0.321 x 9.8 x (0.8 + d) sin35 - 20 d^2 /2 = 0 - 0.321 x 5.08^2 /2
1.44 + 1.80d - 10 d^2 = - 4.14
10 d^2 - 1.80d - 5.58 = 0
d = 0.84 m
(D) 0.321 v^2 /2 - 4.14 = 1.44
v = 5.90 m/s
(E) w = 18.8 rad/s
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