A 0.338 kg basketball with radius r= 18.8 cm is left at the top of an inclined plane that makes an angle of e = 30° with respect to the horizontal. The ball starts from rest and rolls without slipping a distance d = 3.84 m reaching an angular speed of o = 3.00 rad/s. What is the ball's moment of inertia? (q48) a) 02.83 kg m? b) 04.24 kg m2 c) 0 0.00796 kg m2 d) O14 kg m2

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A 0.338 kg basketball with radius r= 18.8 cm is left at the top of an inclined plane that makes an angle of 0 = 30° with respect to the horizontal. The ball starts from rest and rolls without slipping a
distance d = 3.84 m reaching an angular speed of o 3.00 rad/s. What is the ball's moment of inertia? (q48)
a) O2.83 kg m
b) 04.24 kg m
c) O 0.00796 kg m
d) O14 kg m2
Transcribed Image Text:A 0.338 kg basketball with radius r= 18.8 cm is left at the top of an inclined plane that makes an angle of 0 = 30° with respect to the horizontal. The ball starts from rest and rolls without slipping a distance d = 3.84 m reaching an angular speed of o 3.00 rad/s. What is the ball's moment of inertia? (q48) a) O2.83 kg m b) 04.24 kg m c) O 0.00796 kg m d) O14 kg m2
Expert Solution
Step 1

When a ball traveled distance 'd' on the inclined plane, the height of the slope at that instant will be given as,

h=dsin30°=3.84 msin30°=1.92 m

The total potential energy stored in the ball at this height can be given as,

P=mgh=0.338 kg9.8 m/s21.92 m=6.3598 J

Step 2

The velocity of the ball is,

v=rω=18.8 cm3 rad/s=0.188 m3 rad/s=0.564 m/s

The total energy of the ball at the bottom is,

E=12mv2+12Iω2

Here, I denotes the moment of inertial of the ball.

E=120.338 kg0.564 m/s2+12I3 rad/s2

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